10.2 Quantum Field Theory in a Nanoshell

The “classical” quantum theory discussed in this book has major difficulties describing really relativistic effects such as particle creation and destruction. Einstein’s $E=mc^2$ allows particles to be destroyed as long as their mass times the square speed of light shows up as energy elsewhere. They can also be created when enough energy is available. Indeed, as the Dirac equation, section 9.2, first showed, electrons and positrons can annihilate one another, or they can be created by a very energetic photon near a heavy nucleus.

And quantum field theory is not just for esoteric conditions. The photons of light are routinely created under normal conditions. Still more basic to an engineer, so are their equivalents in solids, the phonons. Then there is the band theory of solids: electrons are “created” within the conduction band, if they pick up enough energy, or “annihilated” when they lose it. And similarly for the real-life equivalent of positrons, holes in the valence band.

Such phenomena are routinely described within the framework of quantum field theory, and almost unavoidably you will run into it in literature, [13,8]. Electron-phonon interactions are particularly important for engineering applications, leading to electrical resistance (along with crystal defects and impurities), and the combination of electrons into Cooper pairs that act as bosons and so give rise to superconductivity. The intention of this section is to explain enough of the ideas so that you can recognize it when you see it. What to do about it after you recognize it is another matter.

Especially the relativistic applications are very involved. To explain quantum field theory in a nutshell takes 500 pages, [27]. You will also need to pick up linear algebra, tensor algebra, and group theory. However, if you are just interested in relativistic quantum mechanics from an intellectual point of view, rather than for practical applications, the answer is all good. Feynman gave a set of lectures on “quantum electrodynamics” for a general audience around 1983, and the text is readily available at low cost. Here, freed from the constraint of his lecture notes to cover a standard fare of material, Feynman truly gets it right. Without doubt, this is the best exposition of the fundamentals of quantum mechanics that has ever been written, or ever will. The subject is reduced to its bare abstract axioms, and no more can be said. If the human race is still around a millennium or so from now, technology may take care of the needed details of quantum mechanics. But those who need or want to understand what it means will still reach for Feynman. The 2006 edition, [7], has a foreword by Zee that gives a few hints how to relate the basic concepts in the discussion to more conventional mathematics like the complex numbers found in this book.

It will not be much help applying quantum field theory to engineering problems, however. In the absence of 1,000 pages and a willing author, the following discussion will truly be quantum field theory in a nanoshell. I thank Wikipedia for the basic approach. As far as the rest is concerned, it has been pieced together from, in order of importance, [[16]], [26], [[3]], [13,8]. Any mistakes in doing so are mine.

10.2.1 Occupation numbers

Consider once more systems of weakly interacting particles like the ones that were studied in section 8.2. The energy eigenfunctions of such a system can be written in terms of whatever are the single-particle energy eigenfunctions $\pes1/{\skew0\vec r}/p//,\pes2/{\skew0\vec r}/p//,\ldots$. A completely arbitrary example of such a system eigenfunction for a system of $I=95$ distinguishable particles is:

$$\psi^I_q = \pes7/{\skew0\vec r}_1/p/1/ \pes66/{\skew0\ve... ...kew0\vec r}_5/p/5/ \ldots \pes59/{\skew0\vec r}_{95}/p/95/ %$$ (10.19)

This system eigenfunction has an energy that is the sum of the 95 single-particle eigenstate energies involved:

$${\vphantom' E}^I_q = {\vphantom' E}^p_{7} + {\vphantom' E}... ...40} + {\vphantom' E}^p_{7} + \ldots + {\vphantom' E}^p_{59}$$

Figure 10.1: Graphical depiction of an arbitrary system energy eigenfunction for 95 distinguishable particles.

Instead of writing out the example eigenfunction mathematically as done in (10.19), it can be graphically depicted as in figure 10.1. In the figure the different types of single-particle states are shown as boxes, and the particles that are in those particular single-particle states are shown inside the boxes. In the example, particle 1 is inside the $\pes7//p//$ box, particle 2 is inside the $\pes66//p//$ one, etcetera. It is just the reverse from the mathematical expression (10.19): the mathematical expression shows for each particle in turn what the single-particle eigenstate of that particle is. The figure shows for each single-particle eigenstate in turn what particles are in that eigenstate.

Figure 10.2: Graphical depiction of an arbitrary system energy eigenfunction for 95 identical bosons.

However, if the 95 particles are identical bosons, (like photons or phonons), the example mathematical eigenfunction (10.19) and corresponding depiction figure 10.1 is unacceptable. Eigenfunctions for bosons must be unchanged if two particles are swapped. As chapter 4.7 explained, in terms of the mathematical expression (10.19) it means that all wave functions that can be obtained from (10.19) by swapping particle numbers must be combined together equally into a single wave function. There may not actually be $95!$ of them, but there will be a lot, and there is no way to actually list such a massive mathematical expression here. It is much easier in terms of the graphical depiction figure 10.1: graphically all these countless system eigenfunctions differ only with respect to the numbers in the particles. And since in the final eigenfunction, all particles are present in exactly the same way, then so are their numbers within the particles; the numbers no longer add distinguishing information and can be left out. That makes the graphical depiction of the correct example eigenfunction for a system of identical bosons as in figure 10.2.

Figure 10.3: Graphical depiction of an arbitrary system energy eigenfunction for 31 identical fermions.

For a system of identical fermions, (like electrons, protons, or neutrons,) the eigenfunctions must change sign if two particles are swapped. As chapter 4.7 showed, that means that you cannot create an eigenfunction for a system of 95 fermions from the example eigenfunction (10.19) and the swapped versions of it. Various single-particle eigenfunctions appear multiple times in (10.19), like $\pes7//p//$, which is occupied by particles 1, 5, and 48. A system eigenfunction for 95 identical fermions requires 95 different single-particle eigenfunctions. Graphically, the example figure 10.2, which is fine for a system of identical bosons, is completely unacceptable for a system of identical fermions, because there cannot be more than one particle in a given type of single-particle eigenstate. A depiction of an arbitrary energy eigenfunction that is acceptable for a system of 31 identical fermions is in figure 10.3.

As explained in chapter 4.7, a neat way of writing down the system energy eigenfunction of the pictured example is to form a Slater determinant from the “occupied states”

$$\pes{1}//p//, \pes{2}//p//, \pes{3}//p//, \ldots, \pes{48}//p//, \pes{59}//p//, \pes{63}//p//.$$

It is good to meet old friends again, isn’t it?

Now consider what happens in relativistic quantum mechanics. For example, suppose an electron and positron annihilate each other. What are you going to do, leave holes in the argument list of your wave function, where the electron and positron used to be? Or worse, what if a photon with very high energy hits an heavy nucleus and creates an electron-positron pair in the collision from scratch? Are you going to scribble in a set of additional arguments for the new particles into your mathematical wave function? Scribble in another row and column in the Slater determinant for your electrons? That is voodoo mathematics.

And if positrons are too weird for you, consider photons, the particles of electromagnetic radiation, like ordinary light. As section 8.14.5 showed, the electrons in hot surfaces create and destroy photons readily when the thermal equilibrium shifts. Moving at the speed of light, with zero rest mass, photons are as relativistic as they come. Good luck scribbling in trillions of new arguments for the photons into your wave function when your black box heats up. Then there are solids; as section 8.14.6 showed, the phonons of crystal vibrational waves are the equivalent of the photons of electromagnetic waves.

One of the key insights of quantum field theory is to do away with classical mathematical forms of the wave function such as (10.19) or the Slater determinants. Instead, the graphical depictions, such as the examples in figures 10.2 and 10.3, are captured in terms of mathematics. How do you do that? By listing how many particles are in each type of single-particle state, in other words, by listing the single-state “occupation numbers.”

Consider the example bosonic eigenfunction of figure 10.2. The occupation numbers for that state would be

$$\vec\imath = \left\vert 2,3, 1,3,3,1,3, 2,1,1,3,1,2,2, 0,3,2,2,2,1,0,2,\ldots\right\rangle$$

indicating that there are two bosons in single-particle state $\pes1//p//$, three in $\pes2//p//$, one in $\pes3//p//$, etcetera. Knowing those numbers is completely equivalent to knowing the classical system energy eigenfunction; it could be reconstructed from them. Similarly, the occupation numbers for the example fermionic eigenfunction of figure 10.3 would be

$$\vec\imath = \left\vert 1,1, 1,1,1,1,1, 1,1,1,1,0,1,1, 1,0,1,1,1,1,1,0,1,\ldots\right\rangle$$

Such sets of occupation numbers are called “Fock states.” Each describes one system energy eigenfunction.

The most general wave function for a set of $I$ particles is a linear combination of all the Fock states whose occupation numbers add up to $I$. In relativistic applications like photons in a box, there is no constraint on the number of particles and all states are possible. The set of all possible wave functions that can be formed from linear combinations of all the Fock states regardless of number of particles is called the “Fock space.”

How about the case of distinguishable particles as in figure 10.1? In that case, the numbers inside the particles also make a difference, so where do they go?? The answer of quantum field theory is to deny the existence of generic particles that take numbers. There are no generic particles in quantum field theory. There is a field of electrons, there is a field of protons, (or quarks, actually), there is a field of photons, etcetera, and each of these fields is granted its own set of occupation numbers. There is no way to describe a generic particle using a number. For example, if there is an electron in a single particle state, in quantum field theory it means that the “electron field” has a one-particle excitation at that energy state; no particle numbers are involved.

Some physicist feel that this is a strong point in favor of believing that quantum field theory is the way nature really works. In the classical formulation of quantum mechanics, the (anti)symmetrization requirements are an additional ingredient, added to explain the data. In quantum field theory, it comes naturally: particles that are not indistinguishable simply cannot be described by the formalism. Still, our convenience in describing it is an uncertain motivator for nature.

The successful analysis of the blackbody spectrum in section 8.14.5 already testified to the usefulness of the Fock space. If you check the derivations leading to it, they were all conducted based on occupation numbers. A classical wave function for a system of photons was never written down; in fact, that cannot be done.

Figure 10.4: Example wave functions for a system with just one type of single particle state. Left: identical bosons; right: identical fermions.

There is a lot more involved in quantum field theory than just the blackbody spectrum, of course. To explain some of the basic ideas, simple examples can be helpful. The simplest example that can be studied involves just one type of single-particle state, say just a single-particle ground state. The graphical depiction of an arbitrary example wave function is then as in figure 10.4. In nonrelativistic quantum mechanics, it would be a completely trivial quantum system. In the case of identical bosons, all $I$ of them would have to go into the only type of state there is. In the case of identical fermions, there can only be one fermion, and it has to go into the only state there is.

But when particles can be created or destroyed, things get more interesting. When there is no given number of particles, there can be any number of identical bosons within that single particle state. That allows $\vert\rangle$ (no particles,) $\vert 1\rangle$ (1 particle), $\vert 2\rangle$ (2 particles), etcetera. And the general wave function can be a linear combination of those possibilities. It is the same for identical fermions, except that there are now only the states $\vert\rangle$ (no particles) and $\vert 1\rangle$ (1 particle).

A relativistic system with just one type of single-particle state does seems very artificial, raising the question how esoteric the example is. But there are in fact two very well established classical systems that behave just like this:

1. The one-dimensional harmonic oscillator has energy levels that happen to be exactly equally spaced. It can pick up an energy above the ground state that is any whole multiple of $\hbar\omega$, where $\omega$ is its angular frequency. If you are willing to accept the “particles” to be quanta of energy of size $\hbar\omega$, then it provides a model of a bosonic system with just one single-particle state. Any whole number particles can go into that state, each contributing energy $\hbar\omega$ to the system. (The half particle representing the ground state energy is in this interpretation considered to be a build-in part of the single-particle-state box in figure 10.4.) Reformulating the results of chapter 2.6.2 in quantum field theory terms: the harmonic oscillator ground state $h_0$ is the state $\vert\rangle$ with zero particles, the excited state $h_1$ is the state $\vert 1\rangle$ with one particle $\hbar\omega$, the excited state $h_2$ is the state $\vert 2\rangle$ with two particles $\hbar\omega$, etcetera. The general wave function, either way, is a linear combination of these states, expressing an uncertainty in energy. Oops, excuse very much, an uncertainty in the number of these energy particles!
2. A single electron has exactly two spin states. It can pick up exactly one unit $\hbar$ of $z$-momentum above the spin-down state. If you accept the “particles” to be single quanta of $z$-momentum of size $\hbar$, then it provides an example of a fermionic system with just one single-particle state. There can be either zero or one quantum $\hbar$ of angular momentum in the single-particle state. The general wave function is a linear combination of the state with one angular momentum “particle” and the state with no angular momentum “particle”. This example admittedly is quite poor, since normally when you talk about a particle, you talk about an amount of energy, like in Einstein’s mass-energy relation. If it bothers you, think of the electron as being confined inside a magnetic field; then the spin-up state is associated with a corresponding increase in energy.

While the above two examples of “relativistic” systems with only one single-particle state are obviously made up, they do provide a very valuable sanity check on any relativistic analysis.

Not only that, the two examples are also very useful to understand the difference between a zero wave function and the so-called “vacuum state”

$$\fbox{$\displaystyle \vert\vec 0\rangle \equiv \vert,0,0,\ldots\rangle $} %$$ (10.20)

in which all occupation numbers are zero. The vacuum state is a normalized, nonzero, wave function just like the other possible sets of occupation numbers; it describes that there are no particles with certainty. You can see it from the two examples above: for the harmonic oscillator, the state $\vert\rangle$ is the ground state $h_0$; for the electron-spin example, it is the spin-down state. These are completely normal eigenstates that the system can be in. They are not zero wave functions, which would be unable to give any probabilities.

10.2.2 Annihilation and creation operators

The key to relativistic quantum mechanics is that particles can be annihilated or created. So it may not be surprising that it is very helpful to define operators that “annihilate” and “create” particles .

To keep the notations relatively simple, it will initially be assumed that there is just one type of single particle state. Graphically that means that there is just one single particle state box, like in figure 10.4. However, there can be an arbitrary number of particles in that box.

10.2.2.1 Definition

The desired actions of the creation and annihilation operators are sketched in figure 10.5. An annihilation operator $\widehat a$ turns a state $\big\vert i\big\rangle$ with $i$ particles into a state $\big\vert i-1\big\rangle$ with $i-1$ particles, and a creation operator $\widehat a^\dagger$ turns a state $\big\vert i\big\rangle$ with $i$ particles into a state $\big\vert i+1\big\rangle$ with $i+1$ particles.

Figure 10.5: Annihilation and creation operators for a system with just one type of single particle state. Left: identical bosons; right: identical fermions.

Mathematically, the operators are defined by the relations

$$\widehat a\big\vert i\big\rangle = \alpha _i \big\vert i-1... ... a^\dagger \big\vert 1\big\rangle = 0 \mbox{ for fermions} %$$ (10.21)

where the $\alpha _i$ and $\alpha^\dagger _i$ are numerical constants still to be chosen. To avoid having to write the special cases separately each time, $\alpha _0$ will be defined to be zero and, if it is a fermionic system, so will $\alpha^\dagger _1$; then you do not really need to worry about the fact that $\big\vert{-}1\big\rangle$ and $\big\vert 2\big\rangle$ do not exist.

Note that it is mathematically perfectly OK to define linear operators by specifying what they do to the basis states of a system. But you must hope that they will turn out to be operators that are mathematically helpful. To help achieve that, you want to chose the numerical constants appropriately. Consider what happens if the operators are applied in sequence:

$$\widehat a^\dagger \widehat a\big\vert i\big\rangle = \alpha^\dagger _{i-1}\alpha _i \big\vert i\big\rangle$$

Reading from right to left, the order in which the operators act on the state, first $\widehat a$ destroys a particle, then $\widehat a^\dagger$ restores it again. It gives the same state back, except for the numerical factor $\alpha^\dagger _{i-1}\alpha _i$. That makes every state $\big\vert i\big\rangle$ an eigenvector of the operator $\widehat a^\dagger \widehat a$ with eigenvalue $\alpha^\dagger _{i-1}\alpha _i$.

If the constants $\alpha^\dagger _{i-1}$ and $\alpha _i$ are chosen to make the eigenvalue a real number, then the operator $\widehat a^\dagger \widehat a$ will be Hermitian. More specifically, if they are chosen to make the eigenvalue equal to $i$, then $\widehat a^\dagger \widehat a$ will be the “particle number operator” whose eigenvalues are the number of particles in the single-particle state. The most logical choice for the constants to achieve that is clearly

$$\alpha _i=\sqrt{i} \qquad \alpha^\dagger _{i-1}=\sqrt{... ... \mbox{ except } \alpha^\dagger _1 = 0\mbox{ for fermions} %$$ (10.22)

This choice of constants is particularly convenient since it makes the operators $\widehat a$ and $\widehat a^\dagger$ Hermitian conjugates. That means that if you take them to the other side in an inner product, they turn into each other:

$$\Big\langle \big\vert{\underline i}\big\rangle \Big\vert \... ...ne i}\big\rangle \Big\vert \big\vert i\big\rangle \Big\rangle$$

To see why, first note that states with different occupation numbers are taken to be orthonormal in Fock space. If the total number of particles is given, that follows from the classical form of the wave function. And the simple harmonic oscillator and spin examples of the previous subsection illustrate that it still applies when particles can be created or destroyed: these examples were just rewrites of orthonormal wave functions.

It follows that in the first equality above, the inner products are only nonzero if $i={\underline i}+1$: after lowering the particle number with $\widehat a$, or raising it with $\widehat a^\dagger$, the particle numbers must be the same at both sides of the inner product. When $i={\underline i}+1$, $\alpha _i=\alpha^\dagger _{\underline i}=\sqrt{i}$ so the equality still applies. The second equality is just the complex conjugate of the first, with a change in notations. It remains true for fermions, despite the fact that $\alpha^\dagger _{f1}=0$ instead of $\sqrt{2}$, because there is no $\big\vert 2\big\rangle$ state for which it would make a difference. Also, if it is true for the basis states, it is true for any combination of them.

You may well wonder why $\widehat a^\dagger \widehat a$ is the particle count operator; why not $\widehat a\widehat a^\dagger$? The reason is that $\widehat a\widehat a^\dagger$ would not work for the state $\big\vert\big\rangle$ unless you took $\alpha^\dagger _0$ or $\alpha _1$ zero, and then they could no longer create or annihilate the corresponding state. Still, it is interesting to see what the effect of $\widehat a\widehat a^\dagger$ is; according to the chosen definitions, for bosons

$$\widehat a_b\widehat a^\dagger _b \big\vert i\big\rangle = \sqrt{i+1}\sqrt{i+1} \big\vert i\big\rangle$$

So the operator $\widehat a_b\widehat a^\dagger _b$ has eigenvalues one greater than the number of particles. That means that if you subtract $\widehat a_b\widehat a^\dagger _b$ and $\widehat a^\dagger _b\widehat a_b$, you get the unit operator that leaves all states unchanged. And the difference between $\widehat a_b\widehat a^\dagger _b$ and $\widehat a^\dagger _b\widehat a_b$ is by definition the commutator of $\widehat a_b$ and $\widehat a^\dagger _b$, indicated by square brackets:

$$\fbox{$\displaystyle [\widehat a_b,\widehat a^\dagger _b... ...a^\dagger _b - \widehat a^\dagger _b \widehat a_b = 1 $} %$$ (10.23)

Isn’t that cute! Of course, $[\widehat a_b,\widehat a_b]$ and $[\widehat a^\dagger _b,\widehat a^\dagger _b]$ are zero since everything commutes with itself.

It does not work for fermions, because $\alpha^\dagger _{f1}=0$ instead of $\sqrt{2}$. But for fermions, the only state for which $\widehat a_{\!f}\widehat a^\dagger _{\!f}$ produces something nonzero is $\big\vert\big\rangle$ and then it leaves the state unchanged. Similarly, the only state for which $\widehat a^\dagger _{\!f}\widehat a_{\!f}$ produces something nonzero is $\big\vert 1\big\rangle$ and then it leaves that state unchanged. That means that if you add $\widehat a_{\!f}\widehat a^\dagger _{\!f}$ and $\widehat a^\dagger _{\!f}\widehat a_{\!f}$ together, it reproduces the same state state whether it is $\big\vert\big\rangle$ or $\big\vert 1\big\rangle$ (or any combination of them). The sum of $\widehat a_{\!f}\widehat a^\dagger _{\!f}$ and $\widehat a^\dagger _{\!f}\widehat a_{\!f}$ is by definition called the “anticommutator” of $\widehat a_{\!f}$ and $\widehat a^\dagger _{\!f}$ and is indicated by curly brackets:

$$\fbox{$\displaystyle \{\widehat a_{\!f},\widehat a^\dagg... ...\!f} + \widehat a^\dagger _{\!f} \widehat a_{\!f} = 1 $} %$$ (10.24)

Isn’t that neat? Note also that $\{\widehat a_b,\widehat a_b\}$ and $\{\widehat a_b,\widehat a_b\}$ are zero since applying either operator twice ends up in a non-existing state.

How about the Hamiltonian? Well, for noninteracting particles the energy of $i$ particles is $i$ times the single particle energy ${\vphantom' E}^p$. And since the operator that gives the number of particles is $\widehat a^\dagger \widehat a$, that is ${\vphantom' E}^p\widehat a^\dagger \widehat a$. So, the total Hamiltonian for noninteracting particles becomes:

$$\fbox{$\displaystyle H = {\vphantom' E}^p\widehat a^\dagger \widehat a+ E_{ve} $} %$$ (10.25)

where $E_{ve}$ stands for the vacuum, or ground state, energy of the system where there are no particles. This then allows the Schrödinger equation to be written in terms of occupation numbers and creation and annihilation operators.

10.2.2.2 The caHermitians

It is important to note that the creation and annihilation operators are not Hermitian, and therefore cannot correspond to physically measurable quantities. But since they are Hermitian conjugates, it is easy to form Hermitian operators from them:

$$\widehat P \equiv {\textstyle\frac{1}{2}}(\widehat a^\dagg... ...style\frac{1}{2}} {\rm i}(\widehat a^\dagger - \widehat a) %$$ (10.26)

Conversely, the creation and annihilation operators can be written as

$$\widehat a= \widehat P + {\rm i}\widehat Q \qquad \widehat a^\dagger = \widehat P - {\rm i}\widehat Q %$$ (10.27)

In lack of a better name that the author knows of, this book will call $\widehat{P}$ and $\widehat{Q}$ the caHermitians.

For bosons, the following commutators follow from the ones for the creation and annihilation operators:

$$[\widehat P_b,\widehat Q_b]= {\textstyle\frac{1}{2}}{\rm i}\q... ...\dagger _b\widehat a_b,\widehat Q_b] = {\rm i}\widehat P_b %$$ (10.28)

Therefore $\widehat{P}_b$ and $\widehat{Q}_b$ neither commute with each other, nor with the Hamiltonian. It follows that whatever physical variables they may stand for will not be certain at the same time, and will develop uncertainty in time if initially certain.

The Hamiltonian (10.25) for noninteracting particles may be written in terms of the caHermitians using (10.27) and (10.28), to give

$$H = {\vphantom' E}^p\left(\widehat P_b^2 + \widehat Q_b^2 -{\textstyle\frac{1}{2}}\right) + E_{ve} %$$ (10.29)

Often the energy turns out to be simply proportional to $\widehat{P}_b^2+\widehat{Q}_b^2$; then the vacuum energy must be half a particle.

For fermions, the following useful relations follow from the anticommutators for the creation and annihilation operators

$$\widehat P_{\!f}^2 = {\textstyle\frac{1}{4}} \qquad \widehat Q_{\!f}^2 = {\textstyle\frac{1}{4}} %$$ (10.30)

The Hamiltonian then becomes

$$H = {\vphantom' E}^p\left({\rm i}[\widehat P_{\!f},\widehat Q_{\!f}] + {\textstyle\frac{1}{2}}\right) + E_{ve} %$$ (10.31)

10.2.2.3 Examples

It is interesting to see how these ideas work out for the two example systems with just one single-particle state as described at the end of subsection 10.2.1.

Consider first the example of bosons that are energy quanta of a one-dimensional harmonic oscillator. The following discussion will derive the harmonic oscillator solution from scratch using the creation and annihilation operators. It provides an alternative to the much more algebraic derivation of chapter 2.6 and its note {A.12}.

The classical Hamiltonian can be written, in the notations of chapter 2.6,

$$H = \frac{1}{2m}{\widehat p}^2 + {\textstyle\frac{1}{2}} m\omega^2{\widehat x}^2$$

or in terms of $\hbar\omega$:

$$H = \hbar\omega \left(\frac{{\widehat p}^2}{2\hbar m\omega} + \frac{m\omega{\widehat x}^2}{2\hbar}\right)$$

From comparison with (10.29), it looks like maybe the caHermitians are

$$\widehat P = \sqrt{\frac{m\omega}{2\hbar}}x \qquad \widehat Q = \sqrt{\frac{1}{2\hbar m\omega}}{\widehat p}$$

For now just define them that way; also define $E_{ve}={\textstyle\frac{1}{2}}\hbar\omega$ and

$$\widehat a= \widehat P + {\rm i}\widehat Q \qquad \widehat a^\dagger = \widehat P - {\rm i}\widehat Q$$

It has not yet been shown that $\widehat a$ and $\widehat a^\dagger$ are annihilation and creation operators. Nor that the Hamiltonian can be written in terms of them, instead of using $\widehat{P}$ and $\widehat{Q}$.

However, the commutator $[\widehat{P},\widehat{Q}]$ is according to the canonical commutator ${\rm i}{\textstyle\frac{1}{2}}$, which is just as it should be. That in turn implies that $[\widehat a,\widehat a^\dagger ]=1$. Then the Hamiltonian can indeed be written as

$$H = \hbar\omega (\widehat a^\dagger \widehat a+ {\textstyle\frac{1}{2}})$$

To see whether $\widehat a^\dagger$ is a creation operator, apply the Hamiltonian on a state $\widehat a^\dagger \big\vert i\big\rangle$ where $\big\vert i\big\rangle$ is some arbitrary energy state whose further properties are still unknown:

$$H\widehat a^\dagger \big\vert i\big\rangle = \hbar\omega(\... ...tstyle\frac{1}{2}}\widehat a^\dagger )\big\vert i\big\rangle$$

But $\widehat a\widehat a^\dagger$ can be written as $\widehat a^\dagger \widehat a+1$ because of the unit commutator, so

$$H\widehat a^\dagger \big\vert i\big\rangle = \hbar\omega\w... ...angle + \hbar\omega\widehat a^\dagger \big\vert i\big\rangle$$

The first term is just $\widehat a^\dagger$ times the Hamiltonian applied on $\big\vert i\big\rangle$, so this term multiplies $\widehat a^\dagger \big\vert i\big\rangle$ with the energy $E_i$ of the original $\big\vert i\big\rangle$ state. The second term multiplies $\widehat a^\dagger \big\vert i\big\rangle$ also by a constant, but now $\hbar\omega$. It follows that $\widehat a^\dagger \big\vert i\big\rangle$ is an energy eigenstate like $\big\vert i\big\rangle$, but with an energy that is one quantum $\hbar\omega$ higher. That does assume that $\widehat a^\dagger \big\vert i\big\rangle$ is nonzero, because eigenfunctions may not be zero. Similarly, it is seen that $\widehat a\big\vert i\big\rangle$ is an energy eigenstate with one quantum $\hbar\omega$ less in energy, if nonzero. So $\widehat a^\dagger$ and $\widehat a$ are indeed creation and annihilation operators.

Keep applying $\widehat a$ on the state $\big\vert i\big\rangle$ to lower its energy even more. This must eventually terminate in zero because the energy cannot become negative. (That assumes that the eigenfunctions are mathematically reasonably well behaved, as the solution of chapter 2.6 verifies they are. That can also be seen without using these solutions, so it is not cheating.) Call the final nonzero state $\big\vert\big\rangle$. Solve the fairly trivial equation $\widehat a\big\vert\big\rangle =0$ to find the lowest energy state $\big\vert\big\rangle =h_0(x)$ and note that it is unique. (And that is the same one as derived in chapter 2.6.) Use $\widehat a^\dagger$ to go back up the energy scale and find the other energy states $\big\vert i\big\rangle =h_i$ for $i=1$ ,2, 3, ...Verify using the Hamiltonian that going down in energy with $\widehat a$ and then up again with $\widehat a^\dagger$ brings you back to a multiple of the original state, not to some other state with the same energy or to zero. Conclude therefore that all energy states have now been found. And that their energies are spaced apart by whole multiples of the quantum $\hbar\omega$.

While doing this, it is convenient to know not just that $\widehat a\big\vert i\big\rangle$ produces a multiple of the state $\big\vert i-1\big\rangle$, but also what multiple $\alpha _i$ that is. Now the $\alpha _i$ can be made real and positive by a suitable choice of the normalization factors of the various energy eigenstates $\big\vert i\big\rangle$. Then the $\alpha^\dagger _i$ are positive too because $\alpha^\dagger _{i-1}\alpha _i$ produces the number of energy quanta in the Hamiltonian. The magnitude can be deduced from the square norm of the state produced. In particular, for $\widehat a\big\vert i\big\rangle$:

$$\alpha _i^*\alpha _i \Big\langle \widehat a\big\vert i\b... ...at a^\dagger \widehat a\big\vert i\big\rangle \Big\rangle = i$$

the first because the Hermitian conjugate of $\widehat a$ is $\widehat a^\dagger$ and the latter because $\widehat a^\dagger \widehat a$ must give the number of quanta. So $\alpha _i=\sqrt{i}$, and then $\alpha^\dagger _i=\sqrt{i+1}$ from the above. The harmonic oscillator has been solved. This derivation using the ideas of quantum field theory is much neater than the classical one; just compare the very logical story above to the algebraic mess in note {A.12}.

It should be noted, however, that a completely equivalent derivation can be given using the classical description of the harmonic oscillator. Many books do in fact do it that way, e.g. [11]. In the classical treatment, the creation and annihilation operators are called the “ladder” operators. But without the ideas of quantum field theory, it is difficult to justify the ladder operators by anything better than as a weird mathematical trick that just turns out to work.

If you have read the advanced section on angular momentum, the example system for fermions is also interesting. In that model system, the Hamiltonian is a multiple of the angular momentum in the $z$-direction of an electron. The state $\big\vert\big\rangle$ is the spin-down state ${\downarrow}$ and the state $\big\vert 1\big\rangle$ is the spin-up state ${\uparrow}$. Now the annihilation operator must turn ${\uparrow}$ into ${\downarrow}$ and ${\downarrow}$ into zero. In terms of the so-called Pauli spin matrices of section 9.1.9, the operator that does that is ${\textstyle\frac{1}{2}}(\sigma_x-{\rm i}\sigma_y)$. Similarly, the creation operator is ${\textstyle\frac{1}{2}}(\sigma_x+{\rm i}\sigma_y)$. That makes the caHermitians ${\textstyle\frac{1}{2}}\sigma_x$ and $-{\textstyle\frac{1}{2}}\sigma_y$. The commutator ${\rm i}[P,Q]$ that appears in the Hamiltonian is then ${\textstyle\frac{1}{2}}\sigma_z$, which is a multiple of the angular momentum in the $z$-direction as it should be.

10.2.2.4 More single particle states

Now consider the case that there is more than one type of single-particle state. Graphically there is now more than one particle box, as in figures 10.2 and 10.3. Then an annihilation operator $\widehat a_n$ and a creation operator $\widehat a^\dagger _n$ must be defined for each type of single-particle state $\pes{n}//p//$. In other words, there is one for each occupation number $i_n$. The mathematical definition of these operators for bosons is

$$\fbox{$\displaystyle \begin{array}{l} \displaystyle\st... ...ots,i_{n-1},i_n+1,i_{n+1},\ldots\rangle \end{array} $} %$$ (10.32)

The commutation relations are

$$\fbox{$\displaystyle \left[\widehat a_{b,n},\widehat a_{... ..._{b,{\underline n}}\right] = \delta_{n{\underline n}} $} %$$ (10.33)

Here $\delta_{n{\underline n}}$ is the Kronecker delta, equal to one if $n={\underline n}$, and zero in all other cases. These commutator relations apply for $n\ne{\underline n}$ because then the operators do unrelated things to different single-particle states, so it does not make a difference in which order you apply them. For example, $\widehat a_{b,n}\widehat a_{b,{\underline n}}=\widehat a_{b,{\underline n}}\widehat a_{b,n}$, so the commutator $\widehat a_{b,n}\widehat a_{b,{\underline n}}-\widehat a_{b,{\underline n}}\widehat a_{b,n}$ is zero. For $n={\underline n}$, they are unchanged from the case of just one single-particle state.

For fermions it is a bit more complex. The graphical representation of the example fermionic energy eigenfunction figure 10.3 cheats a bit, because it suggests that there is only one classical wave function for a given set of occupation numbers. Actually, there are two logical ones, based on how the particles are ordered; the two are the same except that they have the opposite sign. Suppose that you create a particle in a state $n$; classically you would want to call that particle 1, and then create a particle in a state ${\underline n}$, classically you would want to call it particle 2. Do the particle creation in the opposite order, and it is particle 1 that ends up in state ${\underline n}$ and particle 2 that ends up in state $n$. That means that the classical wave function will have changed sign, but the occupation-number wave function will not unless you do something. What you can do is define the annihilation and creation operators for fermions as follows:

$$\fbox{$\displaystyle \begin{array}{l} \displaystyle\st... ...ots,i_{n-1},1,i_{n+1},\ldots\rangle = 0 \end{array} $} %$$ (10.34)

The only difference from the annihilation and creation operators for just one type of single-particle state is the potential sign changes due to the $(-1)^{\ldots}$. It adds a minus sign whenever you swap the order of annihilating/creating two particles in different states. For the annihilation and creation operators of the same state, it may change both their signs, but that does nothing much: it leaves the important products such as $\widehat a^\dagger _n\widehat a_n$ and the anticommutators unchanged.

Of course, you can define the annihilation and creation operators with whatever sign you want, but putting in the sign pattern above may produce easier mathematics. In fact, there is an immediate benefit already for the anticommutator relations; they take the same form as for bosons, except with anticommutators instead of commutators:

$$\fbox{$\displaystyle \left\{\widehat a_{f,n},\widehat a_... ...{f,{\underline n}}\right\} = \delta_{n{\underline n}} $} %$$ (10.35)

These relationships apply for $n\ne{\underline n}$ exactly because of the sign change caused by swapping the order of the operators. For $n={\underline n}$, they are unchanged from the case of just one single-particle state.

The Hamiltonian for a system of non interacting particles is now found by summing over all types of single-particle states:

$$\fbox{$\displaystyle H = \sum_n {\vphantom' E}^p_n \widehat a^\dagger _n \widehat a_n + E_{ve,n} $} %$$ (10.36)

This Hamiltonian conserves the number of particles in each state; particles destroyed by the annihilation operator are immediately restored by the creation operator. Phrased differently, this Hamiltonian commutes with the operator $\widehat a^\dagger _{\underline n}\widehat a_{\underline n}$ giving the number of particles in a state $\pes{\underline n}//p//$, so energy eigenstates can be taken to be states with given numbers of particles in each single-particle state. Such systems can be described by classical quantum mechanics. But the next subsection will give an example of particles that do interact, making quantum field theory essential.

10.2.3 Quantization of radiation

In the discussion of the emission and absorption of radiation in chapter 5.3.3, the electromagnetic field was the classical Maxwell one. However, that cannot be right. According to the Planck-Einstein relation, the electromagnetic field comes in discrete quanta of energy $\hbar\omega$ called photons. A classical electromagnetic field cannot explain that.

The electromagnetic field must be described by operators that act nontrivially on a wave function that includes photons. To identify these operators will involve two steps. First a more careful look needs to be taken at the classical electromagnetic field, and in particular at its energy. By comparing that with the Hamiltonian in terms of creation and annihilation operators, as given in the previous section, the operators corresponding to the electromagnetic field can then be inferred.

To make the process more intuitive, it helps to initially assume that the electromagnetic field is not in the form of a traveling wave, but of radiation confined to a suitable box of volume $V$.

10.2.3.1 Classical energy

The discussion on emission and absorption of radiation in chapter 5.3.3 assumed the electromagnetic field to be the single traveling wave

$$\vec E = {\hat k}E_0 \cos(\omega t -\phi - k y) \qquad c \vec B = {\hat\imath}E_0 \cos(\omega t - \phi - k y)$$

where $\vec E$ and $\vec B$ are the electric and magnetic fields, respectively, $\omega$ is the frequency, $k$ is the wave number $\omega/c$, and $\phi$ is an arbitrary constant phase angle. This wave is moving in the positive $y$ direction with the speed of light $c$. To get a standing wave, add a second wave going the other way:

$$\vec E = -{\hat k}E_0 \cos(\omega t -\phi + k y) \qquad c \vec B = {\hat\imath}E_0 \cos(\omega t - \phi + k y).$$

If these are added together, using the trig formula for separating the cosines into time-only and space-only sines and cosines, the result is

$$\vec E = {\hat k}e_k(t) \sin(k y) \qquad c \vec B = {\hat\imath}b_k(t) \cos(k y) %$$ (10.37)

$$e_k(t) = 2 E_0 \sin(\omega t -\phi) \qquad b_k(t) = 2 E_0 \cos(\omega t -\phi) %$$ (10.38)

Alternatively, just plug the assumption (10.37) directly into Maxwell’s equations. That produces

$$\frac{{\rm d}e_k}{{\rm d}t} = \omega b_k \qquad \frac{{\rm d}b_k}{{\rm d}t} = - \omega e_k %$$ (10.39)

The solution to those equations are the coefficients (10.38) above.

Such a standing wave solution is appropriate for a box with perfectly conducting walls at $y=0$ and $y=\ell_y$, where $\sin(k\ell_y)=0$. On perfectly conducting walls the electric field $\vec E$ must be zero on behalf of Ohm’s law. For the other surfaces of the box, just assume periodic boundary conditions over some chosen periods $\ell_x$ and $\ell_z$, their length does not make a difference here.

Next, it is shown in basic textbooks on electromagnetics that the energy in an electromagnetic field is

$$E_V = {\textstyle\frac{1}{2}}\epsilon_0 \int_V \vec E^2 + c^2\vec B^2 {\,\rm d}^3{\skew0\vec r} %$$ (10.40)

where $\epsilon_0=8.85\,10^{-12}$ C$^2$/J m is the permittivity of space. This energy is typically derived by comparing the energy obtained from discharging a condenser with the electric field it holds when charged, and from a coil compared to its magnetic field. Note that this expression implies that the energy of the electric field of a point charge is infinite.

As an aside, if the energy $E_V$ is differentiated with respect to time, substituting in Maxwell’s equations to get rid of the time derivatives, and cleaning up, the result is

$$\frac{{\rm d}E_V}{{\rm d}t} = - \epsilon_0c \int_V \nabla \cdot( \vec E \times c \vec B) {\,\rm d}^3{\skew0\vec r}$$

From the divergence theorem it can now be seen that the flow rate of electromagnetic energy is given by the “Poynting vector”

$$\epsilon_0c^2 \vec E \times \vec B %$$ (10.41)

This was included because other books have Poynting vectors and you would be very disappointed if yours did not.

The important point here is that in terms of the coefficients $e_k$ and $b_k$, the energy is found to be

$$E_V = \frac{\epsilon_0V}{4} \left(e_k^2+b_k^2\right) %$$ (10.42)

10.2.3.2 Quantization

Following the Planck-Einstein relation, electromagnetic radiation should come in photons, each with one unit $\hbar\omega$ of energy. This indicates that the energy in the electromagnetic field is not a classical value, but corresponds to the discrete eigenvalues of some as yet unknown Hamiltonian operator. The question then is, what is that Hamiltonian?

Unfortunately, there is no straightforward way to deduce quantum mechanics operators from mere knowledge of the classical approximation. Vice-versa is not a problem: given the operators, it is fairly straightforward to deduce the corresponding classical equations for a macroscopic system. It is much like at the start of this book, where it was postulated that the momentum of a particle corresponds to the operator $\hbar\partial/{\rm i}\partial{x}$. That was a leap of faith. However, it was eventually seen, in chapter 5, that it did produce the correct classical momentum for macroscopic systems, as well as correct quantum results like the energy levels of the hydrogen atom, in chapter 3.2. A similar leap of faith will be needed to quantize the electromagnetic field.

Whatever the details of the Hamiltonian, it is clear that the appropriate mathematical tool is here quantum field theory. After all, photons are not conserved particles. Atoms readily absorb them or radiate new ones when they heat up. Now the system considered here involves only one mode of radiation; therefore the wave function can be indicated by the simple Fock space ket $\big\vert i\big\rangle$, where $i$ is the number of photons present in the mode. Also, since the single-photon energy is $\hbar\omega$, the quantum field Hamiltonian operator (10.29) becomes

$$H = \hbar\omega\left(\widehat P^2 + \widehat Q^2 -{\textstyle\frac{1}{2}}\right) + E_{ve}$$

Somehow then, the caHermitian operators $\widehat{P}$ and $\widehat{Q}$ must be identified. They must produce the classical equations of motion in the macroscopic limit, including the classical energy (10.42),

$$E_V = \frac{\epsilon_0V}{4} \left(e_k^2+b_k^2\right)$$

Comparing the two under macroscopic conditions in which $\frac12\hbar\omega$ and the ground state energy $E_{ve}$ can be ignored, the very simplest assumption is that the caHermitians are scaled versions of the coefficients $e_k$ and $b_k$:

$$e_k \to 2 \varepsilon_p P = \varepsilon_p (\widehat a^\dag... ...n_p Q = \varepsilon_p {\rm i}(\widehat a^\dagger -\widehat a)$$ (10.43)

where the scaling factor must be

$$\varepsilon_p = \sqrt{\frac{\hbar\omega}{\epsilon_0V}} %$$ (10.44)

This scaling factor can be thought of as the square root of the nominal mean square electric field per photon.

If this association of mode coefficients with caHermitian operators is indeed correct even under non macroscopic conditions, one immediate consequence is that the ground state energy must be equal to that of half a photon. And that is just for the single mode considered here. Since there are infinitely many modes of radiation, the total vacuum energy is infinite.

While that is certainly counterintuitive, it may be noted that even classically, the energy in the electromagnetic field is infinite, assuming that electrons are indeed point charges. On the other hand, the caHermitians are the Hermitian components of the very logically and simply defined creation and annihilation operators, and you would really expect them to be physically meaningful. They certainly were for the harmonic oscillator and spin system examples of subsection 10.2.2.

Therefore, the assumption will be made that the scaled caHermitians appear in the quantized electromagnetic field where the measurable quantities $e_k$ and $b_k$ appear in the classical electromagnetic field:

$$\skew4\widehat{\vec E}({\skew0\vec r}) = {\hat k}\varepsil... ...ilon_p \cos(k y) {\rm i}(\widehat a^\dagger -\widehat a) %$$ (10.45)

This process of replacing the coefficients of the modes by operators is called “second quantization.” No, there was no earlier quantization of the electromagnetic field involved. The word “second” is there for historical reasons: physicists have historically found it hysterical to confuse students.

Note that just like the time-dependent momentum of a classical particle $p(t)$ becomes the time-independent operator $\hbar\partial/{\rm i}\partial{x}$, the creation and annihilation operators are taken to be time-independent. In quantum mechanics, the time dependence is in the wave function, not the operators:

$$\big\vert\Psi\big\rangle = \sum_{i} c_i e^{{\rm i}E_i t/\hbar} \big\vert i\big\rangle$$

where the energy $E_i$ of the state with $i$ photons is $(i+{\textstyle\frac{1}{2}})\hbar\omega$. (The Heisenberg picture absorbs the time dependence in the operator; that is particularly convenient for relativistic applications. However, true relativity is beyond the scope of this book and this discussion will stay as close to the normal Schrödinger picture as possible.)

To see whether this quantization of the electromagnetic field does indeed make sense, its immediate consequences will now be explored. First consider the Hamiltonian according to the Newtonian (or is that Maxwellian?) analogy:

$$H = {\textstyle\frac{1}{2}}\epsilon_0 \int_V \skew4\wide... ...E}^2 + c^2 \skew4\widehat{\vec B}^2 {\,\rm d}^3{\skew0\vec r}$$

Substituting in the quantized electromagnetic field, (10.44) and (10.45), and integrating,

$$H = {\textstyle\frac{1}{2}}\epsilon_0 \frac{\hbar\omega}{\... ...extstyle\frac{1}{2}} V(\widehat a^\dagger -\widehat a)^2\Big)$$

Multiplying out gives

$$H = {\textstyle\frac{1}{4}} \hbar\omega \Big(2\widehat a^\dagger \widehat a+ 2\widehat a\widehat a^\dagger \Big)$$

and using the commutator $\widehat a\widehat a^\dagger -\widehat a^\dagger \widehat a=1$ to get rid of $\widehat a\widehat a^\dagger$ gives

$$H = \hbar\omega \Big(\widehat a^\dagger \widehat a+{\textstyle\frac{1}{2}}\Big)$$

which is just as it should be.

Also consider the equation for the expectation value $\langle{P}\rangle$:

$$\frac{{\rm d}\langle P\rangle}{{\rm d}t} = \frac{{\rm i}}{\hbar} \langle[H,P]\rangle = \omega \langle Q\rangle$$

using (10.28) and (10.29). Similarly for the expectation value $\langle{Q}\rangle$:

$$\frac{{\rm d}\langle Q\rangle}{{\rm d}t} = \frac{{\rm i}}{\hbar} \langle[H,Q]\rangle = - \omega \langle P\rangle$$

Those are the same equations as satisfied by $e_k$ and $b_k$. They are also the equations for the position and linear momentum of a harmonic oscillator, when suitably scaled. It is often said that the mode amplitudes of the electromagnetic field are mathematically modelled as quantum harmonic oscillators.

Now suppose there are exactly $i$ photons, what is the expectation value of the electric field at a given position and time? Well, the wave function will be

$$\big\vert\Psi\big\rangle = c_ie^{{\rm i}(i+\frac12)\omega t}\big\vert i\big\rangle$$

and so

$$\langle \vec E\rangle({\skew0\vec r},t) = {\hat k}\varepsi... ...g\vert \widehat a^\dagger + \widehat a\big\vert i \big\rangle$$

That is zero because $\widehat a^\dagger$ and $\widehat a$ applied on $\big\vert i\big\rangle$ create states $\big\vert i+1\big\rangle$ and $\big\vert i-1\big\rangle$ that are orthogonal to the $\big\vert i\big\rangle$ in the left side of the inner product. The same way the expectation magnetic field is zero!

Oops, not quite as expected. In fact, the previous subsection pointed out that the caHermitians do not commute with the Hamiltonian. If the number of photons, hence the energy, is certain, then the electromagnetic field is not. And the caHermitians also do not commute with each other; if the electric field is certain, then the magnetic field is not, and vice-versa.

To get something resembling a classical electric field, there must be uncertainty in energy. In particular, if the coefficients of multiple energy states are nonzero, then the expectation values of the electric and magnetic fields become

$$\begin{eqnarray*} \langle \vec E\rangle({\skew0\vec r},t) & = & {\hat k} \va... ...1}e^{{\rm i}\omega t} - c_ic_{i-1}^*e^{-{\rm i}\omega t}\right) \end{eqnarray*}$$

To check this, observe that for any $i$, $\big\langle{i}\big\vert\widehat a^\dagger \big\vert{\underline i}\big\rangle$ is only nonzero if ${\underline i}=i-1$ and the same for its complex conjugate $\big\langle{{\underline i}}\big\vert\widehat a\big\vert i\big\rangle$. Renotating

$$\sum_i\sqrt{i} c_ic^*_{i-1} \equiv {\rm i}C_1e^{{\rm i}\phi_1}$$

where $C_1$ and $\phi_1$ are real constants produces

$$\begin{eqnarray*} \langle \vec E\rangle({\skew0\vec r},t) & = & {\hat k} 2 \... ...t\imath} 2 \varepsilon_p C_1 \sin(ky) \cos(\omega t - \phi_1) \end{eqnarray*}$$

That is in the form of the classical electromagnetic field (10.37).

Similarly it may be found that

$$\begin{eqnarray*} \langle E^2\rangle({\skew0\vec r},t) & = & 2 \varepsilon_p^2... ... {\textstyle\frac{1}{2}} - C_2 \sin(2\omega t - 2\phi_2)\right] \end{eqnarray*}$$

where the expectation number of photons and the constants are defined by

$$\sum_i \vert c_i\vert^2 i \equiv \langle i\rangle \qquad... ... c_{i-2}^* \sqrt{i(i-1)} \equiv {\rm i}C_2 e^{{\rm i}2\phi_2}$$

Note that the mean square expectation electric field is $\varepsilon_p^2$ per photon, with half a photon left in the ground state.

Consider now a “photon packet” in which the numbers of photons with nonzero probabilities are restricted to a relatively narrow range. However, assume that the range is still large enough so that the coefficients can vary slowly from one number of photons to the next, except for possibly a constant phase difference:

$$c_{i-1} \approx c_i e^{-{\rm i}(\phi-\frac{\pi}{2})}$$

Then

$$C_1 \approx \sqrt{\langle i\rangle} \quad C_2 \approx \lan... ..._1 \approx \phi \quad 2 \phi_2 \approx 2 \phi + \frac{\pi}{2}$$

In that case, the expectation electromagnetic field above becomes the classical one, with an energy of about $\langle{i}\rangle$ photons.

10.2.3.3 Photon spin

If photons act as particles, they should have a value of spin. Physicists have concluded that the appropriate spin operators are

$${\widehat S}_x = -{\rm i}\hbar({\hat\jmath}{\hat k}{\cdot}... ...imath}{\hat\jmath}{\cdot}-{\hat\jmath}{\hat\imath}{\cdot}) %$$ (10.46)

Note the dots. These operators need to act on vectors and then produce vectors times dot products. (In terms of linear algebra, the unit vectors above are multiplied as tensor products.)

If the above operators are correct, they should satisfy the fundamental commutation relations (4.20). They do. For example:

$$[{\widehat S}_x,{\widehat S}_y]= -\hbar^2 ({\hat\jmath}{\ha... ...) ({\hat\jmath}{\hat k}{\cdot}-{\hat k}{\hat\jmath}{\cdot})$$

Since the unit vectors are mutually orthogonal, when multiplying out the first term only the dot product ${\hat k}\cdot{\hat k}=1$ is nonzero, and the same for the second term. So

$$[{\widehat S}_x,{\widehat S}_y]= -\hbar^2 {\hat\jmath}{\hat\i... ...hat\imath}{\hat\jmath}{\cdot} = {\rm i}\hbar {\widehat S}_z$$

The next question is how to apply them, given that the wave function of photons is a Fock space ket without a precise physical interpretation. However, following the ideas of quantum mechanics, presumably photon wave functions can be written as linear combinations of wave functions with definite electric fields. All the ones for the single mode considered here have an electric field that is proportional to ${\hat k}$, and the operators above can be applied on that. But unfortunately, ${\hat k}$ is not an eigenvector of any of the spin components above.

However, even given the direction of wave propagation and wave number, there are still two different modes: the electric field could be fluctuating in the $x$-direction instead of the $z$-direction. (Fluctuating in an oblique direction is just a linear combination of these two independent modes, and not another possibility.) In short, any of the considered electromagnetic modes can be rotated 90 degrees around the $y$-axis to give a second mode with an electric field proportional to ${\hat\imath}$. If these modes are combined pairwise in the combination ${\hat k}+{\rm i}{\hat\imath}$ it produces an eigenstate of ${\widehat S}_y$ with spin $\hbar$, while ${\hat k}-{\rm i}{\hat\imath}$ produces an eigenstate with spin $-\hbar$. That can easily be checked by direct substitution in the eigenvalue problem.

Note that there are only two independent states, so that is it. There is no third state with spin zero in the $y$-direction, the direction of wave propagation. The missing state reflects the classical limitation that the electric field cannot have a component in the direction of wave propagation. However, it can be seen using the analysis of chapter 9.1 that suitable combinations of equal amounts of spin forward and backward in $y$ can produce photons with zero spin in a direction normal to the direction of propagation.

One thing should still be checked: that the magnetic field does not conflict. Now, if the electric field is rotated from ${\hat k}$ to ${\hat\imath}$, the magnetic field rotates from ${\hat\imath}$ to $-{\hat k}$. So the eigenstates have magnetic fields proportional to ${\hat\imath}-{\rm i}{\hat k}$ and ${\hat\imath}+{\rm i}{\hat k}$. That is just $-{\rm i}$, respectively ${\rm i}$ times the vectors of the electric field, so even including the magnetic field the states remain eigenstates.

10.2.3.4 Traveling waves

To get the quantized electromagnetic field of traveling waves, the quickest way to get there is to take the standing wave apart using

$$\sin{ky} = \frac{e^{{\rm i}ky} - e^{-{\rm i}ky}}{2i} \qquad \cos{ky} = \frac{e^{{\rm i}ky} + e^{-{\rm i}ky}}{2}$$

Then the parts that propagate forwards in $y$ must be of the form ${\widehat a}e^{{{\rm i}}ky}$ or ${\widehat a^\dagger }e^{-{{\rm i}}ky}$. To see why, just look at the expectation values of these terms for the generic wave function $\sum_ic_ie^{-{{\rm i}}i{\omega}t}\big\vert i\big\rangle$.

A single wave of wave number $k$ moving in the positive $y$-direction and polarized in the $z$-direction therefore takes the form

$$\widehat{\vec E}({\skew0\vec r}) = {\hat k}\varepsilon_p' ... ...ehat a^\dagger e^{-{\rm i}k y} - \widehat ae^{{\rm i}k y}) %$$ (10.47)

where the scaling constant is

$$\varepsilon_p' = \sqrt{\frac{\hbar\omega}{2\epsilon_0V'}}$$ (10.48)

For a traveling wave, it is more physical to assume that it is in a box that is periodic in the $y$-direction; that is true for a box with twice the length, hence a volume $V'=2V$. The constant $\varepsilon_p'$ can be thought of as the square root of half the mean square electric field per photon.

However, in general there will be a similar wave polarized in the $x$-direction. And then there will be pairs of such waves for different directions of propagation and different wave numbers $k$. To describe all these waves, it is convenient to combine wave number and direction of propagation into a wave number vector ${\vec k}$ that has the magnitude of $k$ and the direction of wave propagation. Then the complete electromagnetic field operators become

$$\widehat{\vec E}({\skew0\vec r}) = \varepsilon_p' \sum_{... ...c k},\mu} e^{{\rm i}{\vec k}\cdot{\skew0\vec r}}) \strut %$$ (10.49)

$$c \widehat{\vec B}({\skew0\vec r}) = \varepsilon_p' \sum... ...c k},\mu} e^{{\rm i}{\vec k}\cdot{\skew0\vec r}}) \strut %$$ (10.50)

where the unit vector ${\hat\imath}_1$ must be chosen normal to the direction of wave propagation and ${\hat\imath}_2={\hat\imath}_1\times{\vec k}/k$.

10.2.4 Spontaneous emission

In this subsection, the spontaneous emission rate of excited atoms will be derived. It may be recalled that this was done in chapter 5.3.10 following an argument given by Einstein. However, that was cheating: it peeked at the answer for blackbody radiation. This section will verify that a quantum treatment gives the same answer.

Like in chapter 5.3, consider the interaction of an atom with electromagnetic radiation, but this time, do it right, using the quantized electromagnetic field instead of the classical one. The approach will again be to consider the interaction for a two state system involving a single electromagnetic wave and two energy levels of the atom. The total effects should then again follow from summation over all waves and energy levels.

The most appropriate energy states are now

$$\psi_1 = \psi_L\big\vert i+1\big\rangle \qquad \psi_2 = \psi_H\big\vert i\big\rangle$$ (10.51)

In state $\psi_1$ the atom is in the lower energy state and there are $i+1$ photons in the wave; in state $\psi_2$, the atom is excited, but one photon has disappeared. Note that these wave functions depend on both the atom energy level and on the number of photons.

The Hamiltonian is now

$$H = H_0 + \hbar\omega(\widehat a^\dagger \widehat a+{\text... ...) + ez \varepsilon_p' {\rm i}(\widehat a^\dagger -\widehat a)$$ (10.52)

where $H_0$ is the Hamiltonian of the unperturbed atom, the second term is the Hamiltonian of the electromagnetic field, and the final term is the $e\widehat{E}_zz$ interaction energy of the electron with the electric field, but now quantized as in (10.47). It was again assumed that interaction with the magnetic field can be ignored and that the atom is small enough compared to the electromagnetic wave length that it can be taken to be at the origin.

Next the Hamiltonian matrix coefficients are needed. The first one is

$$H_{11} = \big\langle i+1\big\vert\psi_L\Big\vert \Big(H_... ...a^\dagger -\widehat a)\Big) \psi_L\big\vert i+1\big\rangle$$

The first term of the Hamiltonian acts on the spatial state and produces the lower atom energy level. The second term acts on the Fock space ket to produce the electromagnetic energy of $i+1$ photons. The final term produces zero, because of the symmetry of the atom eigenstates. Alternatively, it produces zero because $\widehat a^\dagger$ and $\widehat a$ produce states $\big\vert i+2\big\rangle$ and $\big\vert i\big\rangle$ that are orthogonal to the $\big\vert i+1\big\rangle$ in the left side of the inner product. So

$$H_{11} = E_L + (i+1+{\textstyle\frac{1}{2}})\hbar\omega$$

and similarly

$$H_{22} = E_H + (i+{\textstyle\frac{1}{2}})\hbar\omega$$

The remaining Hamiltonian matrix coefficient is

$$H_{12} = \big\langle i+1\big\vert\psi_L\Big\vert \Big(H_... ...t a^\dagger -\widehat a)\Big) \psi_H\big\vert i\big\rangle$$

Here the first two terms produce zero, because $\psi_1$ and $\psi_2$ are orthonormal eigenstates of these operators. The third term produces

$$H_{12} = {\rm i}\varepsilon_p'\langle\psi_L\vert ez\vert\psi_R\rangle\sqrt{i+1}$$

because the creation operator $\widehat a^\dagger$ turns $\big\vert i\big\rangle$ into $\sqrt{i+1}\big\vert i+1\big\rangle$.

As in chapter 5.3, a Hamiltonian of a simplified two-state system may be defined as

$$\overline{H}_{12} \equiv H_{12} e^{-{\rm i}\int(H_{22}-H_{11}){\rm d}t/\hbar}$$

where from the above

$$H_{22}-H_{11} = E_H - E_L - \hbar\omega = \hbar(\omega_p-\omega)$$

where $\omega_p$ is the frequency of the photon released in a transition from the higher to the lower atom energy state. The simplified two state system becomes

$$\dot{\bar a} = \frac{\varepsilon_p'}{\hbar} \langle\psi_... ...si_R\rangle^*\sqrt{i+1} e^{-{\rm i}(\omega-\omega_p)t} \bar a$$

where $\bar{a}$ and $\bar{b}$ give the probabilities of states $\psi_1$ and $\psi_2$ respectively.

Consider a system that starts out with the atom in the excited state, $a_0=0$ and $b_0=1$. Then, if the perturbation is weak over the time that it acts, $\bar{b}$ can be approximated as one in the first equation, and the transition probability to the lower atom energy state $\psi_2$ is found to be

$$\Big\vert\bar a\Big\vert^2 = {\textstyle\frac{1}{4}} \frac... ...a_p)t}{{\textstyle\frac{1}{2}}(\omega-\omega_p)t} \right)^2$$

For a large number of photons $i$, this is the same as the classical result (5.35), because $4{\varepsilon_p'}^2$ is the peak square electric field per photon.

But now consider the electromagnetic ground state where the number of photons $i$ is zero. The transition probability above is as if there is still one photon of electromagnetic energy left. And as noted in chapter 5.3.10, that is exactly what is needed to explain spontaneous emission using the quantum equations.

Some additional observations may be interesting. While you may think of it as excitation by the ground state electromagnetic field, the actual energy of the ground state was earlier seen to be half a photon, not one photon. And the zero level of energy should not affect the dynamics anyway. According to the analysis here, spontaneous emission is a twilight effect: the Hamiltonian coefficient $H_{12}$ is the energy if the atom is not excited if the atom is excited. Think of it in classical common sense terms. There is an excited atom and no photons around it. (Or if you prefer, the number of photons is as low as it can ever get. Classical common sense would make that zero.) Why would things ever change? But in quantum mechanics, the twilight term allows the excited atom to interact with the electromagnetic radiation of the photon that would be there if it was not excited. Sic.

10.2.5 Field operators

As noted at the start of this section, quantum field theory is particularly suited for relativistic applications because the number of particles can vary. However, in relativistic applications, it is often best to work in terms of position coordinates instead of single-particle energy eigenfunctions. Relativistic applications must make sure that matter does not exceed the speed of light, and that coordinate systems moving at different speeds are physically equivalent and related through the Lorentz transformation. These conditions are posed in terms of position and time.

To handle such problems, the annihilation and creation operators can be converted into so-called “field operators” that annihilate or create particles at a given position in space. Now classically, a particle at a given position ${\skew0\vec r}_0$ corresponds to a wave function that is a delta function, $\Psi=\delta({\skew0\vec r}-{\skew0\vec r}_0)$, chapter 5.4. A delta function can be written in terms of the single-particle eigenfunctions $\psi_n$ as $\sum{c}_n\psi_n$. Here the constants can be found from taking inner products; $c_n=\langle\psi_n\vert\Psi\rangle$, and that gives $c_n=\psi^*_n({\skew0\vec r}_0)$ because of the property of the delta function to pick out that value of any function that it is in an inner product with. Since $c_n$ is the amount of eigenfunction $\psi_n$ that must be annihilated/created to annihilate/create the delta function at ${\skew0\vec r}_0$, the field operators become

$$\widehat a({\skew0\vec r}) = \sum_n \psi^*_n({\skew0\vec r... ...c r}) = \sum_n \psi^*_n({\skew0\vec r}) \widehat a^\dagger _n$$ (10.53)

where the subscript zero was dropped from ${\skew0\vec r}$ since it is no longer needed to distinguish from the independent variable of the delta function. It means that ${\skew0\vec r}$ is now the position at which the particle is annihilated/created.

In the case of particles in free space, the energy eigenfunctions are the momentum eigenfunctions $e^{{\rm i}{\skew 4\widehat{\skew{-.5}\vec p}}\cdot{\skew0\vec r}/\hbar}$, and the sums become integrals called the Fourier transforms; see chapter 5.4 and 5.5.1 for more details. In fact, unless you are particularly interested in converting the expression (10.36) for the Hamiltonian, basing the field operators on the momentum eigenfunctions works fine even if the particles are not in free space.

A big advantage of the way the annihilation and creation operators were defined now shows up: their (anti)commutation relations are effectively unchanged in taking linear combinations. In particular

$$\fbox{$\displaystyle \>\Big[\widehat a_b({\skew0\vec r})... ...}')\Big]\> = \delta^3({\skew0\vec r}-{\skew0\vec r}') $} %$$ (10.54)

$$\fbox{$\displaystyle \Big\{\widehat a_{\!f}({\skew0\vec ... ...r}')\Big\} = \delta^3({\skew0\vec r}-{\skew0\vec r}') $} %$$ (10.55)

In other references you might see an additional constant multiplying the three-dimensional delta function, depending on how the position and momentum eigenfunctions were normalized.

The field operators help solve a vexing problem in relativistic quantum mechanics; how to put space and time on equal footing as relativity needs. The classical Schrödinger equation ${\rm i}\hbar\psi_t=H\psi$ treats space and time quite different; the spatial derivatives, in $H$, are second order, but the time derivative is first order. The first-order time derivative allows you to think of the time coordinate as simply a label on different spatial wave functions, one for each time, and application of the spatial Hamiltonian produces the change from one spatial wave function to the next one, a time ${\rm d}{t}$ later. Of course, you cannot think of the spatial coordinates in the same way; even if there was only one spatial coordinate instead of three: the second order spatial derivatives do not represent a change of wave function from one position to the next.

As section 9.2 discussed, for spinless particles, the Schrödinger equation can be converted into the Klein-Gordon equation, which turns the time derivative to second order by adding the rest mass energy to the Hamiltonian, and for electrons, the Schrödinger equation can be converted into the Dirac equation by switching to a vector wave function, which turns the spatial derivatives to first order. But there are problems; for example, the Klein-Gordon equation does not naturally conserve probabilities unless the solution is a simple wave; the Dirac equation has energy levels extending to minus infinity that must be thought of as being already filled with electrons to prevent an explosion of energy when the electrons fall down those states. Worse, filling the negative energy states would not help for bosons, since bosons do not obey the exclusion principle.

The field operators turn out to provide a better option, because they allow both the spatial coordinates and time to be converted into labels on annihilation and creation operators. It allows relativistic theories to be constructed that treat space and time in a uniform way.

10.2.6 An example using field operators

This example exercise from Srednicki [22, p. 11] compares quantum field theory to the classical formulation of quantum mechanics. The objective is to convert the classical spatial Schrödinger equation for $I$ particles,

$${\rm i}\hbar \frac{\partial \Psi}{\partial t} = \left[ ... ...ew0\vec r}_i-{\skew0\vec r}_{\underline i}) \right] \Psi %$$ (10.56)

into quantum field form. The classical wave function has the positions of the numbered particles and time as arguments:

$$\mbox{classical quantum mechanics:}\quad \Psi=\Psi({\ske... ...skew0\vec r}_2,{\skew0\vec r}_3,\ldots,{\skew0\vec r}_I;t) %$$ (10.57)

where ${\skew0\vec r}_1$ is the position of particle 1, ${\skew0\vec r}_2$ is the position of particle 2, etcetera.

In quantum field theory, the wave function for exactly $I$ particles takes the form

$$\vert\Psi\rangle = \int_{{\rm all\ }{\skew0\vec r}_1}\ld... ...\,\rm d}^3{\skew0\vec r}_1\ldots{\rm d}^3{\skew0\vec r}_I %$$ (10.58)

where ket $\vert\Psi\rangle$ is the wave function expressed in Fock space kets, and plain $\Psi(\ldots)$ is to be shown to be equivalent to the classical wave function above. The Fock space Hamiltonian is

$$H = \int_{{\rm all\ }{\skew0\vec r}} \widehat a^\dagger ({\skew0\ve... ...kew0\vec r}) \right] \widehat a({\skew0\vec r}) {\,\rm d}^3{\skew0\vec r}$$

$$+ {\textstyle\frac{1}{2}} \int_{{\rm all\ }{\skew0\vec r}}\int_... ...skew0\vec r}) {\,\rm d}^3{\skew0\vec r}{\rm d}^3{\underline{\skew0\vec r}}%$$ (10.59)

It is to be shown that the Fock space Schrödinger equation for $\vert\Psi\rangle$ produces the classical Schrödinger equation (10.56) for function $\Psi(\ldots)$, whether it is a system of bosons or fermions.

Before trying to tackle this problem, it is probably a good idea to review representations of functions using delta functions. As the simplest example, a wave function $\Psi(x)$ of just one spatial coordinate can be written as

$$\Psi(x) = \int_{{\rm all\ }{\underline x}} \; \under... ...nderline x}){\rm d}{\underline x}$}} _{{\rm basis\ states}}$$

The way to think about the above integral expression for $\Psi(x)$ is just like you would think about a vector in three dimensions being written as $\vec{v}=v_1{\hat\imath}+v_2{\hat\jmath}+v_3{\hat k}$ or a vector in 30 dimensions as $\vec{v}=\sum_{i=1}^{30}v_i{\hat\imath}_i$. The $\Psi({\underline x})$ are the coefficients, corresponding to the $v_i$-components of the vectors. The $\delta(x-{\underline x}){\rm d}{\underline x}$ are the basis states, just like the unit vectors ${\hat\imath}_i$. If you want a graphical illustration, each $\delta(x-{\underline x}){\rm d}{\underline x}$ would correspond to one spike of unit height at a position ${\underline x}$ in figure 1.3, and you need to sum (integrate) over them all, with their coefficients, to get the total vector.

Now $H\Psi(x)$ is just another function of $x$, so it can be written similarly:

$$\begin{eqnarray*} H \Psi(x) & = & \int_{{\rm all\ }{\underline x}} H\Psi... ... \right] \delta(x - {\underline x}) {\,\rm d}{\underline x} \end{eqnarray*}$$

Note that the Hamiltonian acts on the coefficients, not on the basis states. You may be surprised by this since if you straightforwardly apply the Hamiltonian, in terms of $x$, on the integral expression for $\Psi(x)$, you get

$$H \Psi(x) = \int_{{\rm all\ }{\underline x}} \Psi({\un... ...delta(x - {\underline x}) \right] {\,\rm d}{\underline x}$$

in which the Hamiltonian acts on the basis states, not the coefficients. However, the two expressions are indeed the same. (You can see that using a couple of integrations by parts in the latter, after recognizing that differentiation of the delta function with respect to $x$ or ${\underline x}$ is the same save for a sign change. Much better, make the change of integration variable $u={\underline x}-x$ before applying the Hamiltonian to the integral.)

The bottom line is that you do not want to use the expression in which the Hamiltonian is applied to the basis states, because derivatives of delta functions are highly singular objects that you should not touch with a ten foot pole. Still, there is an important observation here: you might either know what an operator does to the coefficients, leaving the basis states untouched, or what it does to the basis states, leaving the coefficients untouched. Either one will tell you the final effect of the operator, but the mathematics is different.

Now that the general terms of engagement have been discussed, it is time to start solving Srednicki’s problem. First consider the expression for the wave function

$$\vert\Psi\rangle = \int_{{\rm all\ }{\skew0\vec r}_1}\ld... ...3{\skew0\vec r}_I} _{{\rm Fock\ space\ basis\ state\ kets}}$$

The ket $\vert\vec0\rangle$ is the vacuum state, but the preceding creation operators $\widehat a^\dagger$ create particles at positions ${\skew0\vec r}_1,{\skew0\vec r}_2,\ldots$. So the net state becomes a Fock state where a particle called 1 is in a delta function at a position ${\skew0\vec r}_1$, a particle called 2 in a delta function at position ${\skew0\vec r}_2$, etcetera. The classical wave function $\Psi(\ldots)$ determines the probability for the particles to actually be at those states, so it is the coefficient of that Fock state ket. The integration gives the combined wave function $\vert\Psi\rangle$ as a ket state in Fock space.

Note that Fock states do not know about particle numbers. A Fock basis state is the same regardless what the classical wave function calls the particles. It means that the same Fock basis state ket reappears in the integration above at all swapped positions of the particles. (For fermions read: the same except possibly a sign change, since swapping the order of application of any two $\widehat a^\dagger$ creation operators flips the sign, compare subsection 10.2.2.) This will become important at the end of the derivation.

As far as understanding the Fock space Hamiltonian, for now you may just note a superficial similarity in form with the expectation value of energy. Its appropriateness will follow from the fact that the correct classical Schrödinger equation is obtained from it.

The left hand side of the Fock space Schrödinger equation is evaluated by pushing the time derivative inside the integral as a partial:

$$\begin{array}{l} \displaystyle {\rm i}\hbar\frac{{\rm d}... ...{\skew0\vec r}_1\ldots{\rm d}^3{\skew0\vec r}_I \end{array}$$

so the time derivative drops down on the classical wave function in the normal way.

Applying the Fock-space Hamiltonian (10.59) on the wave function is quite a different story, however. It is best to start with just a single particle:

$$H \vert\Psi\rangle = \int_{{\rm all\ }{\skew0\vec r}}\in... ...0\rangle {\,\rm d}^3{\skew0\vec r}_1{\rm d}^3{\skew0\vec r}$$

The field operator $\widehat a({\skew0\vec r})$ may be pushed past the classical wave function $\Psi(\ldots)$; $\widehat a({\skew0\vec r})$ is defined by what it does to the Fock basis states while leaving their coefficients, here $\Psi(\ldots)$, unchanged:

$$H \vert\Psi\rangle = \int_{{\rm all\ }{\skew0\vec r}}\in... ...0\rangle {\,\rm d}^3{\skew0\vec r}_1{\rm d}^3{\skew0\vec r}$$

It is now that the (anti)commutator relations become useful. The fact that for bosons $[\widehat a({\skew0\vec r})\widehat a^\dagger ({\skew0\vec r}_1)]$ or for fermions $\{\widehat a({\skew0\vec r})\widehat a^\dagger ({\skew0\vec r}_1)\}$ equals $\delta^3({\skew0\vec r}-{\skew0\vec r}_1)$ means that you can swap the order of the operators as long as you add a delta function term:

$$\widehat a_b({\skew0\vec r})\widehat a^\dagger _b({\skew0\... ...{\skew0\vec r}) + \delta^3({\skew0\vec r}-{\skew0\vec r}_1)$$

But when you swap the order of the operators in the expression for $H\vert\Psi\rangle$, you get a factor $\widehat a({\skew0\vec r})\vert\vec0\rangle$, and that is zero, because applying an annihilation operator on the vacuum state produces zero, figure 10.5. So the delta function is all that remains:

$$H \vert\Psi\rangle = \int_{{\rm all\ }{\skew0\vec r}}\in... ...0\rangle {\,\rm d}^3{\skew0\vec r}_1{\rm d}^3{\skew0\vec r}$$

Integration over ${\skew0\vec r}_1$ now picks out the value $\Psi({\skew0\vec r},t)$ from function $\Psi({\skew0\vec r}_1,t)$, as delta functions do, so

$$H \vert\Psi\rangle = \int_{{\rm all\ }{\skew0\vec r}} ... ...w0\vec r};t) \vert\vec 0\rangle {\,\rm d}^3{\skew0\vec r}$$

The creation operator $\widehat a^\dagger ({\skew0\vec r})$ can be pushed over the coefficient $H\Psi({\skew0\vec r};t)$ of the vacuum state ket for the same reason that $\widehat a({\skew0\vec r})$ could be pushed over $\Psi({\skew0\vec r}_1;t)$; these operators do not affect the coefficients of the Fock states, just the states themselves.

Then, renotating ${\skew0\vec r}$ to ${\skew0\vec r}_1$, the grand total Fock state Schrödinger equation for a system of one particle becomes

$$\begin{eqnarray*} \lefteqn{\int_{{\rm all\ }{\skew0\vec r}_1} {\rm i}\hbar\f... ...skew0\vec r}_1)\vert\vec 0\rangle {\,\rm d}^3{\skew0\vec r}_1 \end{eqnarray*}$$

It is now seen that if the classical wave function $\Psi({\skew0\vec r}_1;t)$ satisfies the classical Schrödinger equation, the Fock-space Schrödinger equation above is also satisfied. And so is the converse: if the Fock-state equation above is satisfied, the classical wave function must satisfy the classical Schrödinger equation. The reason is that Fock states can only be equal if the coefficients of all the basis states are equal, just like vectors can only be equal if all their components are equal.

If there is more than one particle, however, the latter conclusion is not justified. Remember that the same Fock space kets reappear in the integration at swapped positions of the particles. It now makes a difference. The following example from basic vectors illustrates the problem: yes, $a{\hat\imath}=a'{\hat\imath}$ implies that $a=a'$, but no, $(a+b){\hat\imath}=(a'+b'){\hat\imath}$ does not imply that $a=a'$ and $b=b'$; it merely implies that $a+b=a'+b'$. However, if additionally it is postulated that the classical wave function has the symmetry properties appropriate for bosons or fermions, then the Fock-space Schrödinger equation does imply the classical one. In terms of the example from vectors, $(a+a){\hat\imath}=(a'+a'){\hat\imath}$ does imply that $a=a'$.

Operator swapping like in the derivation above also helps to understand why the Fock-space Hamiltonian has an appearance similar to an energy expectation value. For example, consider the effect of placing the one-particle Hamiltonian between $\langle\vec0\vert\widehat a({\underline{\skew0\vec r}}_1)\Psi^*({\underline{\skew0\vec r}}_1;t)$ and $\Psi({\skew0\vec r}_1;t)\widehat a^\dagger ({\skew0\vec r}_1)\vert\vec0\rangle$ and integrating over all ${\underline{\skew0\vec r}}_1$ and ${\skew0\vec r}_1$.

So the problem has been solved for a system with one particle. Doing it for $I$ particles will be left as an exercise for your mathematical skills.