D.33 Quantum field derivations

This derivation will find the properties of a system described by a Hamiltonian of the form:

$\parbox{404pt}{\hspace{15pt}\hfill$\displaystyle H = {\vphantom' E}^{\rm p}\left(\widehat P^2 + \widehat Q^2\right) + E_{\rm{ref}} $\hfill(1)}$

Here $\widehat{P}$ and $\widehat{Q}$ are Hermitian operators with commutator

$\parbox{404pt}{\hspace{15pt}\hfill$\displaystyle \left[\widehat P,\widehat Q\right] = -{\textstyle\frac{1}{2}} {\rm i} $\hfill(2)}$

and ${\vphantom' E}^{\rm p}$ and $E_{\rm {ref}}$ are constants with units of energy.

First note that the commutator (2) directly implies the uncertainty relationship, chapter 4.5.2 (4.46):

$\parbox{404pt}{\hspace{15pt}\hfill$\displaystyle \sigma_P \sigma_Q \mathrel{\raisebox{-1pt}{$\geqslant$}}{\textstyle\frac{1}{4}} $\hfill(3)}$

Also note that the evolution equations for the expec­tation values of $P$ and $Q$ follow directly from chapter 7.2 (7.4). The commutator appearing in it is readily worked out using the commutator (2) and the rules of chapter 4.5.4. Since energy eigen­states are stationary, according to the evolution equations in such states the expec­tation values of $P$ and $Q$ will have to be zero.

The equality of the $\widehat{P}$ and $\widehat{Q}$ terms in the Hamiltonian is a simple matter of symmetry. Nothing changes if you swap $\widehat{P}$ and $\widehat{Q}$, adding a minus sign for one. Then unavoidably the two terms in the Hamiltonian must be equal; it is shown below that the eigen­functions are unique.

The commutator (2) also implies that $\widehat{P}$, $\widehat{Q}$, and all their combin­ations, do not commute with the Hamiltonian. So they are not conserved quantities of the system. However, there are two combin­ations,

$\parbox{404pt}{\hspace{15pt}\hfill$\displaystyle \widehat a\equiv \widehat P ... ... \qquad \widehat a^\dagger \equiv \widehat P + {\rm i}\widehat Q $\hfill(4)}$

whose commutator with the Hamiltonian gives back a multiple of the same thing:

$$[H,\widehat a]= -{\vphantom' E}^{\rm p}\widehat a\qquad [H,\widehat a^\dagger ] = {\vphantom' E}^{\rm p}\widehat a^\dagger$$

In other words, $\widehat a$ and $\widehat a^\dagger$ are commutator eigen­operators of the Hamiltonian. The above relations are readily checked using the given commutator (2) and the rules of chapter 4.5.4.

To see why that is important, multiply both sides of the eigenvalue problems above with a system energy eigen­function of energy $E$:

$$[H,\widehat a]\psi_E = -{\vphantom' E}^{\rm p}\widehat a\psi_... ...er ] \psi_E = {\vphantom' E}^{\rm p}\widehat a^\dagger \psi_E$$

After writing out the definitions of the commutators, recognizing $H\psi_E$ as $E\psi_E$, and rearranging, that gives

$$H (\widehat a\psi_E) = (E-{\vphantom' E}^{\rm p})(\widehat... ...si_E) = (E+{\vphantom' E}^{\rm p})(\widehat a^\dagger \psi_E)$$

These results can be compared to the definition of an energy eigen­function. Then it is seen that $\widehat a\psi_E$ is an energy eigen­function with one unit ${\vphantom' E}^{\rm p}$ less energy than $\psi_E$. And $\widehat a^\dagger \psi_E$ is an energy eigen­function with one unit ${\vphantom' E}^{\rm p}$ more energy than $\psi_E$. So apparently $\widehat a$ and $\widehat a^\dagger$ act as annihil­ation and creation operators of quanta of energy ${\vphantom' E}^{\rm p}$. They act as shown to the left in figure A.6.

There are however two important caveats for these statements. If $\widehat a\psi_E$ or $\widehat a^\dagger \psi_E$ is zero, it is not an energy eigen­function. Eigen­functions must be non­zero. Also, even if the states $\widehat a\psi_E$ or $\widehat a^\dagger \psi_E$ are not zero, they will not normally be normalized states.

To get a better under­standing of these issues, it is helpful to first find the Hamiltonian in terms of $\widehat a$ and $\widehat a^\dagger$. There are two equivalent forms,

$\parbox{404pt}{\hspace{15pt}\hfill$\displaystyle H = {\vphantom' E}^{\rm p}\w... ...at a+ {\textstyle\frac{1}{2}} {\vphantom' E}^{\rm p}+E_{\rm{ref}} $\hfill(5)}$

These expressions can be verified by plugging in the definitions (4) of $\widehat a$ and $\widehat a^\dagger$ and using the commutator (2). (Note that subtracting the two expressions gives the commutator of $\widehat a$ and $\widehat a^\dagger$ to be 1.)

Now look at the first Hamiltonian first. If $\widehat a^\dagger \psi_E$ would be zero for some state $\psi_E$, then that state would have energy $E_{\rm {ref}}-{\textstyle\frac{1}{2}}{\vphantom' E}^{\rm p}$. But that is not possible. If you look at the original Hamiltonian (1), the energy must at least be $E_{\rm {ref}}$; square Hermitian operators are non­negative.

(To be more precise, if you square a Hermitian operator, you square the eigen­values, making them non­negative. It is said that the square operator is “positive definite,” or, if there are zero eigen­values, positive semi-definite. And such an operator produces non­negative expec­tation values. And the expec­tation values of the operators in the Hamiltonian do add up to the total energy; just take an inner product of the Hamiltonian eigenvalue problem with the wave function. See chapter 4.4 for more infor­mation on expec­tation values.)

It follows that $\widehat a^\dagger \psi_E$ is never zero. This operator can be applied indefinitely to find states of higher and higher energy. So there is no maximum energy.

But there is a possi­bility that $\widehat a\psi_E$ is zero. As the second form of the Hamiltonian in (5) shows, that requires that the energy of state $\psi_E$ equals

$$E_0 = {\textstyle\frac{1}{2}}{\vphantom' E}^{\rm p}+ E_{\rm {ref}}$$

Now if you start from any energy state $\psi_E$ and apply $\widehat a$ sufficiently many times, you must eventually end up at this energy level. If not, you could go on lowering the energy forever. That would be incon­sistent with the fact that the energy cannot be lower than $E_{\rm {ref}}$. It follows that the above energy is the lowest energy that a state can have. So it is the ground state energy.

And any other energy must be a whole multiple of ${\vphantom' E}^{\rm p}$ higher than the ground state energy. Otherwise you could not end up at the ground state energy by applying $\widehat a$. Therefore, the energy eigen­states can be denoted more meaningfully by $\big\vert i\big\rangle$ rather than $\psi_E$. Here $i$ is the number of quanta ${\vphantom' E}^{\rm p}$ that the energy is above the ground state level.

Now assume that the ground state is unique. In that case, there is one unique energy eigen­function at each energy level. That is a consequence of the fact that if you go down a unit in energy with $\widehat a$ and then up a unit again with $\widehat a^\dagger$, (or vice versa), you must end up not just at the same energy, but at the same state. Otherwise the state would not be an eigen­function of the Hamiltonian in one of the forms given in (5). Repeated appli­cation shows that if you go down any number of steps, and then up the same number of steps, you end up at the same state. Since every state must end up at the unique ground state, every state must be the result of applying $\widehat a^\dagger$ to the ground state sufficiently many times. There is just one such state for each energy level.

If there are two independent ground states, applying $\widehat a^\dagger$ on each gives two separate sets of energy eigen­states. And similar if there are still more ground states. Additional symbols will need to be added to the kets to keep the different families apart.

It was already mentioned that the states produced by the operators $\widehat a$ and $\widehat a^\dagger$ are usually not normalized. For example, the state $\widehat a\big\vert i\big\rangle$ will have a square magnitude given by the inner product

$$\vert\widehat a\big\vert i\big\rangle \vert^2 = \Big\langl... ...rangle \Big\vert \widehat a\big\vert i\big\rangle \Big\rangle$$

Now if you take $\widehat a$ or $\widehat a^\dagger$ to the other side of an inner product, it will change into the other one; the ${\rm i}$ in the definitions (4) will change sign. So the square magnitude of $\widehat a\big\vert i\big\rangle$ becomes

$$\vert\widehat a\big\vert i\big\rangle \vert^2 = \Big\langl... ...idehat a^\dagger \widehat a\big\vert i\big\rangle \Big\rangle$$

From the second form of the Hamiltonian in (5), it is seen that $\widehat a^\dagger \widehat a$ gives the number of energy quanta $i$. And since the state $\big\vert i\big\rangle$ is normalized, the square magnitude of $\widehat a\big\vert i\big\rangle$ is therefore $i$. That means that

$$\widehat a\big\vert i\big\rangle = c \sqrt{i}\big\vert i{-}1\big\rangle$$

where $c$ is some number of magnitude 1. Similarly

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

But note that you can always change the definition of an energy eigen­function by a constant of magnitude 1. That allows you, while going up from $\big\vert\big\rangle$ using $\widehat a^\dagger$, to redefine each state so that $d$ is 1. And if $d$ is always one, then so is $c$. Otherwise $\widehat a^\dagger \widehat a$ would not be $i$.

In the ground state, the expec­tation values of $P$ and $Q$ are zero, while the expec­tation values of $P^2$ and $Q^2$ are equal to the minimum $\frac14$ allowed by the uncertainty relation (3). The derivations of these statements are the same as those for the harmonic oscillator ground state in {D.13}.