A.24 Quantum version of spontaneous emission

Chapter 7.8 explained the general inter­action between atoms and electro­magnetic fields. However, sponta­neous emission of radiation was found using a dirty trick due to Einstein. He peeked at the solution for blackbody radiation. This addendum will give a proper quantum description. Warning: while this addendum tries to be reasonably self-contained, to really appreciate the details you may have to read some other addenda too.

The problem with the descriptions of emission and absorption of radiation in chapter 7.7 and 7.8 is that they assume that the electro­magnetic field is given. The electro­magnetic field is not given; it changes by one photon. That is rather important for sponta­neous emission, where it changes from no photons to one photon. To account for that correctly requires that the electro­magnetic field is properly quantized. That is done in this note.

To keep it simple, it will be assumed that the atom is a hydrogen one. Then there is just one electron to worry about. (The general analysis can be found in {A.25}). The hydrogen atom is initially in some high energy state $\psi_{\rm {H}}$. Then it emits a photon and transitions to a lower energy state $\psi_{\rm {L}}$. The emitted photon comes out in a state with energy

$$E_\gamma = \hbar\omega \approx E_{\rm {H}} - E_{\rm {L}}$$

Recall that the photon energy is given in terms of its frequency $\omega$ by the Planck-Einstein relation. This photon energy is approximately the difference between the atomic energies. It does not have to be exact; there can be some energy slop, chapter 7.6.1.

Only a single photon energy state needs to be considered at a time. At the end of the story, the results can be summed over all possible photon states. To allow for stimulated emission, it will be assumed that initially there may already be $i$ preexisting photons present. For sponta­neous emission, $i$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. The initial system state will be indicated as:

$$\psi_1 = \psi_{\rm {H}} \big\vert i\big\rangle$$

Here the so-called Fock space ket $\big\vert i\big\rangle$ is simply a concise way of indicating that there are $i$ photons in the considered photon quantum state.

In the final state the atom has decayed to a lower energy state $\psi_{\rm {L}}$. In doing so it has released 1 more photon into the considered photon state. So the final wave function is

$$\psi_2 = \psi_{\rm {L}} \big\vert i{+}1\big\rangle$$

The key to the emission process is now the set of Hamiltonian coefficients, chapter 7.6,

$$\langle E_1\rangle = \langle\psi_1\vert H \psi_1\rangle ... ...uad \langle E_2\rangle = \langle\psi_2\vert H \psi_2\rangle$$

Here $H$ is the Hamiltonian. All that really needs to be done in this note is to identify these coefficients, and in particular the so-called matrix element $H_{21}$. With the matrix element known, Fermi’s golden rule can be used to find the precise transition rate, chapter 7.6.1.

To identify the Hamiltonian coefficients, first the Hamiltonian must be identified. Recall that the Hamiltonian is the operator of the total energy of the system. It will take the form

$$H = H_{\rm {atom}} + H_\gamma + H_{\rm {atom,\gamma}}$$

The first term in the right hand side is the inherent energy of the hydrogen atom. This Hamiltonian was written down way back in chapter 4.3. However, its precise form is of no interest here. The second term in the right hand side is the energy in the electro­magnetic field. Electro­magnetic fields too have inherent energy, about $\hbar\omega$ per photon in fact. The third term is the energy of the inter­action between the atomic electron and the electro­magnetic field.

Unlike the first term in the Hamiltonian, the other two are inherently relativistic: the number of photons is hardly a conserved quantity. Photons are readily created or absorbed by a charged particle, like the electron here. And it turns out that Hamiltonians for photons are intrinsi­cally linked to operators that annihilate and create photons. Mathemati­cally, at least. These operators are defined by the relations

$$\widehat a\big\vert i\big\rangle = \sqrt{i} \big\vert i{-}... ...ig\vert i{-}1\big\rangle = \sqrt{i} \big\vert i\big\rangle %$$ (A.166)

for any number of photons $i$. In words, the annihil­ation operator $\widehat a$ takes a state of $i$ photons and turns it into a state with one less photon. The creation operator $\widehat a^\dagger$ puts the photon back in. The scalar factors $\sqrt{i}$ are a matter of convenience. If you did not put them in here, you would have to do it elsewhere.

The Hamiltonian that describes the inherent energy in the electro­magnetic field turns out to be, {A.23},

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

As a sanity check, this Hamiltonian can be applied on a state of $i$ photons. Using the definitions of the annihil­ation and creation operators given above,

$$H_\gamma \big\vert i\big\rangle = {\textstyle\frac{1}{2... ...hbar\omega(i+{\textstyle\frac{1}{2}}) \big\vert i\big\rangle$$

The factor in front of the final ket is the energy eigenvalue. So the energy in the field increases by one unit $\hbar\omega$ for each photon added, exactly as it should. The additional half photon is the ground state energy of the electro­magnetic field. Even in its ground state, the electro­magnetic field has some energy left. That is much like a one-di­mensional harmonic oscillator still has ${\textstyle\frac{1}{2}}\hbar\omega$ of energy left in its ground state, chapter 4.1.

Finally the energy of the inter­action between the electron and electro­magnetic field is needed. This third part of the total Hamiltonian is the messiest. To keep it as simple as possible, it will assumed that the transition is of the normal electric dipole type. In such transitions the electron inter­acts only with the electric part of the electro­magnetic field. In addition, just like in the analysis of chapter 7.7.1 using a classical electro­magnetic field, it will be assumed that the electric field is in the $z$-direction and propagates in the $y$-direction. (The general multipole analysis can be found in {A.25}).

Now recall that in quantum mechanics, observable properties of particles are the eigen­values of Hermitian operators, chapter 3.3. For example, the observable values of linear momentum of an electron in the $y$-direction are the eigen­values of the linear momentum operator ${\widehat p}_y$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\hbar\partial$$\raisebox{.5pt}{$/$}$${\rm i}\partial{y}$. This operator acts on the electron wave function.

Similarly, the electric field ${\cal E}_z$ that the electron inter­acts with is an observable property of the corre­sponding photons. So the observable values of the electric field must be the eigen­values of a Hermitian electric field operator $\skew4\widehat{\cal E}_z$. And this operator acts on photon wave functions.

In the analysis using a classical electro­magnetic field, the energy of inter­action between the electron and the electro­magnetic field was taken to be approximately $e{\cal E}_zz$. That is similar to the $mgh$ potential of a particle due to gravity. The electron electric charge $\vphantom0\raisebox{1.5pt}{$-$}$$e$ takes the place of the mass $m$, the electric field ${\cal E}_z$ that of the accel­eration of gravity $g$, and $z$ that of the height $h$. Using the quantized electric field, there is no given classical field ${\cal E}_z$, and instead the operator $\skew4\widehat{\cal E}_z$ must be used:

$$H_{\rm {atom,\gamma}} = e \skew4\widehat{\cal E}_z z$$

The operator $\skew4\widehat{\cal E}_z$ acts on the ket part of the combined atom-photon wave function. (And, although you may not think of it that way, the factor $z$ is really an operator that acts on the electron wave function part. That is true even in the analysis using the classical field.)

The electric field operator $\skew4\widehat{\cal E}_z$ can be identified from the appro­priate photon wave function. The photon wave function here is assumed to have its linear momentum in the $y$-direction and its unobservable electric field in the $z$-direction. The corre­sponding normalized wave function and unobservable electric field were given in {A.21.6} (A.95):

$$\skew3\vec A_\gamma^{\rm n}= \frac{\varepsilon_k}{{\rm i}k... ...\varepsilon_k = \sqrt{\frac{\hbar\omega}{\epsilon_0{\cal V}}}$$

Here $\epsilon_0$ is the permit­tivity of space. Also ${\cal V}$ is the volume of the large periodic box in which the entire system will be assumed to be located. In truly infinite space the analysis would be extremely messy, littered with ugly delta functions.

The rules to get the operator of the observable electric field were discussed in addendum {A.23}. First the unobservable electric field above is multi­plied by the annihil­ation operator, then the Hermitian conjugate of that product is added, and the sum is divided by $\sqrt{2}$:

$$\skew4\widehat{\cal E}_z = \frac{\varepsilon_k}{\sqrt{2}}(\widehat ae^{{\rm i}ky} + \widehat a^\dagger e^{-{\rm i}ky})$$

(Note that for the usual Schrödinger approach followed here, time dependence is described by the wave function. Most sources switch here to a Heisenberg approach where the time-dependence is pushed into the operators. There is however no particular need to do so.)

In the electric dipole approxi­mation, it is assumed that the atom is so small compared to the wave length of the photon that $ky$ can be assumed to be zero. Therefore

$$\skew4\widehat{\cal E}_z = \frac{\varepsilon_k}{\sqrt{2}}(\widehat a+ \widehat a^\dagger )$$

The combined Hamiltonian is then

$$H = H_{\rm {atom}} + H_\gamma + e\frac{\varepsilon_k}{\sqrt{2}}(\widehat a+\widehat a^\dagger )z$$

with the first two terms as described earlier.

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

$$\langle E_1\rangle = \langle \psi_1\vert H\psi_1\rangle ... ...hat a^\dagger )z \Big)\psi_{\rm {H}}\big\vert i\big\rangle$$

Now the atomic part of the Hamiltonian produces a mere factor $E_{\rm {H}}$ when it acts on the atomic part of the right hand wave function. Further, as discussed above, the electro­magnetic Hamiltonian produces a factor $(i+\frac12)\hbar\omega$ when it acts on the right hand ket. Finally the inter­action part of the Hamiltonian does not produce a contribution. One way to see that is from the atomic inner product. The atomic inner product is zero because negative values of $z$ integrate away against positive ones. Another way to see it is from the electro­magnetic inner product. The operators $\widehat a$ and $\widehat a^\dagger$ produce states $\big\vert i{-}1\big\rangle$ respectively $\big\vert i{+}1\big\rangle$ when they act on the right hand ket. And those are orthogonal to the left hand ket; inner products between kets with different numbers of photons are zero. Kets are by definition ortho­normal.

All together then

$$\langle E_1\rangle = E_{\rm {H}} + (i+{\textstyle\frac{1}{2}})\hbar\omega$$

The same way

$$\langle E_2\rangle = E_{\rm {L}} + (i+1+{\textstyle\frac{1}{2}})\hbar\omega$$

Finally the matrix element:

$$H_{21} = \langle \psi_2\vert H\psi_1\rangle = \big\langl... ...hat a^\dagger )z \Big)\psi_{\rm {H}}\big\vert i\big\rangle$$

In this case the atomic part of the Hamiltonian produces zero. The reason is that this Hamiltonian produces a simple scalar factor $E_{\rm {H}}$ when it acts on the right hand state. It leaves the state $\psi_{\rm {H}}$ itself unchanged. And this state produces zero in an inner product with the atomic state $\psi_{\rm {L}}$; energy eigen­states are ortho­normal. Similarly, the electro­magnetic Hamiltonian produces zero. It leaves the ket $\big\vert i\big\rangle$ in the right hand wave wave function unchanged, and that is orthogonal to the left hand $\big\langle i+1\big\vert$. However, in this case the inter­action Hamiltonian produces a non­zero contribution:

$$H_{21} = \frac{\varepsilon_k\sqrt{i+1}}{\sqrt{2}} \big\langle \psi_{\rm {L}}\big\vert e z \psi_{\rm {H}}\big\rangle$$

The reason is that the creation operator $\widehat a^\dagger$ acting on the right hand ket produces a multiple $\sqrt{i+1}$ times the left hand ket. The remaining inner product $\big\langle\psi_{\rm {L}}\big\vert ez\psi_{\rm {H}}\big\rangle$ is called the “atomic matrix element,” as it only depends on what the atomic states are.

The task laid out in chapter 7.6.1 has been accom­plished: the relativistic matrix element has been found. A final expression for the sponta­neous emission rate can now be determined.

Before doing so, however, it is good to first compare the obtained result with that of chapter 7.7.1. That section used a classical given electro­magnetic field, not a quantized one. So the comparison will show up the effect of the quant­ization of the electro­magnetic field. The section defined a modified matrix element

$$\overline{H}_{21} = H_{21} e^{{\rm i}(\langle E_2\rangle-\langle E_1\rangle)t/\hbar}$$

This matrix element determined the entire evolution of the system. For the quantized electric field discussed here, this coefficient works out to be

$$\overline{H}_{21} = \frac{\varepsilon_k\sqrt{i+1}}{\sqrt... ...\varepsilon_k = \sqrt{\frac{\hbar\omega}{\epsilon_0{\cal V}}}$$ (A.167)

where $\omega_0$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(E_{\rm {H}}-E_{\rm {L}})$$\raisebox{.5pt}{$/$}$$\hbar$.

That is essentially the same form as for the classical field. Recall that the second term in (7.44) for the classical field can be ignored. The first term is the same as above, within a constant. To see the real difference in the constants, note that the transition proba­bility is propor­tional to the square magnitude of the matrix element. The square magnitudes are:

$$\mbox{quantized: } \vert\overline{H}_{21}^2\vert = \frac... ... \psi_{\rm {L}}\big\vert e z \psi_{\rm {H}}\big\rangle\vert^2$$

Now if there is a large number $i$ of photons in the state, the two expressions are approximately the same. The electro­magnetic energy of the wave according to classical physics, $\epsilon_0{\cal E}_{\rm {f}}^2{\cal V}$$\raisebox{.5pt}{$/$}$2, {A.23}, is then approximately the number of photons $i$ $\vphantom0\raisebox{1.1pt}{$\approx$}$ $i+1$ times $\hbar\omega$.

But for sponta­neous emission there is a big difference. In that case, classical physics would take the initial electro­magnetic field ${\cal E}_{\rm {f}}$ to be zero. And that then implies that the atom stays in the excited state $\psi_{\rm {H}}$ for always. There is no electro­magnetic field to move it out of the state. So there is no sponta­neous emission.

Instead quantum mechanics takes the initial field to have $i$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 photons. But note the square matrix element above. It is not zero! The matrix element is as if there is still a full photon left in the electro­magnetic field. So sponta­neous emission can and does occur in the quantized electro­magnetic field. Also, as noted in chapter 7.8, one full photon is exactly what is needed to explain sponta­neous emission. Einstein’s $A$ coefficient has been found using pure quantum analysis. Without peeking at the black body spectrum.

That can also be seen without detouring through the messy analysis of chapter 7.7 and 7.8. To find the sponta­neous emission rate directly, the matrix element above can be plugged into Fermi’s Golden Rule (7.38) of chapter 7.6.1. The density of states needed in it was given earlier in chapter 6.3 (6.7) and 6.19. Do note that these modes include all directions of the electric field, not just the $z$-direction. To account for that, you need to average the square atomic matrix element over all three Cartesian directions. That produces the sponta­neous transition rate

$$\frac{\omega^3}{\pi\hbar c^3\epsilon_0} \frac{\vert\lang... ...\langle\psi_{\rm {L}}\vert ez\psi_{\rm {H}}\rangle\vert^2}{3}$$

The above result is the same as Einstein’s, (7.47) and (7.48). (To see why a simple average works in the final term, first note that it is obviously the right average for photons with axial linear momenta and fields. Then note that the average is independent of the angular orien­tation of the axis system in which the photons are described. So it also works for photons that are axial in any rotated coordinate system. To verify that the average is independent of angular orien­tation does not really require linear algebra; it suffices to show that it is true for rotation about one axis, say the $z$-axis.)

Some additional observ­ations may be inter­esting. You might think of the sponta­neous emission as caused by excitation from the ground state electro­magnetic field. But as seen earlier, the actual energy of the ground state is 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, chapter 5.3. The Hamiltonian coefficient $H_{21}$ is the energy if the atom is not excited and there is a photon if the atom is excited and there is no photon. In quantum mechanics, the twilight term allows the excited atom to interact with the photon that would be there if it was not excited. Sic.