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# 6.1: Transitions Between Stationary States

## Energy Conservation

We stated that even the flawed Bohr model correctly predicts the energy levels of the hydrogen atom, which in turn correctly predicts the emission and absorption spectrum of hydrogen. Specifically, a quantum of light produced by a transition of a hydrogen atom from a higher energy state $$\left|\;E_n\;\right>$$ to a lower energy state $$\left|\;E_n'\;\right>$$ must – to conserve energy – have an energy equal to $$E_n-E_n'$$, which relates directly to its frequency/wavelength, as we stated in Equation 5.1.8.

But even in classical physics, energy conservation is only a necessary condition, it isn't a sufficient one. For example, energy conservation is not violated by a pile of hot mashed potatoes on the floor suddenly cooling off slightly, such that the loss of thermal energy is converted into mechanical energy, causing the potatoes to suddenly fly off the floor and back onto the plate of a buffet patron too stunned to even contemplate the 5-second rule. In other words, we need to explore the mechanism by which these transitions between energy states can operate.

## Transitional States

Let's start by refreshing our memory that a general quantum state can be written – using the completeness relation – as a linear combination of stationary states (Equation 2.4.14):

$\left|\;\Psi\left(t\right)\;\right> = \sum\limits_n C_n\left|\;E_n\;\right>\;e^{-i\frac{E_n}{\hbar}t}$

Technically, we need to include all the quantum numbers in the linear combination. That is, a general state can also mix angular momentum states. But this adds extra clutter that will just obscure the message, so they will be suppressed for now. We will come back to the role they play in photon emission and absorption shortly.

Let's now suppose that the state is a superposition of just two states, which we will refer to as "initial" and "final," labeled with "$$i$$" and "$$f$$," respectively. This can be thought of as the intermediate state between the energy eigenstate that the hydrogen atoms begins with, to the energy eigenstate where it ends. This state involves only two terms from the sum:

$\left|\;\Psi_{i\;f}\left(t\right)\;\right> = C_i\left|\;E_i\;\right>\;e^{-i\frac{E_i}{\hbar}t}+C_f\left|\;E_f\;\right>\;e^{-i\frac{E_f}{\hbar}t}$

We can convert this into a time-dependent wave function:

$\Psi_{i\;f}\left(\overrightarrow r,\;t\right) = \left<\;\overrightarrow r\;|\;\Psi_{i\;f}\left(t\right)\;\right> = C_i\;\psi_i\left(\overrightarrow r\right)e^{-i\frac{E_i}{\hbar}t}+C_f\;\psi_f\left(\overrightarrow r\right)\;e^{-i\frac{E_f}{\hbar}t}\;,$

where $$\psi_i$$ and $$\psi_f$$ are the initial and final stationary-state wave functions, $$\left<\;\overrightarrow r\;|\;E_i\;\right>$$ and $$\left<\;\overrightarrow r\;|\;E_f\;\right>$$, respectively.

We can determine the probability density of this transitional state:

$\left|\Psi_{i\;f}\right|^2 = \Psi^*_{i\;f}\;\Psi_{i\;f} = \left|C_i\right|^2\;\left|\psi_i\right|^2+\left|C_f\right|^2\;\left|\psi_f\right|^2 + \left(C^*_i\;C_f\right)\psi^*_i\;\psi_f \;e^{i\frac{E_i-E_f}{\hbar}t}+ \left(C^*_f\;C_i\right)\psi^*_f\;\psi_i \;e^{-i\frac{E_i-E_f}{\hbar}t}$

Next we invoke a very important concept. A probability density for a single particle, when applied to many particles, simply becomes a particle density. Densities of properties like mass and charge are proportional to particle densities (when the particles all have the same mass or charge). We can therefore define the charge density of a single quantum-mechanical particle by multiplying its charge by its probability density. Looking at the probability density of the transitional state of the electron, we see that its charge density varies harmonically over time, with the frequency of oscillation equaling:

$\omega_{i\;f} = \dfrac{E_i-E_f}{\hbar}$

From E&M, we know that the frequency of oscillation of light coming from oscillating electric charge matches the frequency of oscillation of that charge. So it is clear how such a transition gives rise to a photon emitted (or absorbed) with the frequency that we know it must have from energy conservation. The next question is how a hydrogen atom that is in a stationary state gets into this transitional state in the first place.

## Transitions are Triggered by Perturbations

Okay, so now we need to determine how a hydrogen atom, happily residing in an excited state, enters into a transitional state and eventually falls to a lower energy state, emitting a photon in the process. The answer to this puzzle is that in fact our ideal description of the perfectly static coulomb-only potential is not possible, and even the slightest imperfection that creeps into the hamiltonian over time provides an opportunity for a transition. We'll look at this effect in a moment, but it is interesting to discuss what some possible sources of this change in the hamiltonian over time might be.

The most obvious way to affect the potential energy of the electron is to introduce an additional electromagnetic field. We could, for example, put the hydrogen atom between two capacitor plates, and fluctuate the external electric field, but surely we can be more creative than that! Well, light is an electromagnetic wave, so light that happens by the atom will introduce a time-varying tweak to the hamiltonian.

Another source of a small hamiltonian glitch is more fundamental, and is the reason that even a hydrogen atom by itself in empty space will still radiate and drop to a lower energy level from a higher one spontaneously. It turns out that even the vacuum itself is not immune to quantum-mechanical effects! We tend to think of the vacuum as being a state of zero energy, but in quantum mechanics, there is no such thing – the lowest we can ever go is the non-zero ground state! The oscillations of the vacuum wave function (called vacuum fluctuations) occur everywhere, including in the vicinity of a hydrogen atom that is all by itself. And these tweak the hamitonian of the hydrogen atom enough to induce transitions between (otherwise) stationary states, and with each transition comes a spontaneous emission of a photon.

Let's see how a small, time-dependent tweak (usually referred to as a perturbation) to the coulomb hamiltonian enables the hydrogen atom to change between energy levels. Start once again with the completeness relation, which allows us to write a general solution to the Schrödinger equation in terms of its stationary states. For the hydrogen atom hamiltonian:

$\left|\;\Psi\left(t\right)\;\right> = \sum\limits_n C_n\left|\;E_n\;\right>\;e^{-i\frac{E_n}{\hbar}t}\;,\;\;\;\;\;\text{where:}\;\;\;H_{hydrogen}\left|\;E_n\;\right>=E_n\left|\;E_n\;\right>$

The time dependence of the state expressed in Equation 6.1.6 is completely determined by the hydrogen hamiltonian. Notice that the "recipe" of how much each eigenstate contributes to the state does not change over time. The probability that a measurement of the energy in this mixture of states will be $$E_{n'}$$ does not change with time:

$\left|\left<\;E_{n'}\;|\;\Psi\left(t\right)\;\right>\right|^2 = \left[\sum\limits_n C_n\left<\;E_{n'}\;|\;E_n\;\right>\;e^{-i\frac{E_n}{\hbar}t}\right]^* \left[\sum\limits_n C_n\left<\;E_{n'}\;|\;E_n\;\right>\;e^{-i\frac{E_n}{\hbar}t}\right] = \left[\sum\limits_n C_n\;\delta_{n'n}\;e^{-i\frac{E_n}{\hbar}t}\right]^* \left[\sum\limits_n C_n\;\delta_{n'n}\;e^{-i\frac{E_n}{\hbar}t}\right] = C^*_{n'}C_{n'}$

Now let's suppose that the electron in the hydrogen atom is subjected to a small additional effect, characterized by a changing part of the hamiltonian. Then the total hamiltonian looks like:

$H_{total} = H_{hydrogen} + H\left(t\right)$

The function $$H\left(t\right)$$ can represent the effects of vacuum fluctuations, a passing photon, etc. Suppose $$H\left(t\right)=0$$ at $$t=0$$, so initially only the hydrogen hamiltonian is in effect. When the perturbation kicks-in, the new hamiltonian means we have to write a new expansion above, because the "recipe" of eigenstates will have changed. If we assume the energy eigenvalues don't change very much from those of the hydrogen atom (it is a small perturbation), then we can account for the recipe change by allowing the $$C_n$$ coefficients to be time-dependent, and leaving all the rest of the expansion the same. For our new "total" hamiltonian, we write:

$\left|\;\Psi\left(t\right)\;\right> = \sum\limits_n C_n\left(t\right)\left|\;E_n\;\right>\;e^{-i\frac{E_n}{\hbar}t}\;$

where the $$E_n$$'s are still the eigenvalues of energy for the (unperturbed) hydrogen atom. Okay, now let's plug this into Schrödinger's equation for the total hamiltonian:

$H_{total} \left|\;\Psi\left(t\right)\;\right>= i\hbar\dfrac{\partial}{\partial t}\left|\;\Psi\left(t\right)\;\right>$

We'll start with the right side of the equation. The derivative acts upon the time-dependent coefficients as well as the exponential, with a product rule. In the case of the exponential, the $$i\hbar$$ is canceled-out, leaving only an additional factor of $$E_n$$:

$i\hbar\dfrac{\partial}{\partial t}\left|\;\Psi\left(t\right)\;\right> = i\hbar\dfrac{\partial}{\partial t}\sum\limits_n C_n\left(t\right)\left|\;E_n\;\right>\;e^{-i\frac{E_n}{\hbar}t} = i\hbar\sum\limits_n \dfrac{dC_n}{dt}\left|\;E_n\;\right>\;e^{-i\frac{E_n}{\hbar}t} + \sum\limits_n C_n\left(t\right)\;E_n\;\left|\;E_n\;\right>\;e^{-i\frac{E_n}{\hbar}t}$

Now for the left side of Equation 6.1.10. We know how the hydrogen part of the hamiltonian acts on its energy eigenstates, so we get:

$\left[H_{hydrogen}+H\left(t\right)\right]\sum\limits_n C_n\left(t\right)\left|\;E_n\;\right>\;e^{-i\frac{E_n}{\hbar}t} = \sum\limits_n C_n\left(t\right) E_n \left|\;E_n\;\right>\;e^{-i\frac{E_n}{\hbar}t} + \sum\limits_n C_n\left(t\right)H\left(t\right)\left|\;E_n\;\right>\;e^{-i\frac{E_n}{\hbar}t}$

Reconstructing the Schrödinger equation by setting these last two equations equal, we see that there is a term common to both sides that cancel, leaving us with:

$i\hbar\sum\limits_n \dfrac{dC_n}{dt}\left|\;E_n\;\right>\;e^{-i\frac{E_n}{\hbar}t} = \sum\limits_n C_n\left(t\right)H\left(t\right)\left|\;E_n\;\right>\;e^{-i\frac{E_n}{\hbar}t}$

We can solve for the derivative by taking the inner product of both sides with $$\left<\;E_{n'}\;\right|$$:

$\begin{array}{r} i\hbar\sum\limits_n \dfrac{dC_n}{dt}\left<\;E_{n'}\;|\;E_n\;\right>\;e^{-i\frac{E_n}{\hbar}t} & = & \sum\limits_n C_n\left(t\right)\left<\;E_{n'}\;|\;H\left(t\right)|\;E_n\;\right>\;e^{-i\frac{E_n}{\hbar}t} \\ i\hbar\sum\limits_n \dfrac{dC_n}{dt}\;\delta_{n'n}\;e^{-i\frac{E_n}{\hbar}t} & = & \sum\limits_n C_n\left(t\right)\left<\;E_{n'}\;|\;H\left(t\right)|\;E_n\;\right>\;e^{-i\frac{E_n}{\hbar}t} \\ i\hbar \dfrac{dC_{n'}}{dt}\;e^{-i\frac{E_{n'}}{\hbar}t} & = & \sum\limits_n C_n\left(t\right)\left<\;E_{n'}\;|\;H\left(t\right)|\;E_n\;\right>\;e^{-i\frac{E_n}{\hbar}t} \\ \dfrac{dC_{n'}}{dt} & = & \dfrac{1}{i\hbar}\sum\limits_n C_n\left(t\right)\left<\;E_{n'}\;|\;H\left(t\right)|\;E_n\;\right>\;e^{i\frac{E_{n'}-E_n}{\hbar}t}\end{array}$

Note that on the right side of the equation, the operator $$H\left(t\right)$$ is between the two states, so the bra and ket don't meet, and no kronecker delta results. We are interested in how the hydrogen atom gets from some initial stationary state to a lower state, so let's just call the initial state $$\left|\;E_i\;\right>$$, and since it is 100% in this state to start, the coefficient $$C_i$$ simply equal to one, and there is only one term in the summation. The final state we will call (\left|\;E_f\;\right>\), giving us:

$\dfrac{dC_f}{dt} = \dfrac{1}{i\hbar} \left<\;E_f\;|\;H\left(t\right)|\;E_i\;\right>\;e^{i\frac{E_f-E_i}{\hbar}t}$

The state of the particle started with $$C_i=1$$ which means that at $$t=0$$, $$C_f$$ must be zero (the initial and final states are different). The value of $$C_f$$ after the hamiltonian perturbation has acted for a time $$T$$ is therefore given by the integral:

$C_f = \dfrac{1}{i\hbar} \int\limits_0^T \left<\;E_f\;|\;H\left(t\right)|\;E_i\;\right>\;e^{i\frac{E_f-E_i}{\hbar}t} dt$

The probability that the particle has transitioned from the initial to the final state is:

$P\left(i\rightarrow f\right) = C^*_f C_f= \dfrac{1}{\hbar^2}\left| \int\limits_0^T \left<\;E_f\;|\;H\left(t\right)|\;E_i\;\right>\;e^{i\frac{E_f-E_i}{\hbar}t} \; dt\;\right|^2$

Okay, that's a lot of math, so let's interpret the result. This is the probability that the state of the electron which starts in an initial eigenstate of energy will end up in a different final eigenstate of energy after a time $$T$$. The ket and bra represent the starting and ending energy eigenstates, respectively. If not for the perturbed part of the hamiltonian $$H\left(t\right)$$, these two orthogonal "unit vectors" would give a zero dot product, and the probability of the transition occurring would be zero. Note that in fact none of this is unique to the hydrogen atom, except for the part of the hamiltonian we called "$$H_{hydrogen}$$," which we never actually used. The nuances for the hydrogen atom arise when we take into account the other two quantum numbers, which we will do in the next chapter.

## Geometrical Interpretation

Perhaps even with the interpretation given above, Equation 6.1.17 is a bit abstract and daunting. Let's return to our vector basics to see if we can make more sense out of it...

The hydrogen atom starts in a specific stationary state, and it is destined to drop to a new state of lower energy, but these two states are "unit vectors" in Hilbert space, and are therefore perpendicular to each other. If you want to know the probability that a particle will be found in one state when it is known to be in another, you need to take the magnitude-squared of the inner product of their two state vectors, so if left alone, the probability of the hydrogen state changing energy eigenstates is zero, because the inner product of these perpendicular state vectors is zero.

Now let's "tweak" the initial state vector with an operator ($$H\left(t\right)$$), which evolves in time. A static operator would simply change the state and be done with it, but an operator with a time dependence has the effect of essentially causing the state vector to "rotate." As it rotates, the angle the state vector makes with the final energy eigenstate vector changes. The inner product is no longer zero, and the probability of a transition no longer vanishes.

Figure 6.1.1 – A Crude Depiction of How Perturbation Allows Transition

If we let this perturbation run indefinitely (which vacuum polarization does), then a "characteristic" time interval of that particular perturbation can be established. When a collection of atoms are all in an excited state that can decay into lower-energy states, naturally those transitions with the lowest characteristic time interval will be the most commonly witnessed (in the form of photons with that transition's energy difference).