Transition of state

We assume a one-dimensional quantum harmonic oscillator of mass m and charge e. The expression for the potential energy of this system is this of the harmonic oscillator.

${\displaystyle V={\dfrac {1}{2}}kx^{2}}$.

The total wavefunction is denoted by Ψ(x, t) (capital Psi), and the spatial part of the wave function is ψ(x) (lower case psi). As we deal with stationary states, the total wave function is a solution of the Schrödinger equation and reads

${\displaystyle \Psi (x,t)=\psi (x)\exp \left(-i{\dfrac {E}{\hbar }}t\right)}$,

with eigenvalue ${\displaystyle \textstyle i\hbar {\frac {\partial \Psi }{\partial t}}=E\Psi }$.

The probability of transition from the fundamental level labelled 0 to a level labelled 1 under an electromagnetic stimulation is analysed below.

For this situation, we write the total wave function as a linear combination for a two-levels system:

${\displaystyle \Psi (x,t)=c_{0}(t)\Psi _{0}(x,t)+c_{1}(t)\Psi _{1}(x,t)}$

The coefficients c_0,1 are time-dependent. They represent the proportion of the state (0,1) in the total wave function with time, thus they represent the probability of the wave-function to fall in one of the two state when an observer will collapse the wave function.

As we deal with a two-level system, we have the normalisation relation :

${\displaystyle \langle \Psi (x,t)|\Psi (x,t)\rangle =1\Leftrightarrow {\sqrt {|c_{0}(t)|^{2}+|c_{1}(t)|^{2}}}=1}$

The electromagnetic stimulation will be a uniform electric field, oscillating with a frequency ω. This is very similar to the semi-classical analysis of the behaviour of an atom or a molecule under a polarized electromagnetic plane wave.

Thus, potential energy will be the sum of the unperturbed potential and of the perturbation and reads:

${\displaystyle V(x)={\dfrac {1}{2}}kx^{2}+e\epsilon (t)x}$

The Schrödinger equation will be written :

${\displaystyle \left(-{\dfrac {\hbar ^{2}}{2m}}{\dfrac {\partial ^{2}}{\partial x^{2}}}+V(x)\right)\Psi (x,t)=i\hbar {\dfrac {\partial \Psi (x,t)}{\partial t}}}$

The time derivative in the right part of the Schrödinger equation reads:

${\displaystyle i\hbar {\dfrac {\partial \Psi (x,t)}{\partial t}}=i\hbar \left(\psi _{0}\exp \left(-i{\dfrac {E_{0}t}{\hbar }}\right)\left({c_{0}}'(t)-i{\dfrac {E_{0}}{\hbar }}c_{0}(t)\right)+\psi _{1}\exp \left(-i{\dfrac {E_{1}t}{\hbar }}\right)\left({c_{1}}'(t)-i{\dfrac {E_{1}}{\hbar }}c_{1}(t)\right)\right)}$

${\displaystyle i\hbar {\dfrac {\partial \Psi (x,t)}{\partial t}}=i\hbar \left(\Psi _{0}\left({c_{0}}'(t)-i{\dfrac {E_{0}}{\hbar }}c_{0}(t)\right)+\Psi _{1}\left({c_{1}}'(t)-i{\dfrac {E_{1}}{\hbar }}c_{1}(t)\right)\right)}$

On the right part, the total hamiltonian is the sum of the unperturbed hamiltonian (without the external electric field) and the external perturbation. This allows to substitute the eigenvalues of the stationary states in the total hamiltonian. Thus we write:

${\displaystyle {\hat {H}}\Psi (x,t)=\left(E_{0}c_{0}(t)\Psi _{0}(x,t)+E_{1}c_{1}(t)\Psi _{1}(x,t)+e\epsilon (t)x\Psi (x,t)\right)}$

Using the Schrödinger equation above, we end up with

${\displaystyle e\epsilon (t)x\Psi (x,t)=i\hbar (c_{1}'(t)\Psi _{1}(x,t)+c_{0}'(t)\Psi _{0}(x,t))}$

We use now the bra–ket notation to avoid cumbersome integrals. This reads :

${\displaystyle e\epsilon (t)(c_{1}(t)x|\Psi _{1}(x,t)\rangle +c_{0}(t)x|\Psi _{0}(x,t)\rangle =i\hbar (c_{1}'(t)|\Psi _{1}(x,t)\rangle +c_{0}'(t)|\Psi _{0}(x,t)\rangle )}$

Then we multiply by ${\displaystyle \langle \Psi _{1}|}$ and end up with the following

${\displaystyle e\epsilon (t)(c_{1}(t)\langle \Psi _{1}|x|\Psi _{1}\rangle +c_{0}(t)\langle \Psi _{1}|x|\Psi _{0}\rangle )=i\hbar \left(c_{1}'(t)\langle \Psi _{1}|\Psi _{1}\rangle +c_{0}'(t)\langle \Psi _{1}|\Psi _{0}\rangle \right)}$

The two different levels are orthogonal, so ${\displaystyle \langle \Psi _{1}|\Psi _{0}\rangle =0}$. Also we are working with normalized wave functions, so ${\displaystyle \langle \Psi _{1}|\Psi _{1}\rangle =1}$.

Finally,

${\displaystyle e\epsilon (t)\left(c_{1}(t)\langle \Psi _{1}|x|\Psi _{1}\rangle +c_{0}(t)\langle \Psi _{1}|x|\Psi _{0}\rangle \right)=i\hbar c_{1}'(t)}$

This latter equation expresses the time variation of c₁ with time. This is the crux of our calculation, since by then, we can deduce exactly its expression from the differential equation we obtained.

There is no proper way in general to evaluate ${\displaystyle \langle \Psi _{1}|x|\Psi _{0}\rangle }$, unless we have a precise knowledge of the two unperturbed wave function, that is to say unless we can solve the non-perturbed Schrödinger equation. In the case of the harmonic potential, the wave functions solutions of the one-dimensional quantum harmonic oscillator are known as Hermite polynomials.

We made several assumptions to get to the final result. First we suppose that c₁(0) = 0, because at time t = 0, the interaction of the field with the matter did not start. That impose for the total wave function to be normalized that c₀(0) = 1. We use these conditions, and we can write, at t = 0:

${\displaystyle e\epsilon (t)\langle \Psi _{1}|x|\Psi _{0}\rangle =i\hbar c_{1}'(t)}$

Again, in this non-relativistic picture, we remove the time dependence outside.

${\displaystyle e\epsilon (t)\exp \left(-i{\dfrac {E_{0}-E_{1}}{\hbar }}t\right)\langle \psi _{1}|x|\psi _{0}\rangle =i\hbar c_{1}'(t)}$

The quantity ${\displaystyle e\langle \psi _{1}|x|\psi _{0}\rangle }$ is called the transition moment integral. Its dimensions are [charge]·[length] and SI units A·s·m.

It can be measured experimentally, or calculated analytically if one know the expression of the spatial wave function for both the energy levels. It can be the case if we deal with an harmonic oscillator like it's the case here. We will not it :${\displaystyle \mu _{01}}$ as the transition moment from the level 0 to the level 1.

Finally, we end with

${\displaystyle c_{1}'(t)={\dfrac {\mu _{01}}{i\hbar }}\epsilon (t)\exp \left(-i{\dfrac {E_{0}-E_{1}}{\hbar }}t\right)\Rightarrow c_{1}(t')={\dfrac {\mu _{01}}{i\hbar }}\int _{0}^{t'}\epsilon (t)\exp \left(-i{\dfrac {E_{0}-E_{1}}{\hbar }}t\right)\mathrm {d} t}$

The remaining task is to integrate this expression to obtain c₁(t). However, we must recall from the previous approximations we made, we are here at time t = 0. So the solution we obtain from integration will be only valid as long as |c₀(t)|² is still very close to 1, that is to say for very short time after the perturbation began to act.

We suppose that the time dependent perturbation has the following form, to make the computation easier.

${\displaystyle \epsilon (t)=\epsilon _{0}\exp(i\omega t)}$

This is a scalar quantity, as we assumed from the beginning a scalar charged particle and a one dimensional electric field.

So we have to integrate the following expression :

${\displaystyle c_{1}(t')={\dfrac {\mu _{01}\epsilon _{0}}{i\hbar }}\int _{0}^{t'}\mathrm {d} t\exp \left(-i\left({\dfrac {E_{0}-E_{1}}{\hbar }}-\omega \right)t\right)}$

We can write

${\displaystyle c_{1}(t')={\dfrac {\mu _{01}\epsilon _{0}}{i\hbar }}\int _{0}^{t'}\mathrm {d} t\exp \left(-i{\frac {E_{0}-E_{1}}{\hbar }}t\right)\exp({i\omega t})=\int _{-\infty }^{+\infty }\mathrm {d} t\exp \left(-i{\frac {E_{0}-E_{1}}{\hbar }}t\right)H\left({\frac {t}{t'}}-{\frac {1}{2}}\right)\exp \left(i\omega t\right)}$

and doing the variable change ${\displaystyle t\rightarrow -t}$ we obtain the correct form of the Fourier transform :

${\displaystyle c_{1}(t')=\int _{-\infty }^{+\infty }\mathrm {d} t\exp \left(i{\frac {E_{0}-E_{1}}{\hbar }}t\right)H\left(-{\frac {t}{t'}}-{\frac {1}{2}}\right)\exp \left(-i2\pi \nu t\right)}$

where ${\displaystyle H}$ is the rectangular function. We notice from the previous equation that c₁(t) is the Fourier transform of the product of a cosine with a square of width t'. From then, the formalism of Fourier transforms will make the work easier.

We have

${\displaystyle c_{1}(t')=\mathrm {TF} \left[\exp {\left(i{\frac {E_{0}-E_{1}}{\hbar }}t\right)}H\left(-{\frac {t}{t'}}-{\dfrac {1}{2}}\right)\right]=\mathrm {TF} \left[\exp {\left(i{\frac {E_{0}-E_{1}}{\hbar }}t\right)}\right]\otimes \mathrm {TF} \left[H\left(-{\frac {t}{t'}}-{\dfrac {1}{2}}\right)\right]}$

${\displaystyle c_{1}(t')=\delta \left(f-{\dfrac {E_{0}-E_{1}}{h}}\right)\otimes \mathrm {TF} \left[H\left({\frac {t}{t'}}-{\dfrac {1}{2}}\right)\right]}$

${\displaystyle c_{1}(t')=\delta \left(f-{\dfrac {E_{0}-E_{1}}{h}}\right)\otimes (\exp({i\pi ft'})\mathrm {sinc} (t'f))}$

Where sinc is the cardinal sinus function in its normalized form. The convolution with the Dirac distribution will translate the term on the left of the ${\displaystyle \otimes }$ sign.

We obtain finally

${\displaystyle c_{1}(t')=\exp \left({i\pi \left(f-{\dfrac {E_{0}-E_{1}}{h}}\right)t'}\right)\mathrm {sinc} \left(t'\left(f-{\dfrac {E_{0}-E_{1}}{h}}\right)\right)}$

The probability of a transition is given in general for a multi-level system by the following expression:[2]

${\displaystyle P_{k}=\sum _{n\neq k}c_{n}^{*}(t)c_{n}(t)}$

The probability to fall in the 1 state corresponds to ${\displaystyle |c_{1}(t)|^{2}}$. This is really easy to compute from all the tedious calculation we made previously. We observe in the equation that ${\displaystyle |c_{1}(t)|^{2}}$ has a very simple expression. Indeed, the phase factor, varying with t, disappears naturally.

So we obtain the expression

${\displaystyle |c_{1}(t)|^{2}=\mathrm {sinc} ^{2}\left(t\left(f-{\dfrac {E_{0}-E_{1}}{h}}\right)\right)}$

We made the hypothesis that the stimulation was a complex exponential. However a true electric field is real valued. A further analysis should take it in account. Also, we always assume that t is very small. We should keep it in mind before to conclude.