# Time evolution of two orthogonal states in Time Dependent Perturbation Theory

Given the two orthogonal states for $H_0$ , $|n(t)>_I, |m(t)>_I$, in the interaction picture, we want to find the probability of transforming from one to the other after time t, aka:

$\ (1) \ |<n(t)|m(t)>|^2$.

Naively, I would think to do this:

$\ (2) \ |<n(t)|m(t)>|^2 = |<n(0)|U_I^\dagger U_I|m(0)>|^2 = |<n(0)|m(0)>|^2 = \delta_{nm}$

Where

$U_I = exp[\frac{-i}{\hbar} \int_{t_0}^{t} d\tau e^{\frac{i H_0 \tau}{\hbar}}V(\tau)e^{\frac{-i H_0 \tau}{\hbar}}]$

and V is the perturbation (and is hermitian). (U is unitary)

Using the perturbation theory I would get $\delta_{nm}$ for the 0th and 1st order, but for the second order, I get something completely different. Which way is the correct way to go, and why?

• Is your operator $\hat U_I$ a unitary operator? You didn't say much about this operator. – Sofia Feb 2 '15 at 14:03
• $\hat{U_I}$ is the time evolution operator in the interaction picture. I'm guessing that it is unitary, otherwise there would be a whole lot of problems with the preservation of probability (non unitary operators do not conserve norm). – golanor Feb 2 '15 at 14:13
• but, under a unitary transformation, two orthogonal states remain orthogonal. I suppose that it is known to you. Then, what is the question? Where appears the perturbation in your equations? (I just added numbering, for easiness of referring to them) – Sofia Feb 2 '15 at 14:18
• I rephrased - i'm not sure whether or not U is unitary. – golanor Feb 2 '15 at 14:19
• No, if you are not sure about unitarity, there are two possibilities: 1) check by yourself if the unitarity though holds, or use indeed the perturbation calculus. But, in your place, and for your specific problem, I would first check unitarity. – Sofia Feb 2 '15 at 14:24

The confusion lies with the definition of the transition probability.

The transition amplitude between an $H_0$ eigenstate $H_0 \left|m\right> = E_m \left|m\right>$ and another eigenstate $H_0 \left|n\right> = E_n \left|n\right>$ due to a perturbation $V$ after a time $t$ is given by \begin{align} \left< n \right| U(t) \left| m \right> & = \left< n \right| e^{-iH_ot/\hbar} U_I(t) \left| m \right> \\ & = e^{-iE_n t/\hbar} \left< n \right|U_I(t) \left| m \right> \\ & = e^{-iE_n t/\hbar} \left< n \right| U_I(t) \left| m \right> = e^{-iE_n t/\hbar} \langle n | m(t) \rangle_I. \end{align} Up to first order, we find $$\langle n | U_I(t) | m \rangle \simeq \delta_{nm} - \frac{i}{\hbar} \int_0^t dt' \langle n | V_I(t') | m \rangle.$$

The perturbation theory is an approximation. If you can do your calculus exactly, don't use approximations. Your equality (2) can be exact, though, on condition that $\hat U_I$ is a transformation that preserves inner product. The simplest thing is to check whether $\hat U_I$ is unitary, that would be satisfactory, because a unitary transformation has a property that it preserves the inner product. So, two orthogonal states remain orthogonal.

A remark : I don't say that the unitary transformation is the only one that has the property of conserving the inner product. Other transformation may have this property too. I just said the unitary transformation has this property.

NOTE (Truax' theorem) : this is a useful theorem when checking unitarity. The product of two exponential operators $e^{\hat A}$ and $e^{\hat B}$ is not equal, in general, with $e^{\hat A + \hat B}$. This equality holds only if $\hat A$ and $\hat B$ commute. The theorem says,

$$e^{\hat A} e^{\hat B} = e^{\hat A + \hat B + f(\hat A, \ \hat B, \ [\hat A, \hat B])}$$

where $f$ is a function of the mentioned arguments, and one case in which $f = 0$ is when $\hat A$ and $\hat B$ commute.

• As long as it is not clear whether $\hat U$ is unitary, my post is not a sure answer, so I delete it. – Sofia Feb 2 '15 at 14:36
• The question is why is the perturbation theory method incorrect. Where is the problem with it? – golanor Feb 2 '15 at 17:54
• @golanor - I am skeptical about the theory of perturbation being incorrect. If you can show your calculus, we can see what you did. – Sofia Feb 2 '15 at 18:25