1
$\begingroup$

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?

Thanks in advance!

$\endgroup$
  • $\begingroup$ Is your operator $\hat U_I$ a unitary operator? You didn't say much about this operator. $\endgroup$ – Sofia Feb 2 '15 at 14:03
  • $\begingroup$ $\hat{U_I}$ 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). $\endgroup$ – golanor Feb 2 '15 at 14:13
  • $\begingroup$ 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) $\endgroup$ – Sofia Feb 2 '15 at 14:18
  • $\begingroup$ I rephrased - i'm not sure whether or not U is unitary. $\endgroup$ – golanor Feb 2 '15 at 14:19
  • $\begingroup$ 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. $\endgroup$ – Sofia Feb 2 '15 at 14:24
1
$\begingroup$

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 \begin{equation} \langle n | U_I(t) | m \rangle \simeq \delta_{nm} - \frac{i}{\hbar} \int_0^t dt' \langle n | V_I(t') | m \rangle. \end{equation}

$\endgroup$
0
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.