I am currently studying the path integral formulation of quantum mechanics and have done a couple of problems (free particle and simple harmonic oscillator). Now, I am already done calculating the propagator for the simple harmonic oscillator and have shown that in the limit $\omega \rightarrow 0$ this gives the propagator of a free particle. I have stumbled on an article on Wikipedia (linked below) about this where it says that you could extract the energy levels of the h. o. from the propagator by comparing it to another representation of the propagator: \begin{aligned} K(x_{f},t_{f};x_{i},t_{i}) &= \left({\frac {m\omega }{2\pi i\hbar \sin \omega T}}\right)^{\frac {1}{2}}\exp {\left({\frac {i}{\hbar }}{\tfrac {1}{2}}m\omega {\frac {(x_{i}^{2}+x_{f}^{2})\cos \omega T-2x_{i}x_{f}}{\sin \omega T}}\right)} \\ &= \sum _{n=0}^{\infty }\exp {\left(-{\frac {iE_{n}T}{\hbar }}\right)}\psi _{n}(x_{f})^{*}\psi _{n}(x_{i}) \end{aligned}

This got me interested and I tried to follow the argumentation. The article seems to drop some details and I don't see how one arrives at the bottom. I would find it really interesting to see the full expression, as one should then be able to also extract the wave functions of all the excited states from this relation. Can someone maybe give me a hint to better understand what is going on in the end of this section?

Source: https://en.wikipedia.org/wiki/Path_integral_formulation#Simple_harmonic_oscillator

  • $\begingroup$ The second line is a standard expansion of a Green's function in terms of the eigenfunctions. So basically one needs to Fourier transform the first line, but, as far as I remember, it is a lot of work. This paper deals with a similar case, and the solution to the oscillator should be in one of the references: journals.aps.org/prb/abstract/10.1103/PhysRevB.66.035304 $\endgroup$ – Roger Vadim May 14 '20 at 14:26
  • $\begingroup$ Thank you. Unfortunately, I can see neither the article nor the references. Also, in the wikipedia article they don't work with fourier transforms. I think following their argumentation you should end up with an expression for the propagator off which you can read of the wavefunctions. I just can't manage to arrive at this expression. $\endgroup$ – Moeman May 14 '20 at 14:55
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    $\begingroup$ Can you maybe clarify your question a little? are you explicitly asking what does the wikipedia article do in the referenced link? That is, what happens in the section that begins in "One may absorb all terms..." till the end? $\endgroup$ – user245141 May 14 '20 at 15:56
  • $\begingroup$ This is exactly what I'm asking. The way it is written makes me think that you should be able to get a closed expression for K from which you can directly read of the energy levels and the eigenfunctions just by comparing it with the expression in the second line of my picture above (and that this closed expression is just not stated in the article). Please tell me if I am mistaken about this. Also thank you! $\endgroup$ – Moeman May 14 '20 at 16:04
  • $\begingroup$ Note that we use MathJax to typeset mathematics; you can find a good tutorial here. $\endgroup$ – Emilio Pisanty May 14 '20 at 16:32

This is quite straightforward.

$$K(x_f,t_f;x_i,t_i)=\langle x_f|e^{-iHT/\hbar}|x_i\rangle,$$ where $T=t_f-t_i$. Now inserting the completeness relation for the energy eigenstates $1=\sum_n |n\rangle\langle n|$ into the above equation, we obtain

$$K(x_f,t_f;x_i,t_i) =\sum_{m,n}\langle x_f|m\rangle\langle m| e^{-iHT/\hbar}|n\rangle\langle n|x_i\rangle\\ =\sum_{n} e^{-iE_nT/\hbar}\psi_n(x_f)\psi^*_n(x_i).$$

I think there is an error in the Wiki expression (the variables in the wave function and its complex conjugate should be exchanged.)

  • $\begingroup$ Ah I am sorry if this was unclear. I am talking about the end of said section, where they are explicitly extracting the energy levels of the harmonic oscillator. Though, they are only saying 'yielding terms of the form' and not giving the exact expression. Thank you very much nonetheless! $\endgroup$ – Moeman May 14 '20 at 15:12
  • $\begingroup$ I think the question is how one could arrive to this expansion starting from the closed form of the propagator. $\endgroup$ – Roger Vadim May 14 '20 at 15:16
  • $\begingroup$ @Vadim Not really. I am accepting the fact that the two expressions for the propagator are equivalent. I just don't see how to arrive at the last couple of lines of the section. I'm sorry if my question is not clear. $\endgroup$ – Moeman May 14 '20 at 15:24
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    $\begingroup$ @Moeman , Well, it is not trivial: You simply identify the expression with Mehler's 1866 kernel, for the propagator of Hermite functions. The derivation itself, the reverse of what you want, is in the WP article on Hermite polynomials. You don't need me to spell it out for you? $\endgroup$ – Cosmas Zachos May 14 '20 at 18:16
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    $\begingroup$ Yes, Gustav Mehler, the Gymnasium Lehrer from heaven, was a sharp one. $\endgroup$ – Cosmas Zachos May 14 '20 at 18:27

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