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I was reading a wonderful explanation of matrix mechanics on mathpages1. There we see that $\hat{q}_{mn}=q_{mn}\exp(i(E_m-E_n)t/\hbar )$ and consequently using Hamilton's classical equations we arrive at the Hesisenberg formualtion and $[\hat{q},\hat{p}]=i\hbar$.This is a silly question but, why can't an analogous argument be applied to momentum differences, i.e., $\hat{q}_{pp'}=q_{pp'}\exp(i(p-p')x/\hbar )$? Can this somehow be related to $<p|\hat{x}|p'>=\int x \exp(i(p-p')x/\hbar)dx=i\hbar \frac{\partial}{\partial p} \delta(p-p')$?

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Well, yes, you are essentially right that Heisenberg's notation, extended by Dirac's, are idiosyncratic formulations of standard Fourier transforms. Your use of carets is off, however--you seem to be using your first one for time-dependence.

Let me translate your Kevin Brown statements into mainstream Dirac usage.

$$\langle m| e^{i\hat{H}t/\hbar}\hat{q} e^{-i\hat{H}t/\hbar}|n\rangle= \langle m| \hat{q} |n\rangle e^{i(E_m-E_n)t/\hbar} . $$

It thus follows that you may easily change representations at will. The coordinate space wave functions are $\psi_n(x)=\langle x|n\rangle$, while the momentum space ones are $\phi_n(p)=\langle p|n\rangle$. (For the oscillator, these are coincidentally both complete, orthonormal Hermite functions, by dint of the full q-p symmetry of that Hamiltonian.)

It then follows that, given $\langle x|p\rangle=\exp (ipx/\hbar)$, $$ \langle m |\hat {q}|n\rangle= \int dp dp' ~ \langle m | p\rangle \langle p |\hat {q}|p'\rangle \langle p'|n\rangle= \int dp dp'dx ~ \phi_m^*(p) \phi_n(p') \langle p |\hat {q}|x\rangle \langle x|p'\rangle\\ =i\hbar \int dp dp' ~ \phi_m^*(p) \phi_n(p') \partial_p\delta(p-p')= -i\hbar \int dp ~ \partial_p\phi_m^*(p) \phi_n(p') =i\hbar \int dp ~ \phi_m^*(p) \partial_p\phi_n(p), $$ the momentum-space representation of your matrix element.

Of course, life is much easier for $$ \langle m |\hat {p}|n\rangle= \int dp ~ \phi_m^*(p) ~p~\phi_n(p). $$

You then see how the orthonormality and completeness, $\sum_n \phi_n^*(p) \phi_n (p')=\delta(p-p') $, of these functions leads to $$ \sum_n \left ( \langle m |\hat {q}|n\rangle\langle n |\hat {p}|k\rangle -\langle m |\hat {p}|n\rangle\langle n |\hat {q}|k\rangle \right)\\ = i\hbar \sum_n \int dp ~ \left (-\partial_p\phi_m^*(p) \phi_n(p)\phi_n^*(p')p' \phi_k(p') - \phi_m^*(p) p \phi_n(p)\phi_n^*(p')\partial_{p'} \phi_k(p') \right ) \\ = i\hbar \sum_n \int dp ~ ( -\partial_p\phi_m^*(p) p \phi_k(p) - \phi_m^*(p) p \partial_{p} \phi_k(p) )=i\hbar\delta_{mk} , $$ the "Heisenberg" (actually, Born) commutation relation $[\hat{q},\hat{p}]=i\hbar\mathbb{1}$.

This is visibly messier than the direct matrix argument, which may be why matrix mechanics is prized by the sentient.

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  • $\begingroup$ For the energy eigen basis there is an interpretation in terms of photons with frequency equal to the energy difference $E_m - E_n$ being emitted or absorbed, is there such an interpretation for momentum differences? Where it is not frequency but rather wave number. I know this is a bit nebulous but just trying to make an analogy here with the frequency/energy of the photons with the wavenumber/momentum of photons... $\endgroup$ – play Mar 13 '18 at 19:39
  • $\begingroup$ Well, yes and no. Indeed, energy ~ frequency and time are conjugate variables, and there is a time-frequency uncertainty relation in signal processing, much in analogy to the position/wavenumber uncertainty relation of the coordinate/momentum conjugate pair. Their Fourier treatment is very analogous. BUT time is not an operator, naively, unless you start getting careful. It was not clear from your question this is what you were aiming at. I read it as a question on the matrix formulation which is clear and a darling of the cognoscenti.... $\endgroup$ – Cosmas Zachos Mar 13 '18 at 19:46
  • $\begingroup$ So would you say it is fatal to try and construct a picture where changes in the momentum of a particle can be encoded as photons being emitted and absorbed in a manner similar to how it is done for changes in energy states.. $\endgroup$ – play Mar 14 '18 at 0:25
  • $\begingroup$ I'll stick my neck out and declare it fatal, even though it could ascend to a pointless lark... The problem has been settled for 90 years. Unless you had a cute idea that does something very new... $\endgroup$ – Cosmas Zachos Mar 14 '18 at 0:34

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