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Isham, in his Lecture on Quantum Theory, Chapter 7, Unitary Operators in Quantum Theory, Section 7.2.2 Displaced Observers and the Canonical Commutation Relations, mentions on page 137 (bottom) the following.

  1. The final step is to identify the operator $\hat{d}_x$ with $\hat{p}_x/\hbar$, where $\hat{p}_x$ is the momentum along the $x$ direction. This can be done by appealing to the classical limit of the theory, or by requiring consistency with the results of elementary wave mechanics. Thus we get the result that the states assigned by $O_2$ and $O_1$ are related by

$$|\psi\rangle_a=e^{ia\hat{p}_x/\hbar}|\psi\rangle.$$

Question: I don't know how to get this from the "classical limit" or "consistency" argument. Any help?


The following is how Isham has defined the operator $\hat{d}_x$ (and $O_1$ and $O_2$). First he defines the operator $\hat{D}(a)$ (after showing it exists) as the operator which satisfies $|\psi\rangle_a=\hat{D}(a)|\psi\rangle$, where $|\psi\rangle$ is the state of a quantum system as observed by an observer $O_1$ and $|\psi\rangle_a$ is the state of the same system as observed by an observer $O_2$ displaced along the positive $x$ direction by a distance $a$. Then he goes on to show that there exists a self-adjoint operator $\hat{d}_x$ such that $\hat{D}(a)=e^{ia\hat{d}_x}$ for all distances $a$.

Atom
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    What is $\hat{d}_x$? Please try to make questions self-contained by explaining the notation used. – ACuriousMind May 21 '20 at 11:02
  • Are you talking about this? – Cosmas Zachos May 21 '20 at 13:23
  • @CosmasZachos Yes! But I have no idea how to get this $|\psi\rangle_a=e^{ia\hat{p}_x/\hbar}|\psi\rangle$. Can you help? – Atom May 21 '20 at 13:33
  • It would not be feasible to quote the entire section from Isham's books so sorry :(. However, let me try to find a link where I can provide you a copy of it. – Atom May 21 '20 at 13:35
  • @ACuriousMind Please see the above link of the book. There is just too much to make this question self-contained. (There are just ten points in the refereed section, so it would't take much time for you to read what I've referred to.) Thanks for taking time! – Atom May 21 '20 at 13:39
  • This is my first textbook for QM. I've had only introductory lectures before. (I've understood almost everything before this in Isham's book.) – Atom May 21 '20 at 13:41
  • Okay. :) But is there anything you can help me with now? – Atom May 21 '20 at 13:45
  • Here, Isham is just summarizing the general introductory wave mechanics (which we also did in class). But in the text, Isham is doing everything from the first principles, laying out explicitly the assumptions at every stage. In particular, see his comment on the top of page 10 about exactly these equations. – Atom May 21 '20 at 13:51
  • So can you say how to get $\hat{d}_x = \hat{p}_x/\hbar$ from the "classical limit of the theory"? I'd really appreciate your help! :) – Atom May 21 '20 at 13:54
  • Atom, you don't need to quote the entire book or the entire chapter. Just specify what's the definition of the operator $\hat{d}_x$. :) That would make your post obviously self-contained. –  May 21 '20 at 17:55
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    @DvijD.C. I've tried to follow what you said. Looks good? – Atom May 21 '20 at 18:21
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    Looks perfect to me now :) –  May 21 '20 at 18:25

1 Answers1

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OK, to the extent you are not stuck on exponentiation, the "consistency with the results of elementary QM" bit is obvious. QM is predicated on Born's cornerstone relation, your text's (1.16), $$ [\hat x, \hat p]=i\hbar 1\!\! 1 , $$

You may immediately verify that in the x-basis, your (7.37) relation $[\hat d_x, \hat x]=-i~1\!\! 1$ reads $$ [x,- \partial_x] f(x) = f(x), $$ for arbitrary f(x); so $\hat p \mapsto -i \hbar \partial_x$, i.e. $$ \hat x = \int dx' ~~|x'\rangle x' \langle x'|, ~~~~ \hat p = -i\hbar\int dx' ~~ |x'\rangle \partial_{x'} \langle x'|~. $$


But... how does (1.16) comport with the classical limit of Dirac's thesis (summarized in his monumental book)? The (fraught) formal limit of operators $\hat f$ and $\hat g$ to classical versions thereof maps quantum commutators to classical Poisson brackets, $$ [\hat f, \hat g ] \leadsto \frac{\{ f,g\} }{i\hbar} \implies \\ [\hat x , \hat p ] \leadsto \{ x, p\}/i\hbar ~~~\implies \\ \{ x, p\}=1 \leadsto [\hat x , \hat p ] =i\hbar 1\!\! 1, \qquad \hat x \leadsto x , ~~~~~ \hat p \leadsto p, ~~~ 1\!\!1 \leadsto 1 ~. $$ That is to say (1.16) is dictated by Dirac's limit, and, as above, it identifies with (7.37).

  • Many of these issues and limits are best illustrated in phase-space quantization, but this outranges your text and question.

Thanks for the link to your text. Many of these points, namely the essential uniqueness of this representation you are talking about (up to equivalence: basis changes) is detailed in your text's section 7.2.2——where your question came from. This is the heart of the celebrated Stone—von Neumann theorem.

$[\hat x , \hat p -\hbar \hat d_x]=0$ suffices for the identification, up to equivalence, since the difference between the two operators is then, in general, a function of $\hat x$; which can be gauged to zero, as one always does when applying the S-vN theorem: $\hat d _x$ is formally equivalent to $e^{-ig(\hat x)} \hat d_x e^{ig(\hat x)}= \hat d_x + g'(\hat x)$.

Cosmas Zachos
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    Sorry for such a late reply. I removed the link. Here it is again: https://drive.google.com/file/d/14py4099PEubyxH5MYUHt8gsY0eSiFC6Y/view?usp=drivesdk – Atom May 21 '20 at 17:13
  • And, though I don’t get all of what you say, I get some idea. :) Thanks!! – Atom May 21 '20 at 17:14