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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 $a$.

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2 Answers2

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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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In that chapter (excelent book, by the way), the author shows how unitary transformations are used to implement or represent symmetry transformations. For instance, spatial homogeneity implies that quantum mechanical outputs (such as measurements, probabilities and expectation values) must be invariant under a spatial translation. At same time, this spatial translation must be given in terms of a unitary transformation $U$.

An infinitesimal unitary transformation, close to identity, associated to a translation in $x$ is $$U_{\epsilon}(d_x)=\mathbb I+i\epsilon d_x,$$ where $\epsilon$ is an infinitesimal real parameter (the ammount of translation). At this point, $d_x$ is just an operator with dimension $(length)^{-1}$ called generator of the translation. If we apply this transformation on the observable $x$ we get (at first order on $\epsilon$) $$x\rightarrow UxU^\dagger=x-i\epsilon[x,d_x].$$ Since $x$ must go to $x+\epsilon$ under such infinitesimal translation, then $$[x,d_x]=i.$$

That is all we know about the generator $d_x$ up to this point. We need now to recall classical mechanics in order to find out who is this operator.

In classical mechanics, translation symmetry on $x$ implies momentum $p$ is conserved. Moreover, this same conserved quantity can generate spatial translation by means of canonical transformations. For instance, an infinitesimal canonical transformation on the dynamical variable $x$ generated by $p$ is given by $$x\rightarrow x+\delta x=x+\epsilon\{x,p\},$$ where $\{x,p\}$ denotes the Poisson Bracket between $x$ and $p$. Since an infinitesimal translation must read $x\rightarrow x+\epsilon$, the above equation implies into the fundamental Poisson bracket $$\{x,p\}=1.$$ Notice how simmilar are the two quantum and the two classical equations for implementing a translation. We assume that the quantum operator $d_x$ must be the quantum momentum, correspondent to the classical momentum $p$. More precisely, $$d_x=\frac{p}{\hbar},$$ where $\hbar$ is a new constant with dimension of $(energy)(time)$ to make the equation above dimensionally correct. That association between classical position with quantum position operator, classical momentum with quantum momentum operator, classical dynamical variables $f(x,p)$ with quantum observables $f(x,p)$ and Poisson brackets with commutators is what is called correspondence principle. It is one (perhaps the most pragmatic one) way to view quantum mechanics. It works whenever the quantum observable has a classical analog.

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