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I am working through Ch. 3 of Ballentine where he finds the commutator relationships between various operators.

He begins on p.78 with a space displacement

$$\mathbf{x'} = \mathbf{x} + \mathbf{a}$$

Which involves a corresponding displacement of position eigenvectors

$$|\mathbf{x'}\rangle =e^{-i\mathbf{a}\cdot \mathbf{P}/\hbar}|\mathbf{x}\rangle = |\mathbf{x} + \mathbf{a}\rangle .$$

However, I believe it should be instead

$$|\mathbf{x'}\rangle =e^{-i\mathbf{a}\cdot \mathbf{P}/\hbar}|\mathbf{x}\rangle = |\mathbf{x} - \mathbf{a}\rangle .$$

Because, he mentions that we are taking the "active" point of view, where we keep the coordinates the same but instead shift our vectors and operators. In this case, $|\mathbf{x'}\rangle$ is shifted $+\mathbf{a}$ which in turns means that it is equal to unprimed position eigenvector $-\mathbf{a}$.

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  • $\begingroup$ What is the question? $\endgroup$ Jan 11 '20 at 23:48
  • $\begingroup$ Am I correct or is Ballentine. $\endgroup$
    – Jeff
    Jan 12 '20 at 0:00
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It's easy to fudge signs in this business, so why don't you just utilize the coordinate space representation of the momentum operator, $$ \mathbf{P} = \int d^3 \mathbf{x} ~| \mathbf{x}\rangle ( - i \hbar \nabla) \langle \mathbf{x}| \implies e^{-i\mathbf{a}\cdot \mathbf{P}/\hbar}= \int d^3 \mathbf{x}' ~~| \mathbf{x}'\rangle e^{- \mathbf{a}\cdot \nabla '} \langle \mathbf{x}'| ~ . $$ Acting with the latter on $| \mathbf{x}\rangle $ and collapsing the Lagrange-shifted δ function in the integral nets you $$ e^{-i\mathbf{a}\cdot \mathbf{P}/\hbar}| \mathbf{x}\rangle = \int d^3 \mathbf{x}' ~~| \mathbf{x}'\rangle e^{- \mathbf{a}\cdot \nabla '} \langle \mathbf{x}'| \mathbf{x} \rangle \\ = \int d^3 \mathbf{x}' ~~| \mathbf{x}'\rangle \delta(\mathbf{x}'- \mathbf{a} -\mathbf{x}) = | \mathbf{x} + \mathbf{a} \rangle ~. $$

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The matrix element is usually defined (if you disagree let me know and I'll change the answer) $$\langle p | x \rangle = (2\pi \hbar)^{\frac{d}{2}} \exp(-\frac{i}{\hbar}p \cdot x)$$ so that if I now recall that $e^{ia/\hbar\, \cdot \hat{p}}|p\rangle = e^{i a / \hbar \,\cdot p}|p\rangle$ implies that $\langle p | e^{-ia / \hbar\, \cdot \hat{p}} = \langle p | e^{-ia/\hbar \, \cdot p} $ find $$\langle p | e^{- i/\hbar\, a \cdot \hat{p}} | x \rangle = e^{-i/\hbar \, a \cdot p} \langle p | x \rangle = (2\pi \hbar)^{\frac{d}{2}} \exp(-\frac{i}{\hbar}p \cdot( x + a))$$ which shows that $x \rightarrow x + a$.

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  • $\begingroup$ Honestly, I am confused a bit by your answer as it requires knowledge further on. Your answer "looks" correct to me, but I was specifically confused by what $\langle p | x \rangle$ is here. It is confusing because $p$ is referring to a position displacement sometimes and other times it is momentum. $\endgroup$
    – Jeff
    Jan 12 '20 at 1:25
  • $\begingroup$ $\hat{p} $ with a hat is an operator, but without is a variable. $|p\rangle $ is an eigen state of the momentum operator with eigenvalue p $\endgroup$
    – lux
    Jan 12 '20 at 1:29
  • $\begingroup$ I don't know the book you're referring to so my answer is not "well-ordered" with respect to the reference but it should be possible to follow it if you're learning quantum mechanics $\endgroup$
    – lux
    Jan 12 '20 at 1:30
  • $\begingroup$ I am using Quantum Mechanics: A modern Development by Ballentine. In which, he derives all the commutator relations for various operators based on invariances under galilean transformations, before even getting to concrete representations. $\endgroup$
    – Jeff
    Jan 12 '20 at 1:36
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After going through the text again, I wanted to give an answer more motivated by the exposition in the text.

From, p.63, since the laws of nature are invariant under certain space-time transformations, if $A | \phi_n \rangle = a_n | \phi_n$ with $A$ representing an observable, and $\phi_n$ an eigenvector, then after the transformation $A' | \phi_n' \rangle = a_n | \phi_n' \rangle$ because they represent the same observable.

And from p.64 and p.65, given a space-time transformation $U$

$$|\phi_n' \rangle = U|\phi_n \rangle$$

$$A' = UAU^{-1}$$

Now, for a space displacement of the position operator we have $$|\mathbf{x'}\rangle =e^{-i\mathbf{a}\cdot \mathbf{P}}|\mathbf{x}\rangle = |\mathbf{x}+\mathbf{a}\rangle$$ $$\mathbf{Q'} = e^{-i\mathbf{a}\cdot \mathbf{P}}\mathbf{Q}e^{i\mathbf{a}\cdot \mathbf{P}}$$ And for a single direction of Q, $$Q_\alpha'|\mathbf{x'}\rangle = x_\alpha |\mathbf{x'}\rangle $$

Which suggests, $$\mathbf{Q'} = \mathbf{Q} - \mathbf{a}I$$ because for a single direction $$Q_\alpha'|\mathbf{x'}\rangle = x_\alpha |\mathbf{x'}\rangle = (Q_\alpha -a_\alpha I)|\mathbf{x}+\mathbf{a}\rangle$$ $$ = Q_\alpha |\mathbf{x}+\mathbf{a}\rangle - a_\alpha |\mathbf{x}+\mathbf{a}\rangle $$ $$ = x_\alpha + a_\alpha |\mathbf{x}+\mathbf{a}\rangle - a_\alpha |\mathbf{x}+\mathbf{a}\rangle $$ $$ = x_\alpha|\mathbf{x}+\mathbf{a}\rangle$$

So, thinking back to the active/passive view of transformations where I was confused, what is happening is that we are shifting the position operator $Q$ forward by $\mathbf{a}$ which corresponds to an active point of view.

This results in the position eigenvectors for a specific eigenvector to be shifted forward or in other words for the eigenvalues of a given eigenvector to be shifted backwards! All very confusing, but I believe this clarifies my original question more in the spirit of the text exposition.

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  • $\begingroup$ Esteemed Jeff - I was a little surprised to see the change to the stats of this post. Please do consider the effects of un-accepting an answer that you were previously content with to replace it with your own. I'm sure you're aware this site works on the idea of Reputation which is directly related to upvotes, downvotes and accepted answers...... $\endgroup$
    – lux
    Jan 16 '20 at 3:47
  • $\begingroup$ As, I stated, I didn't feel any of the previous answers completely answered the original question as they assumed definition of things that aren't yet defined in Ballentine's treatement. Instead, he is trying to derive the properties of the various commutators from first principles. $\endgroup$
    – Jeff
    Jan 16 '20 at 20:59

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