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Suppose I have two operators $A$ and $B$ which each obey $\langle\psi|A|\psi\rangle = \langle\psi|B|\psi\rangle$ for all elements $|\psi\rangle$ of the space. Does this imply $A=B$? I remember seeing a proof on Stack Exchange way back when but can't seem to dig it up now.

That $\langle\psi_1|A|\psi_2\rangle = \langle\psi_1|B|\psi_2\rangle$ for all elements $|\psi_{1/2}\rangle$ is sufficient to conclude $A=B$ is clear from considering a particular basis as the $|\psi_{1}\rangle$, but obviously this is a stronger condition (at first glance) than the former.

Bonus points (generalizing from pure states): Does $\operatorname{trace}\{A \rho\} = \operatorname{trace}\{B\rho\}$ for all valid (density/state) operators $\rho$ ($\rho$ nonnegative, of unit trace, and self-adjoint) imply that $A=B$? Obviously answering this (in the affirmative) answers the first question, so please feel free to just answer this.

Edited to include an image describing the operator. Image included as I'm not sure how to describe the velocity opreator $\textbf{V}$ which Ballentine defines. enter image description here

EE18
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1 Answers1

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Assume for simplicity that the operators $A,B$ are bounded so we don't run into any domain issues. Note that $$\langle \psi\pm\phi,A(\psi\pm\phi)\rangle = \langle \psi,A\psi\rangle + \langle \phi,A\phi\rangle \pm \langle \psi,A\phi\rangle \pm \langle \phi,A\psi\rangle$$ $$\langle \psi\pm i\phi,A(\psi\pm i\phi)\rangle = \langle \psi,A\psi\rangle + \langle \phi,A\phi\rangle \pm i\langle \psi,A\phi\rangle \mp i \langle \phi,A\psi\rangle$$

A bit of algebra yields that

$$4\langle \psi,A\phi\rangle = \langle \psi+\phi,A(\psi+\phi)\rangle - \langle \psi-\phi,A(\psi-\phi)\rangle $$ $$-i \bigg[\langle \psi+i\phi,A(\psi+i\phi)\rangle - \langle \psi-i\phi,A(\psi-i\phi)\rangle\bigg]$$ This is an example of a so-called polarization identity. Your desired result follows straightforwardly from here.

Bonus points (generalizing from pure states): Does $\operatorname{trace}\{A \rho\} = \operatorname{trace}\{B\rho\}$ for all valid (density/state) operators $\rho$ ($\rho$ nonnegative, of unit trace, and self-adjoint) imply that $A=B$?

Yes, but you have it backwards - this is a weaker statement than the first one. If $\mathrm{Tr}(A\rho)= \mathrm{Tr}(B\rho)$ for all density operators $\rho$, than in particular it holds for all pure states $\rho= |\psi\rangle\langle \psi|$, at which point the argument above takes over. In other words, we don't need to assume the equality holds for all density operators - the set of pure density operators suffices.

J. Murray
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  • Re: the first part of your answer -- thank you! Obviously that identity establishes the "stronger" condition I mentioned in the problem (I say this to convince myself of that fact). Re: the second part -- thank you as well! – EE18 Jan 16 '23 at 02:49
  • Incidentally, I perhaps should have waited with a more general question. The operator in question here (one of them, at least) is the position $X$ operator on Hilbert space. I don't believe it's bounded alhtough I'm not sure it materially changes things (I'm sure there are ways to phrase things in terms of rigged Hilbert spaces?). – EE18 Jan 16 '23 at 02:57
  • @EE18 The position operator is not bounded on $L^2(\mathbb R)$. What is the other operator you're considering? – J. Murray Jan 16 '23 at 03:25
  • Please see the attached image -- I wasn't sure how else to convey meaning. – EE18 Jan 16 '23 at 03:37
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    @EE18 I see. Note that the text is actually just seeking some $\mathbf V$ which has the desired property, and clearly $i[H,\mathbf Q]$ fits the bill. At the physicists level of rigor, this is sufficient to motivate the definition [...] – J. Murray Jan 16 '23 at 04:05
  • [...] but mathematically, one might want to check whether $\mathbf V$ is (essentially) self-adjoint and then attempt to classify its self-adjoint extensions. I am not sure, but my suspicion is that $\mathbf V$ is essentially self-adjoint on the set of smooth, compactly-supported functions and so for all practical purposes you can think of (the closure of) $\mathbf V$ as the unique self-adjoint operator which satisfies your requirements. – J. Murray Jan 16 '23 at 04:06
  • See this for the case of infinite-dimensions and operators defined only on a dense subspace. – Tobias Fünke Jan 16 '23 at 07:23