Suppose one is given a self-adjoint operator $A$ acting on an infinite dimensional separable Hilbert space $\mathcal{H}$. Under what conditions can one find an operator $B$ such that $[A,B] = i$? And if this is possible, how does one construct $B$?

For simplicity, assume that $H \simeq L_2(\mathbb{R})$ and that $A = A(X,P)$ is a function of the position operator $X$ and momentum operator $P$.

What I ultimately want to know is the following. Given a self-adjoint operator $A(X,P)$, by Stone’s theorem we know that $A(X,P)$ generates a one-parameter unitary group $U(t) = e^{-iA(X,P) t}$. What I want to know is if there is a self-adjoint operator (observable) $B(X,P)$, such that $U(t)$ generates translations of $B(X,P)$. And how to find $B(X,P)$ given $A(X,P)$. I know the the condition for $A(X,P)$ to generate translations of $B(X,P)$ is that these operators must satisfy the canonical commutation relation $[A,B] = i$, and hence the reason for phrasing the question as I did. This is analogous to how $P$ generates translations in $X$, that is, $e^{-i Px} \left|x_0\right\rangle = \left|x+x_0\right\rangle$.

  • $\begingroup$ It depends on what you mean by conjugate operator. The definition you wrote is too vague. The answer by @ACuriousMind is valid under some further requirements in the domains of the (symmetric) operators and their essential selfadjointness. If the space is finite dimensional the answer is trivially that no conjugate operators exist. $\endgroup$ – Valter Moretti Nov 1 '18 at 7:24
  • $\begingroup$ Sorry for typos, I'm using my phone... $\endgroup$ – Valter Moretti Nov 1 '18 at 7:30
  • $\begingroup$ @ValterMoretti Thank you for your replies! I've edited the question above, I hope this helps clarify things. I want the domain of the operators to be the state space $S(\mathcal{H}) := \{\rho \in S(\mathcal{H}) | \rho \geq 0 , {\rm tr} \rho = 1 \}$. I suppose in the case of pure states this would correspond to the Schwarz space on $\mathbb{R}$. $\endgroup$ – e4alex Nov 1 '18 at 14:07
  • $\begingroup$ @ValterMoretti I agree that for a finite dimensional Hilbert the answer is trivial, although one could probably reformulate the question and look for approximately conjugate operators. Something like the first equation, in the first column, on the second page of this article by A. Peres titled "Measurment of time by quantum clocks" <aapt.scitation.org/doi/10.1119/1.12061>. $\endgroup$ – e4alex Nov 1 '18 at 14:10

There is a conjugate operator if and only if there is a unitary operator $U$ such that $UAU^\dagger = X$. This is a consequence of the Stone-von Neumann theorem - all representations of the canonical commutation relations are unitarily equivalent.

  • $\begingroup$ Agreed! But given $A$ how does one know if such a $U$ exists? And how does one find $U$? $\endgroup$ – e4alex Oct 31 '18 at 20:57
  • $\begingroup$ @e4alex I've appended the answer. $\endgroup$ – ACuriousMind Oct 31 '18 at 21:01
  • $\begingroup$ Thank you for your answer. So if I understand you, if and only if the spectrum of $A$ is equal to the spectrum of $X$, which coincides with the entire real line $\mathbb{R}$, then there exists such a unitary $U$. This implies that for $B$ to exist then the spectrum of $A$ is $\mathbb{R}$. $\endgroup$ – e4alex Oct 31 '18 at 21:33
  • $\begingroup$ However, consider A = P^2, the spectrum of which coincides with the positive real line $\mathbb{R}_+$, which is not equal to the spectrum of $X$. And yet $A = P^2$ is canonically conjugate to the operator $B = -\frac{1}{4} \left(\frac{1}{P} X + X \frac{1}{P} \right)$. In this case the spectrum of $A$ is not equal to the spectrum of $X$. Also, what if the spectrum of $A$ is degenerate? How is the unitary constructed? $\endgroup$ – e4alex Oct 31 '18 at 21:59
  • $\begingroup$ @e4alex Ah, unbounded operators, they never play nice. My claim only holds for bounded operators... $\endgroup$ – ACuriousMind Oct 31 '18 at 22:05

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