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So lets say we are in a 1d system and in the position basis just for simplicity. The CCR is: $$ [x,p]=i $$ and the momentum operator is $-i\partial_x$. Is this solution unique or are there other operators which could satisfy the commutation relation.

Similarly, in QFT does the commutation relation $[\phi(x),\pi(y)]=i\delta(x-y)$ uniquely specify that (in the eigenbasis of $\phi$) $\pi(x)=-i\frac{\delta}{\delta \phi(x)}$?

Obviously these are a solution to the algebra, but are they unique?

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In my experience this commutation relation usually has a different sign: $[x,p]=i$, but this is a matter of definition. Derivative $\hat{p}=-i\partial_x$ is a possible representation of operator $\hat{p}$ that satisfies this equation, but not the only possible and not necessarily the best. Let us take, e.g., two Bose operators $\hat{a},\hat{a}^\dagger$ that have matrix representation $$\langle m|\hat{a}|n\rangle = \sqrt{n}\delta_{m, n-1}, \langle m|\hat{a}^\dagger|n\rangle = \sqrt{n+1}\delta_{m, n+1},$$ and satisfy commutation relation $[\hat{a},\hat{a}^\dagger]=1$. We can now construct operators $$\hat{x}=\frac{\hat{a}+\hat{a}^\dagger}{\sqrt{2}}, \hat{p}=i\frac{\hat{a}-\hat{a}^\dagger}{\sqrt{2}},$$ which satisfy the required commutation relation $$[\hat{x},\hat{p}]=i.$$ The two representations are; of course, intimately related - one could re-express $\hat{a},\hat{a}^\dagger$ in terms of $x$, $\hat{p}=-i\partial_x$, but one can see this as another solution satisfying this commutation relation.

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  • $\begingroup$ I get that you can do it in other representations. I was really asking that if we stick to the eigenbasis of $x$, is it unique? $\endgroup$ Commented May 3, 2020 at 18:03
  • $\begingroup$ This is another representation only when we talk about a harmonic oscillator. Otherwise, these are completely different operators with the same commutation relation. $\endgroup$
    – Roger V.
    Commented May 3, 2020 at 18:22
  • $\begingroup$ In x-representation it is an integral equation, for which there is definitely a uniqueness theorem. $\endgroup$
    – Roger V.
    Commented May 3, 2020 at 18:26

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