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If someone hands me over the position operator $x$ and the momentum operator $p$ I know their commutator is given by $[x,p]=i$ (natural units). Now say somebody hands me either $x$ or $p$ but not both and then I find some other sets of different operators $\{ \mathcal{O}_m\}, \{\mathcal{W}_n\}$ which obey $[x,\mathcal{O}_m]=[\mathcal{W}_n,p]=i$ for all $n,m$.

What can be said about $\{ \mathcal{O}_m\}, \{\mathcal{W}_n\}$?

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Very little, beyond the obvious $ \mathcal{O}_m = p + f_m(x)+c_m$, where the constants c denote constant matrices, or operators with no connection to p.

Likewise, $\mathcal{W}_n =x+ g_n(p)+ d_n $ .

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  • $\begingroup$ Why do you say that this is very little information? It looks like a significant restriction on the operators to me $\endgroup$ – Joe Apr 29 at 10:19
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    $\begingroup$ Very little beyond. $\endgroup$ – Cosmas Zachos Apr 29 at 10:23
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Suppose you have a pair of operators acting on a Hilbert space, and such that, $$ \left[A,B\right]=iI $$ Then it can be proven that these operators cannot be both bounded. At least one of them must be unbounded. This makes sense, because, $$ \textrm{Tr}\left(AB-BA\right)=0 $$ $$ i\textrm{Tr}I=i\times\infty [???] $$ But there are in principle many solutions where one of them is bounded, but the other is not. So unless you specify the spectra, there is no clear answer, at least to me.

But, if both operators $A$ and $B$ have $\mathbb{R}$ as their respective spectra, then there is a rigorous theorem that guarantees that they are unitarily equivalent to a so-called Schrödinger pair:

  • Multiplication by $x$
  • $-i\frac{d}{dx}$

"Unitarily equivalent" meaning that there exists $U$ unitary, such that, $$ A=UxU^{-1} $$ $$ B=U\left(-i\frac{d}{dx}\right)U^{-1} $$ Mind you: for the same $U$!! References:

  • A. Galindo, P. Pascual: Quantum Mechanics, Vol. I, p. 80
  • C. R. Putnam: Commutation Properties of Hilbert Space Operators and Related Topics

In your case, I'm assuming $\mathcal{O}_{m}$ and $\mathcal{W}_{n}$ have as spectrum $\mathbb{R}$. If that's not the case, the question is much more involved.

From the theorem:

$$ x=UxU^{-1} $$ $$ \mathcal{O}_{m}=U\left(-i\frac{d}{dx}\right)U^{-1} $$ Now, from the 1st equation, we get that $U$ commutes with $x$, and must be the multiplicative operator, $$ U=e^{if\left(x\right)} $$ and, when fed into the second equation, we get, $$ \mathcal{O}_{m}=U\left(-i\frac{d}{dx}\right)U^{-1}\Rightarrow\mathcal{O}_{m}=e^{if\left(x\right)}\left(-i\frac{d}{dx}\right)e^{-if\left(x\right)} $$ Let's check that this is correct and this new operator unitarily equivalent to the canonical $p$ still constitutes a Schrödinger pair with $x$: $$ \left[x,e^{if\left(x\right)}\left(-i\frac{d}{dx}\right)e^{-if\left(x\right)}\right]\varphi=-i\left[x,e^{if\left(x\right)}\left(-if'e^{-if\left(x\right)}+e^{-if\left(x\right)}\frac{d}{dx}\right)\right]\varphi=-i\left[x,\left(-if'+\frac{d}{dx}\right)\right]\varphi= $$ $$ =-i\left[x,\left(-if'+\frac{d}{dx}\right)\right]\varphi=-i\left(-I\right)=iI $$ where I have applied that, $$ \left[x,f'\left(x\right)\right]=0 $$ So there's your answer. The proof for $\mathcal{W}_{m}$ is similar, but I think you get the gist of it.

Edit:

Taking into account the trivial transformation, $$ \mathcal{O}_{m}\mapsto\mathcal{O}_{m}+a\left(x\right) $$ and similarly for $\mathcal{W}_{m}$, we would have the most general freedom for $\mathcal{O}_{m}$ and $\mathcal{W}_{m}$ to be, $$ \mathcal{O}_{m}=e^{if\left(x\right)}\left(-i\frac{d}{dx}\right)e^{-if\left(x\right)}+a\left(x\right) $$ $$ \mathcal{W}_{m}=e^{ig\left(p\right)}\left(i\frac{d}{dp}\right)e^{-ig\left(p\right)}+b\left(p\right) $$

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  • $\begingroup$ I've noticed @CosmasZachos' answer and definitely canonical comm. relations have that freedom too. I'm too tired to think about generalisation now--didn't have much sleep--, but I will edit later. I'm quite sure the theorem is correct. Perhaps someone can help find the most general freedom, or mentioned freedom can be subsumed under unitarily-equivalent freedom. $\endgroup$ – joigus Apr 29 at 8:19

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