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I know that the matrix element of momentum operator in position basis is: $\langle x|p|x'\rangle = \frac{d}{dx}\delta(x-x')$. But it seems not a number, because $\langle x|p|x'\rangle$ is rely on the position you place it in. For example, $\langle\psi|p|\psi\rangle = \sum_x\sum_{x'}\langle\psi|x\rangle\langle x|p|x'\rangle\langle x|\psi\rangle$, here you have to keep $\langle x|p|x'\rangle$ in its position, cannot move it like $\sum_x\sum_{x'}\langle x|p|x'\rangle\langle\psi|x\rangle\langle x|\psi\rangle$.

However, in momentum basis, matrix element of momentum operator $\langle k|p|k\rangle = \hbar k$ it's just a number and one can move it anywhere: $$\langle\psi|p|\psi\rangle= \sum_k\sum_{k'}\langle\psi|k\rangle\langle k|p|k'\rangle\langle k|\psi\rangle= \sum_k\sum_{k'}\langle k|p|k'\rangle\langle\psi|k\rangle\langle k|\psi\rangle.$$

Why momentum operator behave so weird? Is it true that a general matrix elements of an operator can be numbers in one basis but may not be numbers in other basis? If so, why in second quantization it can just place the matrix element in front of any creation/annihilation operators since matrix element can be rely on the position where it acts?

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Samuel Adrian Antz
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Curiousprototype
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When you work with operators then the order of the operators in the expression is important. One cannot change the order without taking the effect of commutations into account. However, when you insert identities resolved in terms of a complete set of states, then you convert the states into wave functions and the operators become kernel functions. Then the order in which you write them down don't mean anything anymore and you can change that order as you like. The operations of the operators are now taken over by the summation or integration process.

To make it explicit, we consider for example $$ \langle\psi|\hat{p}|\psi\rangle = \int \langle\psi|x\rangle \langle x| \hat{p}|x'\rangle \langle x'\psi\rangle\ dx dx' = \int \psi^*(x) p(x,x') \psi(x')\ dx dx' . $$ Here $\psi(x')$ is a wave function and $p(x,x')$ is a kernel function. The order of these functions is unimportant. The same applies if we used the $k$-basis (Fourier basis).

flippiefanus
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  • But since here kernel $p(x, x') = \frac{d}{dx}\delta(x-x')$, So, I think move $p(x, x')$ in front of $\psi^(x)$ will cause derivative x on both $\psi^(x)$ and $\psi(x')$, isn't it? Why orders is unimportant here. – Curiousprototype Nov 17 '22 at 10:34
  • Why $\int \psi^(x) \frac{d}{dx} \delta(x-x') \psi(x) dxdx'$ and $\int \frac{d}{dx} \delta(x-x')\psi^(x) \psi(x) dxdx'$ are equivalent? – Curiousprototype Nov 17 '22 at 10:58
  • Obviously, if your kernel includes a derivative then the derivative must only operator on the rest of the expression in the kernel and not on what follows. You can use brackets to enforce that. – flippiefanus Nov 17 '22 at 12:04