If $\hat{p}$ is the position operator, one has: $$\langle p| \hat{p}|x_{0}\rangle = \overline{\langle x_{0}|\hat{p}|p\rangle} = p \overline{\langle x_{0}|p\rangle} = p \langle p|x_{0}\rangle.$$ And my professor wrote, in its lecture notes, that if the Hamiltonian $\hat{H}$ is $\hat{H}= \frac{1}{2m}\hat{p} + V(x)$ then: $$\langle p| \hat{H}|x_{0}\rangle = \left(\frac{1}{2m}p^{2}+V(x_{0})\right)\langle p| x_{0}\rangle$$ Why does it follow that $\langle p|V(x)|x_{0}\rangle = V(x_{0})\langle p|x_{0}\rangle$? How does one prove it?
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2Usually $\hat p$ is not a position but rather momentum operator. $\hat x$ is position operator, i.e. $\hat x | x_0 \rangle = x_0 | x_0 \rangle$. Then $V(\hat x) | x_0 \rangle = V(x_0) | x_0 \rangle$ – nwolijin Jun 16 '21 at 14:17
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Your first equation is not correct, unless $\hat{p}$ represents the momentum operator. Actually, the result is still correct of the first equation (but this may be coincidentally...) – MrQ Feb 28 '22 at 20:48
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This is the definition of a function of an observable, as for example discussed in Dirac's The Principles of Quantum Mechanics, fourth edition, p. 41-45:
If $A$ is an observable with $A |a\rangle = a\, |a\rangle$, then a function $f$ of $A$ is defined through
$$ f(A) |a\rangle = f(a) |a\rangle \quad . $$
Thus, in the case of a single-particle potential $V(X)$, where $X$ denotes the position operator, we find $$V(X) |x\rangle = V(x)|x\rangle $$ and therefore $$\langle p|V(X)|x\rangle = V(x) \langle p|x\rangle \quad . $$
Tobias Fünke
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3I think one should add that this is only true for analytical functions, since they can be written as a Taylor Series. – Jan2103 Jun 16 '21 at 14:15
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@Jan2103 See for example the comments in this physics SE post/ answer. – Tobias Fünke Jun 16 '21 at 14:18
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Additionally, Dirac also notes this 'problem' in his Principles of Quantum Mechanics, ed. 4, p. 41, equation 34. – Tobias Fünke Jun 16 '21 at 14:20
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