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Why does the potential energy (operator?) commute with the position operator? Is it because we deal only with real potentials?
They commute because they are both multiplication operators (assuming potential depends only on position), and multiplication operators always commute (unless you have some more exotic algebra). It does not matter if the multiplication operators are real or complex.Why does the potential energy (operator?) commute with the position operator? Is it because we deal only with real potentials?
Why does the potential energy (operator?) commute with the position operator? Is it because we deal only with real potentials?
The commutator [itex]\left[\hat{x},\hat{V}\right][/itex] depends on the representation of the operator algebra
[tex]\left[\hat{p},\hat{x}\right] = i[/tex]
1) In the x-representation :
[tex]\hat{x} \rightarrow x[/tex]
[tex]\hat{p} \rightarrow - i \partial_{x}[/tex]
Schrodinger equation is a differential equation
[tex]i \partial_{t} \Psi (x) = \left[ - \frac{\partial^{2}}{\partial x^{2}} + V(x) \right] \Psi (x)[/tex]
where [itex]V(\hat{x}) = V(x)[/itex] is an ordinary function of x. Therefore
[tex]\left[ \hat{x}, \hat{V}\right] = 0[/tex]
2) In the momentum-representation :
[tex]\hat{x} \rightarrow i \frac{\partial}{\partial p}[/tex]
[tex]\hat{p} \rightarrow p[/tex]
Schrodinger equation becomes an integral equation
[tex]i\partial_{t} \Psi (p) = p^{2} \Psi(p) + \int d \bar{p} V( p - \bar{p} ) \Psi(\bar{p})[/tex]
where
[tex]
V( p - \bar{p} ) \equiv \langle p | V | \bar{p} \rangle = \int dx \ e^{ix( p - \bar{p})} V(x)
[/tex]
is an ordinary function of the momentum p. So in this representation, you have
[tex]\left[ \hat{x} , \hat{V} \right] = i \frac{\partial V}{\partial p}[/tex]
regards
sam
This was a mistake. The operator, which is multiplication operator [tex]\hat{V}[/tex] in the spatial space, becomesSchrodinger equation becomes an integral equation
[tex]i\partial_{t} \Psi (p) = p^{2} \Psi(p) + \int d \bar{p} V( p - \bar{p} ) \Psi(\bar{p})[/tex]
where
[tex]
V( p - \bar{p} ) \equiv \langle p | V | \bar{p} \rangle = \int dx \ e^{ix( p - \bar{p})} V(x)
[/tex]
is an ordinary function of the momentum p.
is fine. However, here[tex]
V( p - \bar{p} ) \equiv \langle p | V | \bar{p} \rangle = \int dx \ e^{ix( p - \bar{p})} V(x)
[/tex]
is an ordinary function of the momentum p.
you seem to assume that V was a multiplication operator in the momentum space too. If you calculate the commutator with the integral operator representation of the potential, the commutator vanishes again.So in this representation, you have
[tex]\left[ \hat{x} , \hat{V} \right] = i \frac{\partial V}{\partial p}[/tex]