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syang9
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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?
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
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.
[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]