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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?
syang9 said:Why does the potential energy (operator?) commute with the position operator? Is it because we deal only with real potentials?
syang9 said: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
samalkhaiat said: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.
samalkhaiat said:[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]
In potential theory, commute refers to the property of operators where their order of multiplication does not affect the final result. The relationship between commute and position is that in potential theory, the position operator and the Hamiltonian operator commute with each other, meaning that the order in which they are applied does not affect the final result.
Commute has a significant impact on the calculation of potential energy. In quantum mechanics, the potential energy operator and the Hamiltonian operator do not commute, meaning that the order in which they are applied affects the final result. This is because potential energy is a function of position, while the Hamiltonian also takes into account momentum and kinetic energy.
In quantum mechanics, commute plays a crucial role in determining the observability of physical quantities. Operators that commute with each other represent physical quantities that can be measured simultaneously and accurately. In contrast, operators that do not commute represent physical quantities that cannot be measured simultaneously with precision.
The uncertainty principle states that certain pairs of observables, such as position and momentum, cannot be measured simultaneously with arbitrary precision. Commute is closely related to this principle, as operators that do not commute represent observables that have inherent uncertainty and cannot be measured simultaneously with precision.
Yes, the concept of commute can also be applied to classical mechanics. In classical mechanics, commute refers to the property of physical quantities where their order of operations does not affect the final result. This property allows for the simplification of mathematical calculations and is essential in solving problems in classical mechanics.