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syang9

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- Thread starter syang9
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syang9

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dx

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- #4

jostpuur

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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.

For given measurable function [itex]V:\mathbb{R}^3\to\mathbb{R}[/itex] the potential operator is

[tex]

\hat{V}:D(\hat{V})\to L^2(\mathbb{R}^3),\quad (\hat{V}\psi)(x)=V(x)\psi(x)\quad\forall x\in\mathbb{R}^3

[/tex]

and the position operator (or a component of it, k=1,2,3), is

[tex]

\hat{x}_k:D(\hat{x}_k)\to L^2(\mathbb{R}^3),\quad

(\hat{x}_k\psi)(x)=x_k\psi(x)\quad\forall x\in\mathbb{R}^3

[/tex]

Here the domains [itex]D(\hat{V})[/itex] and [itex]D(\hat{x}_k)[/itex] are some subsets of [itex]L^2(\mathbb{R}^3)[/itex]. For all [itex]x\in\mathbb{R}^3[/itex]

[tex]

(\hat{V}\hat{x}_k\psi)(x)\;=\;V(x)(\hat{x}_k\psi)(x)\;=\;V(x)x_k\psi(x)\;=\;x_kV(x)\psi(x)\; =\;x_k(\hat{V}\psi)(x)\;=\;(\hat{x}_k\hat{V}\psi)(x).

[/tex]

The key step is V(x)x=xV(x), which is justified because x and V(x) are merely numbers. Therefore

[tex]

\hat{V}\hat{x}_k\psi = \hat{x}_k\hat{V}\psi,

[/tex]

whenever [itex]\psi[/itex] is in domain of these composition maps.

Last edited:

- #5

samalkhaiat

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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

Last edited:

- #6

reilly

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Regards,

Reilly Atkinson

- #7

jostpuur

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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.

This was a mistake. The operator, which is multiplication operator [tex]\hat{V}[/tex] in the spatial space, becomes

[tex]

\hat{\psi}\mapsto \mathcal{F}\hat{V}\mathcal{F}^{-1}\hat{\psi},

[/tex]

[tex]

(\mathcal{F}\hat{V}\mathcal{F}^{-1}\hat{\psi})(p) = \frac{1}{2\pi} \int dx\; d\bar{p}\; e^{ix(p-\bar{p})} V(x) \hat{\psi}(\bar{p})

[/tex]

in the momentum space. Usually you cannot write this as an integral operator with an ordinary function as the kernel. For example, if we have [itex]V(x)=x^2[/itex], and insist on writing the potential operator as an integral operator, we need derivatives of the delta function to the kernel.

[tex]

\frac{1}{2\pi} \int dx\; d\bar{p}\; e^{ix(p-\bar{p})} x^2 \hat{\psi}(\bar{p}) = \cdots =

\int d\bar{p}\;\big(-\partial_p^2 \delta(p-\bar{p})\big) \hat{\psi}(\bar{p})

[/tex]

- #8

jostpuur

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[tex]

\frac{1}{2\pi} \int dx\; e^{-ix(p-\bar{p})} x^2 = -\partial_p^2\delta(p-\bar{p}),

[/tex]

then this

[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

So in this representation, you have

[tex]\left[ \hat{x} , \hat{V} \right] = i \frac{\partial V}{\partial p}[/tex]

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.

- #9

reilly

Science Advisor

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!. As is discussed in almost all QM texts, a commutator is an operator, and any operator is independent of basis. So, if an operator is zero in any representation, it is zero in all representations.

2. Define W = xV(x) = V(x)x. W commutes with itself. Show this is so in your approach, and you will find your error.(This is most easily done with Dirac notation.)

Regards,

Reilly Atkinson

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