Potential commutes with position?

[syang9]

Why does the potential energy (operator?) commute with the position operator? Is it because we deal only with real potentials?

 

[Dr Transport]

The potential operator commutes with the position operator when the potential operator is a function of position only, it will not commute if it is a function of the momenta as is seen in nuclear physics.

 

[dx]

You can think of the potential operator and the position operator as measurements. If the potential doesn't depend on the particles motion, then it wouldn't matter in which order you measure them. But if the potential depends on the momentum of the particle, then they don't commute because the measurement of the particles position affects its momentum.

 

[jostpuur]

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.

For given measurable function $V:\mathbb{R}^3\to\mathbb{R}$ the potential operator is

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

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

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

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

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

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

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

whenever $\psi$ is in domain of these composition maps.

 

[samalkhaiat]

Why does the potential energy (operator?) commute with the position operator? Is it because we deal only with real potentials?

The commutator $\left[\hat{x},\hat{V}\right]$ depends on the representation of the operator algebra
$$\left[\hat{p},\hat{x}\right] = i$$

1) In the x-representation :

$$\hat{x} \rightarrow x$$
$$\hat{p} \rightarrow - i \partial_{x}$$

Schrodinger equation is a differential equation

$$i \partial_{t} \Psi (x) = \left[ - \frac{\partial^{2}}{\partial x^{2}} + V(x) \right] \Psi (x)$$

where $V(\hat{x}) = V(x)$ is an ordinary function of x. Therefore

$$\left[ \hat{x}, \hat{V}\right] = 0$$

2) In the momentum-representation :

$$\hat{x} \rightarrow i \frac{\partial}{\partial p}$$
$$\hat{p} \rightarrow p$$

Schrodinger equation becomes an integral equation

$$i\partial_{t} \Psi (p) = p^{2} \Psi(p) + \int d \bar{p} V( p - \bar{p} ) \Psi(\bar{p})$$

where

$$V( p - \bar{p} ) \equiv \langle p | V | \bar{p} \rangle = \int dx \ e^{ix( p - \bar{p})} V(x)$$

is an ordinary function of the momentum p. So in this representation, you have

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


regards

sam

 

[reilly]

The momentum and position representations are connected by a unitary transformation, which always preserves canonical or any commutation relations. So, if V is a function of x, then [x,V]=0 holds necessarily in the momentum representation, or any representation. Hint; write out V in terms of the momentum rep of x.
Regards,
Reilly Atkinson

 

[jostpuur]

(Let us remain extremely civilized and scientific. I'm concerned with the content)

Schrodinger equation becomes an integral equation

$$i\partial_{t} \Psi (p) = p^{2} \Psi(p) + \int d \bar{p} V( p - \bar{p} ) \Psi(\bar{p})$$

where

$$V( p - \bar{p} ) \equiv \langle p | V | \bar{p} \rangle = \int dx \ e^{ix( p - \bar{p})} V(x)$$

is an ordinary function of the momentum p.

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

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

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

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 $V(x)=x^2$, and insist on writing the potential operator as an integral operator, we need derivatives of the delta function to the kernel.

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

 

[jostpuur]

It could I complained about a small thing. If we accept

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

then this

$$V( p - \bar{p} ) \equiv \langle p | V | \bar{p} \rangle = \int dx \ e^{ix( p - \bar{p})} V(x)$$

is an ordinary function of the momentum p.

is fine. However, here

So in this representation, you have

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

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.

 

[reilly]

Operators in Hilbert Space can be multiplied by other operators, and can multiply states, as in O|p>.

!. 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