# Does an operator of x commute with the differential operator with respect to x?

While solving a problem in Quantum Mechanics I got an expression $\frac{d}{dx}V(x)-V(x)\frac{d}{dx}$. The first term is just the derivative of the potential but the second one seems a bit weird. Is there a commutation relationship between the two operators? Can I just write the sum as zero?

• You mean that the operator is $L=V'(x)-V(x)\cdot\frac{d}{dx}$? If it is, it's not zero because if you apply it to a function $\psi(x)$ you get $L\psi=V'(x)\cdot\psi(x)-V(x)\cdot\psi'(x)$ which in general is not zero- – marco trevi Oct 2 '14 at 8:04
• You should tell us how your operator works on your wave functions... – marco trevi Oct 2 '14 at 8:05

In general, in quantum mechanics, but also in the general theory of first-order, linear differential operators, if $\psi \in H$ is a vector in some (possibly Hilbert) vector space $H$ and $A$ and $B$ are operators (not necessarily bounded) on $H$, we take $(AB) \psi \equiv A(B\psi)$. Thus

$(\dfrac{d}{dx} V(x))\psi = \dfrac{d}{dx}(V(x)\psi) = V^\prime(x) \psi + V(x) \psi^\prime, \tag{1}$

whereas

$(V(x)\dfrac{d}{dx})\psi = V(x)\psi^\prime. \tag{2}$

Thus

$[\dfrac{d}{dx}, V(x)]\psi = V^\prime(x) \psi \ne 0 \tag{3}$

in general. The same phenomenon occurs when considering vector fields as first order operators $X$ and functions $f(x)$, where

$X = \sum X_i(x) \dfrac{\partial}{\partial x_i}; \tag{4}$

then an easy calculation shows that

$[X, f]\psi = X(f) \psi; \tag{5}$

formulas such as (5) arise frequently when dealing with vector fields on manifolds etc.

Hope this helps. Cheers,

and as always.

Fiat Lux!!!

• Right, I saw what I was doing wrong. I had forgotten the second term in formula (1). I didn't realise that I have to apply the product rule. Thank you! It's clear now. – Suzanka Oct 3 '14 at 6:05
• Well, I'm glad to help out, especially if it means advancing the cause of theoretical chemistry! ;-) By the way, if you find the answer helpful, you might consider "accepting" it (green check mark). Your decision of course. Hope to see you around MSE. Cheers! – Robert Lewis Oct 3 '14 at 6:22

Let us use an overline $\;\overline{(\cdots)}\;$ to stress an expression $\;(\cdots)\;$ is an operator over the space of functions that are sufficiently regular for everything to make sense. In quantum mechanics, this is the space of wave functions. The expression at hand really means $$\overline{\frac{d}{dx}}\overline{V(x)} - \overline{V(x)}\overline{\frac{d}{dx}}\tag{*1}$$ which is a commutator of two operators:

• $\displaystyle\;\overline{\frac{d}{dx}}\;$ is the operator differentiating expression on its right with respect to $x$.

• $\displaystyle\;\overline{V(x)}\;$ is the operator of multiplying a function on its right by the scalar function $V(x)$.

If you apply this commutator to a function $\phi$, you get another function $\psi$.

$$\left[\overline{\frac{d}{dx}}\overline{V(x)} - \overline{V(x)}\overline{\frac{d}{dx}}\right]\phi = \psi$$

To see what this means, you need to evaluate both side at some real number $s$. You get

\begin{align} \psi(s) = & \left\{\left[\overline{\frac{d}{dx}}\overline{V(x)} - \overline{V(x)}\overline{\frac{d}{dx}}\right]\phi\right\}(s)\\ &= \left.\frac{d}{dx}\left(V(x)\phi(x)\right) - V(x)\frac{d}{dx}\left(\phi(x)\right)\right|_{x=s}\\ &= V'(s) \phi(s)\\ &= \left\{\overline{V'(x)}\phi\right\}(s) \end{align} Since this is true for all $s$, this leads to

$$\psi = \left[\overline{\frac{d}{dx}}\overline{V(x)} - \overline{V(x)}\overline{\frac{d}{dx}}\right]\phi = \overline{V'(x)}\phi$$

Since this is true for all $\phi$, we find

$$\overline{\frac{d}{dx}}\overline{V(x)} - \overline{V(x)}\overline{\frac{d}{dx}} = \overline{V'(x)}$$

i.e. the operator $(*1)$ is equal to the operator $\overline{V'(x)}$ which multiplying the function at its right by the scalar function $V'(x)$.

When one work with this sort of expressions, one will typically abuse the notation and write the whole mess as

$$\frac{d}{dx} V(x) - V(x) \frac{d}{dx} = V'(x)$$

When one see an expression like this in quantum mechanics, one should be careful. It can mean different things on different situations.

No, this is an operator $\displaystyle (\frac{d}{dx}V(x)-V(x)\frac{d}{dx})\psi=(\frac{d}{dx}V(x))\cdot\psi-V(x)\frac{d\psi}{dx}$ or something...

The weird form refers to the Lie bracket $\displaystyle\left[\frac{d}{dx},V\right]$ in Liealgebra.