I'm reading through Nakahara's Geometry, Topology and Physics and I don't understand the following derivation on pg. 41:

$$ \text{Now we find from the commutation relation of } \partial_x \equiv \frac{d}{dx} \text{ and } e^{ikx} \text{ that} \\ \partial_x e^{ikx} = ik e^{ikx} + e^{ikx} \partial_x = e^{ikx} ( ik + \partial_x) $$

Why do we need the second term? $\partial_x$ seems to be just an ordinary derivative so why is the $e^{ikx} \partial_x $ term necesary ?


Note that the commutator of $\partial_x$ and $e^{ikx}$ can be computed to be

$$\large [\partial_x , e^{ikx}] = \partial_x e^{ikx} - e^{ikx} \partial_x = ik e^{ikx} $$

To show this, consider the action of the commutator on a function $F(x)$ i.e,

$$ [\partial_x , e^{ikx}]F(x) = \partial_x e^{ikx} F(x)- e^{ikx}\partial_x F(x) = e^{ikx}\partial_x F(x) + ik e^{ikx}F(x) - e^{ikx} \partial_x F(x)$$

which means that $$ [\partial_x , e^{ikx}]F(x) = ik e^{ikx}F(x)$$

Or in other words,

$$ \partial_x e^{ikx} - e^{ikx} \partial_x = ike^{ikx}$$

Therefore $$\large\partial_x e^{ikx} = ik e^{ikx} + e^{ikx}\partial_x $$


$$\boxed {\partial_x e^{ikx} = e^{ikx} ( ik + \partial_x)}$$

This is exactly what is written above and if this is what is written in that book then it is certainly correct.

  • $\begingroup$ I believe it is not a typo because the author then goes on to use the expression $\partial_x^n e^{ikx} = e^{ikx}( ik + \partial_x)^n$ $\endgroup$ – David Feng Nov 5 '20 at 23:31
  • $\begingroup$ That looks like the product rule applied n times. $\endgroup$ – joseph h Nov 6 '20 at 0:06
  • $\begingroup$ How are they incorrect? $\endgroup$ – joseph h Nov 6 '20 at 0:24
  • $\begingroup$ The second expression should read $$\{\partial_x , e^{ikx} \} = e^{ikx} ( ik +2 \partial_x)$$ $\endgroup$ – my2cts Nov 6 '20 at 0:25

Whenever computing commutators, we must understand the operators (here $\partial_x$ and $e^{ikx}$) to act on everything to the right. So when writing something like $\partial_x e^{ikx}$ it should be understood to mean the operator. That is, the operator such that for any function $f(x)$, it acts as $\partial_x (e^{ikx}f(x))$. Hence the second term you mention is precisely the derivative hitting this $f(x)$.

It is an unfortunate notation, but also rather common throughout physics, particularly when dealing with generators, commutators, and so on. If the derivative was intended to act only on the exponential, it would be written $\partial_x(e^{ikx})$.

The idea behind this is as follows. Suppose we have a vector $\boldsymbol v$ and two matrices $M$ and $N$ and wish to compute $MN\boldsymbol v$. We know that this is equivalent to first computing $N\boldsymbol v$ and then applying $M$ to the result. So here when we are multiplying operators, the idea is to take a "vector" (function) $f(x)$ and first apply $e^{ikx}$ to obtain $e^{ikx}f(x)$. Then we apply $\partial_x$ to this, so necessarily the derivative should hit both factors.


The commutation relation for $\partial_x$ and $e^{ikx}$ is $$\partial_x e^{ikx} - e^{ikx} \partial_x = ike^{ikx}$$ Note that $$\partial_x e^{ikx}$$ here is an operator on some not shown function so the product rule for differentiation is used.


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