# Derivation of momentum operator from kinetic energy operator

Earlier in class today, my professor explained how to derive the momentum operator. His derivation went like this:

$$\hat{KE} = \frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2}$$ and $$KE=\frac{P^2}{2m} \Rightarrow P = \sqrt{2mKE}$$

Hence: $$\hat{P} = \sqrt{2m\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2}} \Rightarrow \hat{P} = -i\hbar\frac{\partial}{\partial x} \hspace{0.5cm}_{\square}$$

I understood how he got the $$-i\hbar$$ part, but I can't understand how he wrote $$\sqrt{\frac{\partial^2}{\partial x^2}}$$ as $$\frac{\partial}{\partial x}$$.

Is this not a normal derivative operator, hence we can manipulate it as such? Or can $$\frac{\partial}{\partial x}$$ be operated upon as if it were a number?

I do highly doubt the latter case as it doesn't make any sense to me. For example:

$$y = 3x^2 \Rightarrow \frac{dy}{dx}=9x^2 \Rightarrow \frac{d^2y}{dx^2}=18x$$ yet:

$$\sqrt{18x} \neq 9x^2$$

Am I thinking of this more 'number-based' than it should be? Should $$\frac{d}{dx}$$ be thought of as a more 'abstract' operator that we can manipulate as we please?

• How did your professor explain how to derive the operator for kinetic energy? Jan 26 at 1:44

$$y = 3x^2 \Rightarrow \frac{dy}{dx}=9x^2 \Rightarrow \frac{d^2y}{dx^2}=18x$$ yet:

$$\sqrt{18x} \neq 9x^2$$

Here you are comparing $$\sqrt{\frac{d^2f}{dx^2}}$$ to $$\sqrt{(\frac{df}{dx})^2}$$, which are not the same things. The latter is the first derivative, the former is the square root of the second derivative.

I can't understand how he wrote $$\sqrt{\frac{\partial^2}{\partial x^2}}$$ as $$\frac{\partial}{\partial x}$$.

You should start to think of a symbol like: $$\frac{d}{dx}$$ and an operator that operates on everything to its right (on the written page).

Therefore: $$\hat P |f\rangle \to \frac{d}{dx}f(x) \equiv \frac{df}{dx}.$$

Now call $$g(x) = \frac{df}{dx}$$. What is $$\hat P g$$? It is: $$\hat P |g\rangle \to \frac{d}{dx}g(x) \equiv \frac{dg}{dx} = \frac{d^2f}{dx^2}\;,$$ so it is reasonable to symbolically refer to a square root of $$\frac{d^2}{dx^2}$$ as $$\frac{d}{dx}$$.

I understood how he got the $$=-i\hbar$$ part

You should make sure to realize that $$+i\hbar$$ would, in principle, work just as well and that the negative sign is a convention. It is a conventional sign, but it is so entrenched it is effectively the only sign convention you will ever use. Note, if you change the sign convention here, you must change it a lot of other places, like in the definition of the commutator.

Normally (at least as far as I know) $$p = \frac {\hbar} {i} \frac{d} {dx}$$ is an axiom of quantum mechanics (or more generally you could just postulate that $$[x,p] = i \hbar$$ and then use the Stone-von-Neumann theorem to show that after a unitary transformation, in position space p becomes $$\frac {\hbar} {i} \frac{d} {dx}$$) and then you define the Hamiltonian as the Hamilton function known from classical mechanics with every "p" replaced by your p-operator and every "x" replaced with your x-operator (and then in some cases you have to symmetrisize H in order for it to become hermitian).

That being said, I think the key to understanding the derivation by your professor is the definition of the square of an operator. Squaring an operator means applying it two times, so $$A^2 |\psi> = A (A|\psi>)$$ and not $$A^2 |\psi> = (A|\psi>)^2$$. So $$(\frac {d} {dx})^2 \psi(x) = \frac {d} {dx} (\frac {d \psi(x)} {dx}) \ne (\frac {d \psi(x)} {dx})^2$$. So, also $$\sqrt{ \frac{d^2} {dx^2}} = \frac{d} {dx}$$.

By the way, you can also define the square root of an operator in general (for example using a power series, though for unbounded operators there are complications with that; another way for self-adjoint operators is to only define how $$\sqrt{A}$$ acts on their eigenstates and then use that every state can be expressed in those eigenstates), and I am pretty sure that those definitions are consistent with $$\sqrt{A^2} = A$$ for an operator A.

One last comment: In the derivation given by your professor, you could just as well have defined p as $$+ i \hbar \frac {d} {dx}$$. One might think that that just shows that it is not sufficient to start with the definition of the free hamiltonian, but that would only be part-right. Because in fact, all of quantum mechanics can be rewritten with i -> -i without losing anything. So, in a way, it is just convention that $$p = -i \hbar \frac {d} {dx}$$.