# Defined Momentum vs. Defined $k$

In quantum mechanics usually we write the momentum operator $$\hat{p}$$ as: $$\hat{p} = \hbar \hat{k}. \tag{1}$$ with of course: $$\hat{p}|p\rangle = p |p\rangle \tag{2}$$ $$\hat{k}|k\rangle=k|k\rangle \tag{3}$$ But when we investigate the shape of the eigenfunctions with defined momentum $$\psi _p (x)$$ and defined $$k$$ ($$\psi _k(x)$$) we get: $$\langle x|p\rangle=\psi _p(x)=\frac{1}{\sqrt{2\pi\hbar}}\exp\left[i\frac{p}{\hbar}x\right] \tag{4}$$ $$\langle x | k \rangle=\psi _k(x)=\frac{1}{\sqrt{2\pi}}\exp\left[ikx\right] \tag{5}$$

I find this unbearably ugly! In fact since the following relation holds: $$k=\frac{p}{\hbar} \tag{6}$$ it would have been soo nice if both expression shared the same normalization constant, both $$1/\sqrt{2\pi}$$ for example, because if this was the case we could have simply remembered (6) to switch between $$\psi _k(x)$$ and $$\psi _p(x)$$.

I don't really understand why the normalization constant changes, since, in light of (1) and (6), $$\exp{[ikx]}$$ should be the same as $$\exp{[ipx/\hbar]}$$.

Seems that the fact that (4) is not analogous to (5) makes working in the base of $$\hat{k}$$ not the same as working in the base of $$\hat{p}$$, which is strange considering that $$\hat{p}$$ and $$\hat{k}$$ commute and are practically the same operator..

Edit: to better show what's my problem with all this: consider a free particle with the following wave function: $$\psi(x)=\begin{cases}\frac{1}{\sqrt{2a}} \ \ \ \ \ \ |x| and suppose we want to find, for this wave function, the probability distribution for the energy. Our best bet seems to be to perform a change of variable, since the Hamiltonian commutes with $$p$$ and $$k$$, doing this we get (feel free to check my math): $$\psi(k)=\frac{1}{k\sqrt{\pi a}}\sin(ka) \tag{1'}$$ or $$\psi(p)=\sqrt{\frac{h}{\pi a}}\frac{\sin(ap/\hbar)}{p} \tag{2'}$$ and then, since $$E=\frac{p^2}{2m} \tag{3'}$$ $$E=\frac{\hbar ^2 k^2}{2m} \tag{4'}$$ we can fin the probability amplitude $$\psi(E)$$ either by substituting (3') in (1') or (4') in (2'). The problem is the results are not the same! What is going on?

• Do you understand that the difference relies in the units, not in the math... $\hbar$ has units of action or angular momentum if you will, sometimes dimensionless units are nicer to work with, sometimes the opposite. Jun 18 at 16:32
• @ohneVal Problem is: in my edit we see that the probabilities seem to change.. Jun 18 at 16:39
• No, they still don't change! You can calculate $dE= (2E/k)dk$ and $dE= (2E/p)dp$ using change of variables. The probability distribution can be written as $|\psi(k)|^2 dk=|\psi(p)|^2 dp=|\psi(E)|^2 dE$ and everything is still the same. This means that we need, for example, $\psi(E)=\sqrt{p/2E}\psi(p)$ Jun 18 at 18:25
• This isn't a physics question but a maths one - consider for example that the pdf for a Gaussian variable $x$ with width $\sigma$ is $(2\pi \sigma^2)^{-1/2}e^{-x^2/2\sigma^2}$ but if you rescale it to $y=x/a$ you change to $(2\pi a^2\sigma^2)^{-1/2}e^{-x^2/2a^2\sigma^2}$. Jun 18 at 18:28

You can normalize your continuous spectrum (and hence unnormalizable) eigenstates as you like provided you keep the completeness relation correct.

I always normalize my momentum states as $$\langle x|k\rangle= e^{ikx}$$ with no inverse square roots. Then the completeness integral is $${\rm Id}= \int\frac{dk}{2\pi} |k\rangle \langle k|$$ The measure $$dk/2\pi$$ is quite natural because it is the number of momentum states in the interval $$[k,k+dk]$$ per unit volume in $$x$$ space. If you put the $$\sqrt{2\pi}$$ with the state then you lose the ability to easily keep track of the $$2\pi$$'s. My way every momentum-conservation delta function comes with a $$2\pi$$ as $$2\pi \delta(k-k')$$ and so on. Keeping track of $$2\pi$$'s is important because $$(2\pi)^4\approx 1600$$ and losing this factor can really annoy the experimentalists who are looking for your predicted effect.

• This may be nonstandard in some places but I like the reasoning! Jun 19 at 3:06
• Also, I don't think this directly clears up the confusion of the original question. You will still need $\langle x|k\rangle$ to differ from $\langle x|p\rangle$ by a factor of $1/\sqrt{\hbar}$, or else you will have to remember one integration measure is $dk/2\pi$ and the other is $dp/2\pi\hbar$, which is a bit more work in my opinion. Jun 19 at 13:21

This is a question of units. Many physicists choose to use units in which $$\hbar=1$$ such that the two expressions will be the same. Inside of an exponential, the argument must be unitless, which is why we must have $$\exp(i px/\hbar)=\exp(ikx).$$ However, $$|k\rangle$$ and $$|p\rangle$$ have different units; equivalently, $$\psi_k$$ as you have defined it is unitless while $$\psi_p$$ has units of $$1/\sqrt{\mathrm{energy}\times\mathrm{time}}$$.

It turns out that this "normalization constant" does not affect any of the physics involved in using these functions, so we don't often worry about it. I put normalization constant in quotations because momentum eigenstates are not normalized in the usual sense, so the prefactor is mostly a convention that ensures the overlap between two momentum eigenstates is a delta function.

The two expressions are equivalent in the sense that probability distributions are equal: $$|\psi(k)|^2 dk=|\psi(p)|^2 dp.$$

To answer the edit, the three probability distributions are equal: $$|\psi(k)|^2 dk=|\psi(p)|^2 dp=|\psi(E)|^2 dE.$$ One has to use a proper change of variables to make sure, for example, that $$|\psi(p)|^2 dp=|\psi(E)|^2 dE=|\psi(E)|^2 \frac{2E}{p}dp\quad\Leftrightarrow \quad|\psi(p)|^2=|\psi(E)|^2 \frac{2E}{p}.$$

• check my edit if you want Jun 18 at 16:26
• Still no problem - recall that a probability distribution needs to include the measure, such as $dE$. Jun 18 at 18:28