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: $$\psi(x)=\begin{cases}\frac{1}{\sqrt{2a}} \ \ \ \ \ \ |x|<a\\ 0 \ \ \ \ \ \ |x| \geq a \end{cases}$$ 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?

  • $\begingroup$ 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. $\endgroup$
    – ohneVal
    Jun 18 at 16:32
  • $\begingroup$ @ohneVal Problem is: in my edit we see that the probabilities seem to change.. $\endgroup$
    – Noumeno
    Jun 18 at 16:39
  • $\begingroup$ 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, $\psi(E)=\sqrt{p/2E}\psi(p)$ $\endgroup$ Jun 18 at 18:25
  • 1
    $\begingroup$ 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 $(2\pi a^2\sigma^2)^{-1/2}e^{-x^2/2a^2\sigma^2}$. $\endgroup$
    – jacob1729
    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.

  • $\begingroup$ This may be nonstandard in some places but I like the reasoning! $\endgroup$ Jun 19 at 3:06
  • $\begingroup$ 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. $\endgroup$ 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 $$|\psi(p)|^2 dp=|\psi(E)|^2 dE=|\psi(E)|^2 \frac{2E}{p}dp\quad\Leftrightarrow \quad|\psi(p)|^2=|\psi(E)|^2 \frac{2E}{p}.$$

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

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