After finding the eigenfunctions $u_p(x)=Ce^{ipx/\hbar}$ of the momentum operator just like in this UCSD lecture notes, one seeks to normalize them, so one first tries: $$\int\limits_{-\infty}^{\infty} dx \, |C|^2 e^{-ipx/\hbar} e^{ipx/\hbar} = \int\limits_{-\infty}^{\infty} dx \, |C|^2 \rightarrow \infty $$ which diverges unless $C =0$.

Then it is shown that $u_p(x)=\frac{1}{\sqrt{2\pi \hbar}}e^{ipx/\hbar}$ satisfies the normalization condition $\langle p'|p\rangle=\delta(p-p')$

Why does the UCSD page say that the first solutions (the divergent ones) "are not normalizable to one particle"? How does the development that follows relate to many particles?

Does this have something to do with $\int\limits_{-\infty}^{\infty}dp \delta(p-p') =1$? Then all the $p'$ are the other particles?

I have no source to show for this, but how would $\frac{1}{2\pi \hbar}$ particles per unit of length and unit of momentum relate? Is it correct to say that $2\pi \hbar$ is the expected number of times one repeats the measurement of the momentum? Then where would the 'per unit of length' part come from?

  • $\begingroup$ What are the properties of the Dirac delta function again? In particular does $\delta(0)$ have a finite value? $\endgroup$ Jan 21, 2014 at 2:51
  • $\begingroup$ $\delta(0)=\infty$ but is also infinitely narrow, so that $\int\limits_{-\infty}^{\infty}dx \delta(x) =1$... So I think I had a typo above : $\int\limits_{-\infty}^{\infty}dx \delta(p-p') =1$ should have been $\int\limits_{-\infty}^{\infty}dp \delta(p-p') =1$. $\endgroup$ Jan 21, 2014 at 2:58

2 Answers 2


The momentum eigenstate is not normalizable. Suppose $x$ is a periodic variable with period $2\pi R$ -- the periodicity is important as momentum is then a `good' operator. Then one has

  1. the allowed values of momenta are quantized i.e., $p_n = \frac{n \hbar}{(R)}$ with $n\in \mathbb{Z}$.
  2. For the values of $p$ mentioned above, the momentum eigenstates are normalizable with $C=1/\sqrt{2\pi R}$. Let us call this normalized state $\langle x|n\rangle$ in the coordinate basis. Explicitly, one has $$\langle x|n\rangle = \tfrac1{\sqrt{2\pi R}}\ e^{ip_n x/\hbar}\ .$$ (You should treat the remark in the UCSD page about one particle to mean that your normalize your state to one.)
  3. The eigenstates are orthogonal to each other as one can check explicitly. $$ \langle n | m\rangle =\delta_{n,m}\ .$$

The next step is to go from the box normalizable states to the delta function normalizable states. One needs to take $R\rightarrow\infty$ and convert Kronecker delta into the Dirac delta function. One uses the following identification which follows from the properties of the Dirac delta function and standard integration: $$ \sum_m = \frac{R}{\hbar}\int dp \quad \mathrm{and}\quad \delta_{m,n} = \frac{\hbar}{R} \delta(p_n-p_m)\ . $$ With these identifications, we define $$ |p_n\rangle =\sqrt{\tfrac{R}{\hbar}}\ |n\rangle\ . $$ It is now easy to see that the Dirac delta normalization implies that in the $x$-basis, one has $$u_p(x):=\langle x|p_n\rangle = \tfrac1{\sqrt{2\pi \hbar}}\ e^{ip_n x/\hbar}\ .$$

This leads to a simple mnemonic: replace $R$ by $\hbar$ to go from box normalizable states to Delta function normalization. Of course, momentum which was discrete in a box now takes values in the continuum.


@Suresh Doesn't the normalization condition follow directly from the fourier transform relation for delta function? I'm not clear about the need for box normalization and PBC. Sorry I'm writing this as an answer, I don't have enough reputation to comment. Thanks.


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