1
$\begingroup$

I am studying quantum mechanics right now, and I learned that momentum eigenstates are plane waves, that is

$$ \langle x \vert p\rangle = e^{ipx} $$

and

$$ \langle p' \vert p\rangle = \int_{-\infty}^{\infty} e^{i(p-p')x}dx = \delta(p-p') $$

This makes sense when thinking about Fourier transforms, but I'm confused how to think of this integral. If I were to think of this integral as $\sum e^{i(p-p')x}\Delta x$, then I would end up with a bunch of imaginary components in the sum, and I don't see how that ends up being a delta function.

$\endgroup$
3
  • $\begingroup$ How familiar are you with Fourier transforms? The answer is that the integral written above is defined only in a particular sense, but the best way to answer your question depends on your familiarity with the underlying mathematics. $\endgroup$
    – J. Murray
    Dec 11, 2020 at 4:56
  • $\begingroup$ Related: physics.stackexchange.com/q/41880/2451 $\endgroup$
    – Qmechanic
    Dec 15, 2020 at 19:16
  • 2
    $\begingroup$ Your sum notation is not precise enough, hence the confusion. It's simply wrong. You should introduce the index $j=0, \pm 1, \pm 2, \pm 3, ... $, then replace $x \to j \Delta x$ and sum over every possible index value. Then it's clear that the imaginary parts cancel due to $\sin$ being an odd function. $\endgroup$
    – Yuriy S
    Dec 15, 2020 at 20:19

4 Answers 4

3
$\begingroup$

First the quantity $\langle x| p\rangle$ is the probability amplitude that when the particle has the momentum $p$, its position is $x$.

Now coming to the main concern

I'll use the short hand, so I'll suppress the notation a bit and not write $\hbar$ and $k$ for $p$ that just different from the scalar factor.

The delta function is the sampling function since it samples the value of the function at one point $$\int \delta(x-k')f(x)dx=f(k')$$

$$f(x)=\frac{1}{\sqrt{2\pi}}\int e^{-ikx}f(k)dk$$ and its inverse $$f(k')=\frac{1}{\sqrt{2\pi}}\int e^{ik'x}f(x)dx$$ feeding one into other $$f(k')=\int \left(\frac{1}{2\pi}\int dx e^{ix(k'-k)}\right)f(k)dk$$ Comparing it with the first $$\frac{1}{2\pi}\int dx e^{ix(k'-k)}=\delta (k'-k)$$


If you want to work with sum then you need to take the limit in the end so you end up with integral.

$\endgroup$
2
$\begingroup$

$\delta(p-p')$ is often called delta function, but actually it is not function at all. It is a so-called distribution, it does not have values, but it has integrals with functions, these integrals may be functions or distributions. The "delta function" distribution is defined by its behaviour under integration sign:

$$ \int_{-\infty}^{+\infty} \delta(p-p')f(p)dp = f(p'). $$

for any continuous function $f(p)$.

It turns out there are many ways to express this distribution, for example $$ I(p,p') = \frac{1}{2\pi}\int_{-\infty}^{+\infty} e^{i(p-p')x}dx $$ is one of them. Don't agonize over imaginary numbers in the integrand, all of them (except one) can be paired with complex conjugated counter part and their imaginary parts cancel each other:

$$ e^{i(p-p')x} + e^{-i(p-p')x} = 2\cos(px-p'x) $$

The exception is the central contribution due to $e^{i~(p-p')0}$ which is equal to 1.

$\endgroup$
1
$\begingroup$

The mathematical version of the theory is more complicated (you would have to study functional analysis), but there is a very nice explanation in the book Numerical and Analytical Methods for Scientists and Engineers, Using Mathematica. The argument is given in chapter 2 section 2.3.4 on generalized Fourier integrals.

  1. The issue is that the integral $\int_{-\infty}^{\infty} e^{-ikx}dk$ is not convergent, but we still want to use it, make sense of it, and say it's equal to $2\pi\delta(x)$.
  2. We write the integral as a symmetric limit $\lim_{\Delta k\to\infty}\int_{-\Delta k}^{\Delta k} e^{-ikx}dk=\lim_{\Delta k\to\infty}2\sin(\Delta k x)/x$. Note that this limit "does not exist" in the introductory calculus sense, which is why we have to call this a generalized Fourier integral.
  3. We want to see if we can make sense of this, so we look at how it acts on a function if we take an inner product: $\lim_{\Delta k\to\infty}\int_{-L}^{L} f(x)2\sin(\Delta k x)/x dx$. (Here there's another mathematical crime of changing the order of the limit vs the integral. The integration is from $-L$ to $L$ because we assume $f$ is zero outside some region around the origin). Change variables to $y=\Delta k x$ to get $\lim_{\Delta k\to\infty}\int_{-L\Delta k}^{L\Delta k} f(y/\Delta k)2\sin(y)/y dy$.
  4. This integral is convergent, so we should be able to take the limit to get $\int_{-\infty}^{\infty} f(0)2\sin(y)/y dy=2\pi f(0)$.
  5. Conclude that $\int_{-\infty}^{\infty} e^{-ikx}dk=2\pi\delta(x)$.

This argument shows where the factor of $2\pi$ comes from, and also gives some extra parameters $L$ and $\Delta k$ to play around with.

Also note that I usually prefer to work with the normalization $\langle x \vert p\rangle = \frac{1}{\sqrt{2\pi}}e^{ipx}$, but this is just personal preference of how you go about bookkeeping your $2\pi$'s.

$\endgroup$
0
$\begingroup$

< x | p > is the scalar-product of 2 vectors. The resulting scalars can be complex in QM: z = x + i . y = exp(i.alpha), with alpha being the angle of the z-vector with the positive x-axis. The wave function PSI in QM is like a vector containing complex components. Unfortunately, the wave function is sometimes called a probability amplitude. The wave function squared gives a probability density function and an integral of this function gives a probability.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.