# The free particle wave function <x|p>?

1. May 5, 2015

### Xyius

I am following the math of scattering theory in Sakurai, Revised Edition pp.380-381

For a free particle, one can find that the solution is a plane wave that can be written (in position space) as,

$$<x|\phi>=\frac{e^{ip \cdot x}}{(2 \pi \hbar)^{3/2}}$$

However, how does one obtain $<x|p>?$ In the book it has..

$$<x|p>=\frac{e^{ip \cdot x}}{(2 \pi \hbar)^{3/2}}$$

Which is identical to $<x|\phi>$. Why are these two expressions the same? I also don't know what $<x|p>$ means physically. Momentum in the position basis?

Last edited: May 5, 2015
2. May 5, 2015

### fzero

The state $|p\rangle$ is the eigenstate of the momentum operator:

$$\hat{p} | p\rangle = p |p\rangle.$$

The free particle state $|\phi\rangle$ is also a momentum eigenstate, so it should be identified with $|p\rangle$.

In order to compute $\langle x|p\rangle$, we should consider how to expand $|p\rangle$ in the position basis:

$$| p \rangle = \int dx f(p,x) |x\rangle.$$

In the position basis, $\hat{p}=-i\hbar d/dx$, so it must be that

$$\hat{p} | p \rangle = -i\hbar \int dx \frac{df(p,x)}{dx} |x\rangle .$$

But this is an eigenvector with eigenvalue $p$, so it is also the case that

$$\hat{p}| p \rangle = p | p \rangle = p\int dx f(p,x) |x\rangle.$$

Comparing these leads to an equation:

$$-i\hbar \frac{df(p,x)}{dx} =p f(p,x)$$

which has solution

$$f(p,x) = e^{i px/\hbar}.$$

Computing $\langle x |p \rangle$ is then just a matter of recognizing a delta function in

$$\int dx' e^{i px'/\hbar} \langle x |x'\rangle.$$

3. May 5, 2015