# B Position representation

1. Dec 22, 2016

### Physgeek64

I understand that the fourier transform to obtain the representation of a wavefunction in k space is

$$\phi(k) =\frac{1}{2\pi}\int{dx \psi(x)e^{-ikx} }$$
and that $$p=\bar{h} k$$

But why then is $$\phi(p) =\frac{\phi(k)}{\sqrt{\bar{h}}}$$

2. Dec 22, 2016

### Staff: Mentor

Where are you getting this equation from?

3. Dec 22, 2016

### dextercioby

In QM of non-specially relativistic particles, the Fourier transform is usually defined with a $\frac{1}{\sqrt{2\pi\hbar}}$ factor in front of the integral. Thus the OP should be amended by the user, so his (intended) question should automatically find an answer.

4. Dec 22, 2016

### Eye_in_the_Sky

Well, I think you would agree that we must have

dp |ø(p)|2 = dk |ø(k)|2 ,

for the corresponding intervals [p, p + dp] and [k, k + dk], where p=hbark.

But dp = hbar dk; so,

hbar dk |ø(p)|2 = dk |ø(k)|2 ,

and therefore,

|ø(p)|2 = |ø(k)|2/hbar .

5. Dec 24, 2016

### vanhees71

That's simply, because $|\phi(p)|^2$ is a distribution function (namely the probability distribution for momentum). In this sense we can write
$$|\phi(p)|^2 = \frac{\mathrm{d} N}{\mathrm{d} p},$$
where $N$ is the number of particles. Since the wave number and momentum are related by $p=\hbar k$ you get
$$\frac{\mathrm{d} N}{\mathrm{d} p} = \frac{\mathrm{d} N}{\mathrm{d} k} \frac{\mathrm{d} k}{\mathrm{d} p} = \frac{1}{\hbar} \frac{\mathrm{d} N}{\mathrm{d} k}.$$
Now the phase of the wave function is arbitrary, and thus you can conclude from this that
$$\phi(p)=\frac{1}{\sqrt{\hbar}} \tilde{\phi}(k) = \frac{1}{\sqrt{\hbar}} \tilde{\phi}(p/\hbar).$$
Note that you should use a different function symbol for the momentum and the wave-number distribution!