B Understanding Fourier Transform for Wavefunction Representation in K Space

Physgeek64
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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 $$\phi(p) =\frac{\phi(k)}{\sqrt{\bar{h}}} $$

Many thanks in advance :)
 
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Physgeek64 said:
But why then is
$$
\phi(p) =\frac{\phi(k)}{\sqrt{\bar{h}}}
$$

Where are you getting this equation from?
 
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.
 
Physgeek64 said:
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 $$\phi(p) =\frac{\phi(k)}{\sqrt{\bar{h}}} $$
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 .
 
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!
 
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