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My course notes say,

for normalised wave functions $\psi(x,t)$, the function $$\rho(x,t)=|\psi(x,t)|^2=\overline{\psi(x,t)}\psi(x,t)$$ gives the $\color{red}{\text{probability density}}$ for the position of the particle.

I guessed from the bold that something about this is specific to normalised wave functions, but later the notes have

The probability density of $$\psi(x,t)=A\exp(i(kx-\omega t))$$ is $$\rho(x,t)=|\psi|^2=|A|^2.$$ [...] We cannot normalise the wavefunction $\psi$.

So what relationship is there between normalisability and the existence of a probability density function? Can I use $\rho(x,t)=|\psi(x,t)|^2$ for any $\psi$, regardless of whether or not it's normalisable?

mjc
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    This may help - Normalizing the solution to free particle Schrödinger equation – mmesser314 Mar 19 '23 at 18:33
  • @mmesser314 Thanks. I looked at that before posting, and a bit more closely just now, but I don't feel much clearer about what implicit assumptions my course notes might be making in order to write the pdf of a non-normalisable function without further comment. Just under my second quote the notes also write, "We can only talk about the relative probability of finding particle [sic] in two intervals" - but then what is meant by $\rho(x,t)=|\psi|^2=|A|^2$? Something like $$\frac{\rho(x,t){[a_1,b_1]}}{\rho(x,t){[a_2,b_2]}}=|A|^2\text{ for any }a_1,b_1,a_2,b_2\text{ s.t. }b_1-a_1=b_2-a_2?$$ – mjc Mar 19 '23 at 18:51
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    It should really be $\rho/(region)$ = $A^2$. –  Mar 19 '23 at 19:01
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    is it bad that I originally thought this was asking about file types, and I was wondering why only certain files would require normalisable wavefunctions lol – Tristan Mar 20 '23 at 10:19
  • @Tristan Because a .pdf file is in "portable document format", so I wondered if it might exist as a scattering state. – mjc Mar 20 '23 at 17:33

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The probability distribution function is defined as $\rho$ = $\psi^* \psi$. It is integrated from the relevant bounds to find the probability in that region. Probability only makes sense when the probability for any event to occur is 100%. For example, if I roll a die, the probability that I get any number from 1 to 6 is 100%. Likewise, the probability that you find a particle anywhere should be 100%. This is why normalization occurs. When we have non normalizable functions, we can no longer talk about this “absolute” probability. Instead, something that people usually talk about is relative probability.

Let’s take a look at your function. $\rho = A^2\int_{-\infty}^{\infty}dx$ This is clearly divergent. Now what we can do is look at some finite region bounded by points $a$ and $b$. The integral over that region will give us $A^2V$ where $V$ is the volume over that region. Then you divide by that region to get $A^2$.

Now remember this isn’t really a probability distribution function. It’s more of trying to make the next best thing. You are dividing what would be the (non-normalizable!) function by some enclosed volume $V$ to get a finite expression.

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Can I use $\rho(x,t)=|\psi(x,t)|^2$ for any $\psi$, regardless of whether or not it's normalizable?

No. It has to be normalizable.

If you would like to consider a plane wave as a physical/normalizable solution, then the wavefunction can be forcibly normalized by putting the system inside a "box" of length $L$. In that case the normalization condition is $$ 1 = \int_0^L dx |\psi(x)|^2 $$ and you can set $A=\frac{1}{\sqrt{L}}$ in your plane waves and the probability density is constant $$ \rho(x,t) = \frac{1}{L} $$

hft
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Any wave function that describes a particle must be normalizable. $\psi^*(\vec x)\psi(\vec x)$ is the probability density.

$$\int_{region} \psi^*(\vec x)\psi(\vec x) d \vec x$$

is the probability of finding the particle in the region. If the region is all space, the probability of finding it somewhere must be $1$.

A wave function that describes a particle can be built out of basis states that are not themselves suitable for such use. Plane waves are a common example of this.

A plane wave function cannot be normalized, except as a limiting case. The particle can be uniformly spread over all space. The probability of finding it in any finite region is $0$, The probability density at any point is $0$.

Using integration, you can add up infinite numbers of basis states with infinitesimal amplitudes in such a way that the sum is normalizable. The sum can be a wave function of a particle.

I haven't said anything new here. Maybe I have clarified things?

mmesser314
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  • So in your terminology, unlike my course notes, $\psi(x,t)=A\exp(i(kx-\omega t))$ is a 'basis state' but not a 'wave function'? – mjc Mar 19 '23 at 19:07
  • It looks like your notes use $\psi$ for either. That is what I did too. – mmesser314 Mar 19 '23 at 19:17
  • I'm thinking more of the statement "We cannot normalise the wavefunction" than of the use of $\psi$. This seems inconsistent with your "Any wave function must be normalizable.", no? – mjc Mar 19 '23 at 19:21
  • Good point. I clarified my wording. – mmesser314 Mar 19 '23 at 21:50