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This question regards what is stated at page 65, Goodstein, States of Matter.
We want to consider a free particle in one dimension, with movements restricted to a line of lenght $L$; so in practice we have a particle in an infinite square well, am I right? Well, it is then stated that the energy and the wave vector of the particle, in this situation, are quantized; ok, no problem here.
But: then the book goes on stating that: $$k=\frac{2\pi}{L}n \ \ \ ; \ \ \ n=0,\pm 1 , \pm 2 , ... \tag{1}$$ This seems wrong to me. In fact we have a particle with hamiltonian: $$H=\frac{\hbar ^ 2 k^2}{2m}$$ It's wave function should be: $$\psi (x)=Ae^{ikx}+Be^{-ikx}$$ but it's in an infinite square well, with walls at 0 and at $L$, for example, so $\psi(x)$ should be zero at $x=0$ and $x=L$, so $\psi(x)$ (using Euler Formula) should be a $\sin$, to have it be equal to zero in $x=0$: $$\psi(x)=C\sin(kx)$$ now the behaviour of $\psi(x)$ in $x=0$ is all good. We now have to impose that $\psi(L)=0$, so: $$Csin(kL)=0 \ \ \Rightarrow \ \ kL=\pi n$$ because of course the $\sin$ is zero only when its argument ($\theta$) is equal to: $\theta = 0,\pm\pi,\pm2\pi,...$ So we get: $$k=\frac{\pi}{L}n \ \ \ ; \ \ \ n=0,\pm 1 ,\pm 2 ,...$$ So (1) is wrong! But: this way (1) of writing the quantization of the wave vector is present not only in the Goodstein but also in many of my lecture notes, so I am starting to think that I must have made a mistake somewhere and that (1) is indeed correct..
I have heard that maybe all this has something to do with PBC (Periodic Boundary Condition), but Goodstein doesn't talk about this topic and I have no idea on what those are and why they should matter..

Qmechanic
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Noumeno
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1 Answers1

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Periodic boundary conditions state $$\psi(x)=\psi(x+L)\tag{1'}.$$ Goodstein says "[...] it (the particle) has equal probability of being anywhere on the line", which would not be true if we take $\psi(0)=\psi(L)=0$. You can see that with the condition ($1'$), $k$ takes the values that he claims.

Edit:

In this particular case, I don't know what are the motivations to consider PBC appart from what I already said, that Goodstein is asking the particle to be anywhere on the line with the same probability. More information about its use here and here.

As OP says, the wave function is

$$\psi(x)=Ae^{ikx}+Be^{-ikx}.$$

Imposing $(1')$, we have

$$Ae^{ikx}+Be^{-ikx}=Ae^{ikx}e^{ikL}+Be^{-ikx}e^{-ikL}$$ $$Ae^{ikx}\Big(1-e^{ikL}\Big)+Be^{-ikx}\Big(1-e^{-ikL}\Big)=0\tag{2'}.$$

As $e^{ikx}$ and $e^{-ikx}$ are independent functions, if we want ($2'$) to hold for any $A$ and $B$,

$$e^{\pm ikL}=1\,\,\Rightarrow\,\,k=\frac{2\pi}{L}n\,\,\,\text{with }n=0,\pm1,\pm2,...$$

AFG
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