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I am working on fermions in metal, where you need

$\frac{2}{V}\int_{0}^{P_{f}}\frac{L^{3}}{(2\pi\hbar)^{3}}dP$

to get the total number of occupied states, where the coefficient $\frac{L^{3}}{(2\pi\hbar)^{3}}$ comes from the periodic boundary condition that $f(x+L)=f(x)$.

However, I am confused with the two conditions:

  1. $kL = n\pi$, which is the condition you have a stable solution in a cavity, which I think should be the same case you have in the above question.
  2. $kL = 2n\pi$, which come from $f(x+L)=f(x)$.

There is a fact of 2 difference. But I am confused that is the difference between them? where should we use condition 1 and where we should use condition 2?

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    $\begingroup$ Both prescription should give rise to the same macroscopic limit of course. The key thing is that, for PBC, $n$ can take negative values while for Hard boundary conditions, $n$ must be positive. This, in turn, affects the density of state and the energy. See for example my edit here: physics.stackexchange.com/a/787856/232790 and the answer to this: physics.stackexchange.com/q/448882/232790 $\endgroup$
    – Syrocco
    Commented Nov 20, 2023 at 22:50
  • $\begingroup$ The first boundary condition assumes the wavefunctions form standing waves in a cavity, i.e, the boundaries are nodes of the wavefunction. The second boundary condition does not make such an assumption. $\endgroup$ Commented Nov 21, 2023 at 13:03
  • $\begingroup$ @Syrocco Nice explanation! I understand now. I have a following question for you. Why in hard wall boundary condition, you write the wavefunction as sine function, but in periodic boundary condition you write it in complex form? Could we write the wavefunction in hard wall boundary condition also in complex form? Or because we actually have the hard wall boundary condition as $f(0) = f(L) = 0$? $\endgroup$
    – QFT
    Commented Nov 22, 2023 at 17:14
  • $\begingroup$ @QFT i added an answer since it was taking a bit too much place in the comment. Tell me if it answered your questions $\endgroup$
    – Syrocco
    Commented Nov 22, 2023 at 20:32

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As said in the comment, different boundary conditions will not change the macroscopic physics but will change by a factor 2 the energy and the computation of the density of state. Due to the fact that with PBC you can also have negative quantum number. I put again the two links for more informations: 1, 2 and 3!

For your question in the comments: indeed, with hard walls, we require that $f(0)=f(L)=0$ while with PBC, the condition is a bit looser: $f(x)=f(x + L)$. In both cases, the time independent Schrodinger equation is solved by the wavefunction: $\psi(x)=Ae^{ikx}+Be^{−ikx}=C\cos(kx)+D\sin(kx)$. You can chose to represent the wavefunction using sin/cos or exponential, it's the same!

Once you have this proposition of solution (for the particle inside the box), the only thing left to do is imposing the boundary condition.

HBC:

$\psi(L) = \psi(0)\Rightarrow k = n \pi/L$ and $\psi(0) = 0\Rightarrow C = 0$ or equivalently $A = -B$. Thus $\psi(x) \propto \sin(\pi n x/L)$. So it can't be written as a complex form, except the trivial $\psi(x) \propto e^{i\theta}\sin(kx)\propto e^{i\theta}(e^{ikx}-e^{-ikx})$

PBC: $\psi(L) = \psi(x + L)\Rightarrow k = 2n \pi/L$ and that's all. Thus $\psi(x) \propto Ae^{i2\pi nx/L}+Be^{−i2\pi nx/L}=C\cos(2\pi nx/L)+D\sin(2\pi nx/L)$ where you can freely chose $A$ and $B$ or $C$ and $D$. Nonetheless, in most cases, we don't bother with these two solutions and simply write $\psi_n(x) = Ae^{i2\pi nx/L}$.

If you want to get back the "full" solution found when solving the Schrodinger equation$^{1}$, you can simply write: $\psi(x) = \psi_{-n}(x)+\psi_{n}(x)$. This is still an eigenfunction of the Schrodinger equation for the energy $\displaystyle E_{n}={\frac {n^{2}\pi ^{2}\hbar ^{2}}{2mL^{2}}}$. Physically, this is because the term in $-n$ describe a particle with a negative momentum and the term in $n$ describe a particle with same momentum but this time, positive. But the energy of these two particles is the same. Thus,

$^{1}$: The full solution is off course a linear combination of every eigenfunction... Don't get fooled by my misleading language. Here I'm talking about the most general eigenfunction with a given eigenvalue!

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  • $\begingroup$ Excellent. Thanks Syrocco! $\endgroup$
    – QFT
    Commented Nov 27, 2023 at 17:06

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