Integer number of wavelengths after the wavefunction is normalized in a cubic box

One of the steps when learning quantum mechanics in three dimensions is to normalize $\psi$ in a cubic box of side $a$. In math terms $$\int\limits_0^a\int\limits_0^a\int\limits_0^a \psi^*\psi dxdydz=1,$$ so the normalization constant A, must be such that $$A^2 = \frac {1}{a^3}=\frac{1}{V},$$ thus the periodic boundary conditions $$\psi (x+a,y,z)=\psi (x,y,z), etc.,$$ should hold. This means only an integer number of wavelengths fit in a length $a$, right?

Now, to me these boundary conditions simply mean that at $a$ the wavelength must be zero. It can be "after" a full length or a half length. My question is, why can't there be a half integer (say 2,5; 4,5) number of wave lengths in any of the box dimensions?

• Why do you say that the wave function must be periodic? I don't see how the normalization implies that. – garyp Jul 28 '16 at 23:43
• @garyp, because of the normalization constant (please, check the update in the question). – Patrick Jul 28 '16 at 23:57
• Your argument fails to convince me. :) Check out the answer by @Gert. You'll see that the normalization has nothing to do with the periodicity. – garyp Jul 29 '16 at 1:22
• One would at least have to prove the wave function is periodic. Far from all quantum systems produce eigenstates with periodic wave functions. 'Wavy', yes, periodic: not by definition. – Gert Jul 29 '16 at 1:27

Now, to me these boundary conditions simply mean that at $a$ the wavelength must be zero. It can be "after" a full length or a half length. My question is, why can't there be a half integer (say 2,5; 4,5) number of wave lengths in any of the box dimensions?

You really need to solve the time-independent Schrödinger equation for that:

$$\Big[\frac{-\hbar^2}{2m}\nabla^2+V(x,y,z)\Big]\psi(x,y,z)=E\psi(x,y,z)$$

For a single particle in a cubic, zero potential well ($V(x,y,z)=0$) with infinitely high potential walls:

$$\frac{-\hbar^2}{2m}\nabla^2\psi(x,y,z)=E\psi$$

Reworked:

$$\psi_{xx}+\psi_{yy}+\psi_{zz}+k^2\psi=0$$

Where $k^2=\frac{2mE}{\hbar^2}$.

Solve by variable separation($*$), with Ansatz $\psi(x,y,z)=X(x)Y(y)Z(z)$ that gives three ODEs of which the first is:

$$X''(x)+k_x^2X(x)=0$$

$$\implies X(x)=c_1\cos(k_xx)+c_2\sin(k_xx)$$

With $X(0)=0 \implies c_1=0$

$$X(x)=c_2\sin(k_xx)$$ With $X(a)=0$: $$X(a)=c_2\sin(k_xa)=0$$ $$\implies k_xa=n_x\pi$$ With $n_x=1,2,3,...$ $$k_x=\frac{n_x \pi}{a}$$

For $n_x=1$, $a$ is half a wave length, for $n_x=2$, $a$ is a whole wave length, for $n_x=3$, $a$ is one and a half wave lengths, etc.

The two other ODEs for $Y(y)$ and $Z(z)$ are obtained and solved analogously, obtaining $n_y$ and $n_z$ as quantum numbers.

The integration constants $c_2$ are obtained by normalising each equation separately, e.g.:

$$\int_0^aX^2(x)dx=1\implies c_2^2\int_0^a(\sin^2k_xx)dx=1$$

$$c_2=\sqrt{\frac{2}{a}}$$

That's allowed because: $$\int_0^a\int_0^a\int_0^a[X(x)Y(y)Z(z)]^2dxdydz=\int_0^aX^2(x)dx\times\int_0^aY^2(y)dy\times\int_0^aZ^2(z)dz$$

The Normalised wave function is then:

$$\psi(x,y,z)=\Big(\sqrt{\frac{2}{a}}\Big)^3\sin\big(\frac{n_x\pi x}{a}\big)\sin\big(\frac{n_y\pi y}{a}\big)\sin\big(\frac{n_z\pi z}{a}\big)$$

The quantised energy levels are obtained by inserting the wave function into the PDE, this gives:

$$k^2=k_x^2+k_y^2+k_z^2$$

$$\frac{2mE}{\hbar^2}=\frac{\pi^2}{a^2}(n_x^2+n_y^2+n_z^2)$$ $$E_{x,y,z}=\frac{\pi^2 \hbar^2}{2ma^2}(n_x^2+n_y^2+n_z^2)$$

($*$) Insert Ansatz into PDE: $$YZX''+XZY''+XYZ''+k^2XYZ=0$$ Divide by $XYZ$: $$\frac{X''}{X}+\frac{Y''}{Y}+\frac{Z''}{Z}=-k^2$$

$$\frac{X''}{X}=-\frac{Y''}{Y}-\frac{Z''}{Z}-k^2=-k_x^2$$