State in the infinite potential well

[LagrangeEuler]

General state of the infinite potential well is that $L^2[0,L]$, where $L$ is well width, or $C^{\infty}_0(\mathbb{R})$?

 

[BvU]

Please clarify your question

$\$

 

[gentzen]

Please clarify your question

I guess that he wants to know the correct Hilbert space for the solutions of the Schrödinger equations for the infinite potential well.
The tricky part is that $L^2[0,L]$ is the correct Hilbert space with respect to the scalar product, but it cannot express the boundary condition that the wavefunction is zero at 0 and at $L$. Now we could interpret it as a rigged Hilbert space and try to express the boundary condition by using $H^1_0[0,L]$ instead of $H^1[0,L]$. But I have never seen this being done, and I have no idea whether this would be correct, and whether one really cares about expressing the boundary conditions as part of the rigged Hilbert space.

 

[dextercioby]

General state of the infinite potential well is that $L^2[0,L]$, where $L$ is well width, or $C^{\infty}_0(\mathbb{R})$?

The Hilbert space is indeed $L^2 [0,L]$, but it is generally too large for arbitrary states, which are regularly in the domain of self-adjointness of the observables. $C^{\infty}_0(\mathbb{R})$ is generally good enough to describe general normalizable states.
According to the boundary conditions, one can have different realizations of the observables, either self-adjoint (case in which a Sobolev-type of space is needed) or not (for example momentum for "hard-walls").

 

[LagrangeEuler]

The Hilbert space is indeed $L^2 [0,L]$, but it is generally too large for arbitrary states, which are regularly in the domain of self-adjointness of the observables. $C^{\infty}_0(\mathbb{R})$ is generally good enough.
According to the boundary conditions, one can have different realizations of the observables, either self-adjoint (case in which a Sobolev-type of space is needed) or not (for example momentum for "hard-walls").

Thanks. Yes, for instance is it $\psi(x)=Cx^{\frac{1}{2}}(L-x)^{\frac{1}{2}}$ possible state in the well? I think that this function is $L^2[0,L]$, but it is not $C^{\infty}_0(\mathbb{R})$ function. Right? Or to rephrase is it possible to write down
$$\psi(x)=\sum^{\infty}_{n=1}C_n\psi_n(x)$$
where $\psi_n(x)=\sqrt{\frac{2}{L}}\sin \frac{n\pi x}{L}$?

 

[vanhees71]

There's no problem to Fourier transform a piece-wise continuous function. For the functions with the rigid boundary conditions $\psi(0)=\psi(L)$ the given functions $\pi_n$ are a complete set of orthonormalized functions. In this case, since there are no jumps the corresponding Fourier series converges pointwise to the function.

With Mathematica I've found the coefficients (setting $L=2 \pi$ for convenience)
$$C_n=\frac{2 \pi^{3/2}}{n} \text{J}_1(n \pi/2) \sin(n \pi/2).$$

 

[LagrangeEuler]

You did not normalize wave function. After normalization you should get that
$$C_1=\sqrt{3}J_1(\frac{\pi}{2})$$
however that sum
$$|C_1|^2+|C_2|^2+...>1$$.

 

[vanhees71]

Indeed I just used the unnormalized function $\psi(x)=\sqrt{x}(2\pi-x)$. Then $\|\psi\|^2=4 \pi^3/3$, and a numerical evaluation of $\sum_n |c_n|^2$ gives the same. There is no problem with expanding this example wrt. the energy eigenfunctions $\psi_n=1/\sqrt{\pi} \sin(n x/2)$.

Here is a plot comparing the wave function with the expansion using the first 10 and 20 (in fact only 5 an 10, because every other coefficient is 0)