Shankar on constraints and free parameters for a particle in a box

[kmm]

On page 160 in Shankar, he discusses how we get quantized energy levels of bound states - specifically for the particle in a box. We have three regions in space; region I from $\ - \infty, -L/2$, region II from $\ -L/2, L/2$, and region III from $\ L/2, \infty$. For the wavefunction in region I and II, we choose the coefficient of the rising exponential to be zero to get an admissible solution, leaving one free parameter for the falling exponentials in each region. In region II, the wavefunction is a sum with a sine and cosine, with a coefficient for each. This gives a total of four free parameters. Now, at each interface between the three regions at $\ \pm L\2$, for a finite potential V, we require continuity of the wave function and its derivative. Thus, we impose four constraints on the wave function.

What confuses me is that Shankar goes on to say "Thus we impose four conditions on $\psi$ which has only three free parameters. (It may seem that there are four-the coefficients of the two falling exponentials, the sine, and the cosine. However, the overall scale of $\psi$ is irrelevant both in the eigenvalue equation and the continuity conditions, these being linear in $\psi$ and $\psi '$. Thus if say, $\psi '$ does not satisfy the continuity condition at $\ x=L/2$, an overall rescaling of $\psi$ and $\psi '$ will not help.)

It makes sense to me that overall scale doesn't matter in the eigenvalue equation and the continuity conditions, but I don't see how that takes us from apparently four free parameters to three free parameters. That there is one more constraint than there are free parameters is apparently important, because after this Shankar considers a general potential V(x) that binds a particle of energy E, where he slices space into many intervals, and after counting up the total parameters claims that we have one more constraint than we have parameters.

I also don't understand the full significance of having more constraints than free parameters. I don't see what the issue would be if we had more free parameters than constraints.

I appreciate any help!

 

[vanhees71]

I'd not formulate it in this way. There are four free parameters. Three of these parameters are fixed by the conditions derived from the necessity to have the Hamiltonian to be self-adjoint and that the states are normalizable (bound states) or normalizable to a "$\delta$ distribution" (scattering states). This leads to the continuity conditions at the singularities of the potential and that at infinity you must either have a exponential function falling to zero (for $x \rightarrow \pm \infty$) or being oscillating, i.e., going like $\exp(\pm \mathrm{i} k x$ with $k \in \mathbb{R}$. The fourth parameter is fixed by the normalization condition.

 

[kmm]

I'd not formulate it in this way. There are four free parameters. Three of these parameters are fixed by the conditions derived from the necessity to have the Hamiltonian to be self-adjoint and that the states are normalizable (bound states) or normalizable to a "$\delta$ distribution" (scattering states). This leads to the continuity conditions at the singularities of the potential and that at infinity you must either have a exponential function falling to zero (for $x \rightarrow \pm \infty$) or being oscillating, i.e., going like $\exp(\pm \mathrm{i} k x$ with $k \in \mathbb{R}$. The fourth parameter is fixed by the normalization condition.

Thank you, I think this helps clarify things for me. The way I understand it then is, since the fourth parameter is fixed by the normalization condition, and since the overall scale of $\psi$ is irrelevant to the continuity conditions and therefore, to the fact that we get quantized energy states for a bound particle, it is only the three free parameters that are relevant since they are constrained by the continuity conditions. I assume this is why Shankar referred to there being only three free parameters. Does my reasoning seem correct?

But I still don’t understand why he makes the point that there is one more constraint than free parameters. It seems to me that if we have as many constraints as parameters, then we can determine the parameters. Wouldn’t that be sufficient?

 

[vanhees71]

Since the Schrödinger equation is a linear homogeneous partial differential equation for $\psi(t,\vec{x})$, any solution is only determined up to the overall normalization, i.e., you need the normalization condition to fix the overall factor. The same holds for the energy-eigenfunctions ("time-independent Schrödinger equation").

 

[kmm]

Since the Schrödinger equation is a linear homogeneous partial differential equation for $\psi(t,\vec{x})$, any solution is only determined up to the overall normalization, i.e., you need the normalization condition to fix the overall factor. The same holds for the energy-eigenfunctions ("time-independent Schrödinger equation").

Ah right, of course. Thank you for helping me with this!