Non-Normalizability
 

This question came up in quantum mechanics. Basically, it is the last step in an attempted proof that a 1-D system can never have degeneracy.

Let [tex]\psi_1[/tex] be a normalized wavefunction, i.e. [tex]\psi_1[/tex] is a continuous complex-valued function defined on the real line and [tex]\int_{-\infty}^{\infty}|\psi_1(x)|^2dx=1[/tex].

Show that [tex]\psi_2(x)=\psi_1(x)\int_{0}^{x}\frac{1}{\psi_1(u)^2}du[/tex] is [B]not[/B] a normalizable wavefunction.

[Admittedly, this question has a flaw in the case when [tex]\psi_1[/tex] has a root. Let us ignore that issue; say [tex]\psi_1[/tex] has no roots].

Intuitively, my guess is that [tex]\psi_2[/tex] will fail to be normalizable because [tex]\int_{-\infty}^{\infty}|\psi_2(x)|^2dx[/tex] diverges to [tex]\infty[/tex]. This will probably be because as x goes to [tex]\infty[/tex], [tex]\int_{0}^{x}\frac{1}{\psi_1(u)^2}du[/tex] diverges faster than [tex]\psi_1(x)[/tex] converges to 0. But I can't think of how to show that.



Any hints please?

[QUOTE=jinydu;130016]This question came up in quantum mechanics. Basically, it is the last step in an attempted proof that a 1-D system can never have degeneracy.[/QUOTE]Hmm?

It's a long time since I last did any quantum mechanics but I don't remember any such prohibition.

Consider an ensemble of bosons confined to one spatial dimension (a reasonable approximation to the intracavity state of a laser). I can't see any reason why they can't all populate the ground state or, for that matter, any other state.

Perhaps I'm misunderstanding your use of the term "1-D system".


Paul

Doh! nevermind...

The following may need to be tweaked a little because I just noticed that the definition of Psi2 involves the integral of 1/Psi1^2 for 0 to x [rather than over the whole real line], but I think the approach is sound:

Let |.| be shorthand for "norm"; in your example specifically the L^2 norm as defined over the real line.

We have f(x) such that |f| = 1.

Let g(x) = 1/f(x). From the definition of your space, we see that |f*g| = |1| [* means simple scalar multiply here] is not normalizable, since the integral of a nonzero constant over the real line diverges.

By the Cauchy-Schwarz inequality, |f*g| <= |f|*|g|. Since |f| = 1 and |f*g| diverges, |g| diverges as well, i.e. |g| = |1/f| is not normalizable.

Now consider h(x) = C*f(x), with C := |1/f| = |g|. Taking the norm of both sides, we have

|h| = |C*f| = ||g|*f|. |g| [even though it diverges] is just a constant, so can be pulled out, thus

|h| = |g|*|f| = |g|, which diverges. QED

[QUOTE=xilman;130023]
Perhaps I'm misunderstanding your use of the term "1-D system".
[/QUOTE]

Indeed. I meant a 1-D system with 1 particle.

Basically, the equation for [tex]\psi_2[/tex] comes from solving the Schrodinger equation, assuming that I already have a solution, which I call [tex]\psi_1[/tex]. I used the 'reduction of order' technique for second-order linear differential equations.

[QUOTE=ewmayer;130038]The following may need to be tweaked a little because I just noticed that the definition of Psi2 involves the integral of 1/Psi1^2 for 0 to x [rather than over the whole real line], but I think the approach is sound:
[/QUOTE]

Indeed. The integral is in fact meant to give an antiderivative. Any one will do.

Also, in the definition of [tex]\psi_2[/tex], I omitted the absolute value on purpose. It really isn't supposed to be there; that's just the way the reduction of order calculation went. So even if the integral were over the whole real line, it still wouldn't be the [tex]L^2[/tex] norm.

But thanks for trying.

EDIT: On second thought, I might need a stronger result than simply showing that [tex]\psi_2[/tex] is not normalizable. I might need to show that it is does not converge to 0 as x goes to [tex]\infty[/tex].

Here is the overall outline of the argument.

Any stationary state of a 1-particle 1-D quantum system satisfies the time-independent Schrodinger equation:

[tex]-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+V(x)\psi=E\psi[/tex]

The goal is to show that for any fixed value of [tex]E[/tex], there is only one physically realizable state, up to multiplication by a constant. I know from ODE theory that the general equation to that differential equation, for a fixed [tex]E[/tex], will be [tex]\psi(x)=c_1\psi_1(x)+c_2\psi_2(x)[/tex], where [tex]\psi_1[/tex] and [tex]\psi_2[/tex] are two linearly-independent solutions. At first, this would seem to refute my claim that there is only one allowed state; but quantum mechanics imposes an additional constraint: a physically realizable state must be normalized. My idea is to show that if [tex]\psi_1[/tex] is normalized, then the only way to make [tex]\psi[/tex] normalized is to set [tex]c_2=0[/tex]. This will imply the desired result.

Using reduction of order, I already know what [tex]\psi_2[/tex] must be (yes [tex]V(x)[/tex] cancels out in the middle of the calculation). If I can show that it does not decay to 0 at infinity, this will imply that [tex]\psi(x)=c_1\psi_1(x)+c_2\psi_2(x)[/tex] does not decay to zero at infinity (since we know that [tex]\psi_1[/tex] does) whenever [tex]c_2[/tex] is nonzero. I will then be able to use a proof by contradiction.