Non-Normalizability

jinydu · 2008-03-27, 19:43

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

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

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

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

Any hints please?

xilman · 2008-03-27, 20:51

Quote:

Originally Posted by jinydu

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.

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

masser · 2008-03-27, 21:11

Doh! nevermind...

ewmayer · 2008-03-27, 22:09

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

jinydu · 2008-03-27, 22:33

Quote:

Originally Posted by xilman

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

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

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

Quote:

Originally Posted by ewmayer

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:

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

Also, in the definition of $\psi_2$ , 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 $L^2$ norm.

But thanks for trying.

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

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:

$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+V(x)\psi=E\psi$

The goal is to show that for any fixed value of $E$ , 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 $E$ , will be $\psi(x)=c_1\psi_1(x)+c_2\psi_2(x)$ , where $\psi_1$ and $\psi_2$ 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 $\psi_1$ is normalized, then the only way to make $\psi$ normalized is to set $c_2=0$ . This will imply the desired result.

Using reduction of order, I already know what $\psi_2$ must be (yes $V(x)$ 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 $\psi(x)=c_1\psi_1(x)+c_2\psi_2(x)$ does not decay to zero at infinity (since we know that $\psi_1$ does) whenever $c_2$ is nonzero. I will then be able to use a proof by contradiction.

2008-03-27, 19:43	#1
jinydu Dec 2003 Hopefully Near M48 2·3·293 Posts	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 $\psi_1$ be a normalized wavefunction, i.e. $\psi_1$ is a continuous complex-valued function defined on the real line and $\int_{-\infty}^{\infty}\|\psi_1(x)\|^2dx=1$ . Show that $\psi_2(x)=\psi_1(x)\int_{0}^{x}\frac{1}{\psi_1(u)^2}du$ is not a normalizable wavefunction. [Admittedly, this question has a flaw in the case when $\psi_1$ has a root. Let us ignore that issue; say $\psi_1$ has no roots]. Intuitively, my guess is that $\psi_2$ will fail to be normalizable because $\int_{-\infty}^{\infty}\|\psi_2(x)\|^2dx$ diverges to $\infty$ . This will probably be because as x goes to $\infty$ , $\int_{0}^{x}\frac{1}{\psi_1(u)^2}du$ diverges faster than $\psi_1(x)$ converges to 0. But I can't think of how to show that. Any hints please? Last fiddled with by jinydu on 2008-03-27 at 19:45

2008-03-27, 21:11	#3
masser Jul 2003 wear a mask 3214₈ Posts	Doh! nevermind... Last fiddled with by masser on 2008-03-27 at 21:34

2008-03-27, 22:09	#4
ewmayer ∂²ω=0 Sep 2002 República de California 19·613 Posts	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 \|fg\| = \|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, \|fg\| <= \|f\|\|g\|. Since \|f\| = 1 and \|fg\| diverges, \|g\| diverges as well, i.e. \|g\| = \|1/f\| is not normalizable. Now consider h(x) = Cf(x), with C := \|1/f\| = \|g\|. Taking the norm of both sides, we have \|h\| = \|Cf\| = \|\|g\|f\|. \|g\| [even though it diverges] is just a constant, so can be pulled out, thus \|h\| = \|g\|*\|f\| = \|g\|, which diverges. QED