# Wave Function Zero At Infinity?

1. Jun 27, 2008

### GAGS

Its looking quite simple problem but let me explain properly my question.
Wave function as we know is also known as matter wave/field amplitude. Then definitely there is associated a wave with it. Then how can we say that wave amplitude vanished at infinite!

2. Jun 27, 2008

### Fredrik

Staff Emeritus
Because if it isn't,
$$\int |\psi(x)|^2 dx$$
is infinite.

3. Jun 27, 2008

### malawi_glenn

GAGS, for a bound system, wavefunction must go to zero due to the reason Fredrik told you. This is not applicable to free particles.

Recall that the wavefunction is the probability amplitude for finding the particle at a certain location.

4. Jun 27, 2008

### Hurkyl

Staff Emeritus
That doesn't follow. Consider the function:

$$\psi(x) = \begin{cases} 0 & x < 1 \\ 1 & x \in [n, n + n^{-2}) \\ 0 & x \in [n + n^{-2}, n+1) \end{cases}$$

where n ranges over all positive integers. $\psi(x)$ does not converge to zero at $+\infty$. However,

$$\int_{-\infty}^{+\infty} |\psi(x)|^2 \, dx = \sum_{n = 1}^{+\infty} \int_{n}^{n + n^{-2}} 1 \, dx = \sum_{n = 1}^{+\infty} n^{-2} = \pi^2 / 6$$

5. Jun 27, 2008

### malawi_glenn

Hurkyl, is that a continous function? Is that a function an eigenfunction to the Shrodinger equation?

6. Jun 27, 2008

### Hurkyl

Staff Emeritus
It's a wave function, which is enough to demonstrate Fredrick's claim is inadequate.

Doesn't have to be. However, I expect that if I try the same trick with Gaussians, I can construct a potential so that my counterexample is a stationary state.

7. Jun 27, 2008

### malawi_glenn

That is why I added that a bound wave function must approach zero as x goes to infty.

So wave functions does not have to be eigenfunctions to Shrodinger Eq? (in non rel QM)

8. Jun 27, 2008

### Hurkyl

Staff Emeritus
Nope -- only stationary states are eigenfunctions. However, the (generalized) eigenfunctions do form a basis, so that each wavefunction is a (possibly infinite or continuously indexed) linear combination of (generalized) eigenfunctions.

9. Jun 27, 2008

### malawi_glenn

I would really love to see a physical situation where that wavefunction comes out.

10. Jun 27, 2008

### Fredrik

Staff Emeritus
Hurkyl's argument proves that my argument doesn't hold. His argument is valid even if it isn't possible to design an experiment which produces his wave function.

I don't know how to respond. I don't know which wave functions are "valid" and which ones aren't.

11. Jun 27, 2008

### malawi_glenn

Generally that statement of yours doesn't hold, but we must look at the physical situation before making any imposings on the wave function under consideration.

12. Jun 27, 2008

### Domnu

Well, I'm sure if you generated a potential of some sort, you could model the above situation?

13. Jun 27, 2008

### per.sundqvist

You could have eigenvalue problems with compex eigen vaues, which means that the wave outside a meta stable bound region is of traveling wave type (and not equal to zero at infinity). They repesents decay of the particle probablility in time, i.e., the imaginary part of the energy estimates the lifetime of an electron in a metastable state. For example: $$V(x)=A\cdot x+B\cdot \sin(kx)$$.

14. Jun 27, 2008

### ismaili

Sorry I don't quite understand it.
That's why we have complex eigenvalues?

BTW, I have another question.
I know the resonance state corresponding to the complex energy poles
of the scattering matrix.
And the Hamiltonian is Hermitian. (I'm not quite sure?)
Are these complex energy poles here eigenvalues of Hamiltonian?
Or, the complex energy poles are not eigenvalues of the Hamiltonian?
Thanks for any instructions.

15. Jun 27, 2008

### Andy Resnick

Hey... that's cool. Are those functions used for anything?

16. Jun 27, 2008

### Andy Resnick

I think it's required for bound particles, to ensure they stay bound. As Hurkyl points out, there are square-integrable functions that are mischevious, but AFAIK, they do not correspond to physically relevant potentials and Hamiltonians.

17. Jun 28, 2008

### per.sundqvist

No the Hamiltonian is hermitean, but the bondary conditin is of open type. The proof why you get real eigenvalues fails, using Greens first identity when yo get: $$\int\Phi\nabla\Phi\cdot d\vec{S}\neq 0$$. The BC n 1D is: $$d\Psi/dx+ik\Psi=0$$.

18. Jun 28, 2008

### Staff: Mentor

It is indeed applicable to free particles. You're probably thinking of the plane-wave solution $\Psi = \exp [i(px - Et)/\hbar]$ which extends to infinity. But that isn't a physically valid wave function, precisely because it isn't square-integrable. To get a physically valid wave function for a free particle, you have to superpose a collection of waves like that, with different wavelengths, via a Fourier integral. This gives a free-particle wave function that goes to zero as x goes to infinity; and the packet width $\Delta x$ and range of momenta $\Delta p$ satisfy the Heisenberg Uncertainty Principle!

19. Jun 28, 2008

### per.sundqvist

No the hamiltonian is hermitean, but the booundary condition is of open type. The proof of real eigen values fails, using greens first identity, when: $$\int \Psi\nabla\Psi\cdot d\vec{S}\neq 0$$. The BC for this type of problems in 1D is: $$\hat{n}\cdot\nabla\Psi+ik\Psi=0$$.

20. Jun 28, 2008

### Hurkyl

Staff Emeritus
But if you consider a multi-modal wavefunction, such as the one I presented (or a smoothed version of it)....

21. Jun 28, 2008

### ismaili

I don't quite understand it.
We can prove the theorem which states the eigenvalues of a Hermitian operator are real from linear algebra. There is no additional condition for the boundary conditions of the eigenstates. For example, from
$$A|a'\rangle = a'|a'\rangle$$ and $$\langle a''|A = a''^*\langle a''|$$
where $$A$$ is an Hermitian operator and $$a',a''$$ are its eigenvalues.
We times the first equation with $$\langle a''|$$, the second equation with $$|a'\rangle$$, then substract,
$$\Rightarrow (a' - a''^*)\langle a''|a'\rangle = 0$$
now we select $$a' = a''$$, then we conclude that $$a'$$ is real.
So, eigenvalues of a Hermitian operator must be real.
How come the resonance state has complex energy eigenvalues?
My idea is that the complex energy poles of S-matrix corresponding to resonance states are not energy eigenvalues of Hamiltonian, so the complex energy is not the energy of the resonance state.
Where did I got lost? thx

22. Jun 28, 2008

### per.sundqvist

Hmm, here is my derivation:

$$\varphi^*H\varphi &=& \lambda\varphi^*\varphi$$

$$\varphi H^*\varphi^* &=& \lambda^*\varphi\varphi^*$$

Using hermiteacity:
$$H=H^*=-\nabla^2$$

Now integrating the difference, and using Greens second identity (not the first as I wrote):
$$\begin{eqnarray} (\lambda-\lambda^*)\int\varphi^*\varphi dv &=& -\int(\varphi^*\nabla^2\varphi-\varphi\nabla^2\varphi^*)dv= \nonumber \\ &=& -\oint_{\partial S}(\varphi^*\nabla\varphi-\varphi\nabla\varphi^*)\cdot d\vec{S} \neq 0 \end{eqnarray}$$

where dS is the boundary surface, ie two points in 1D. If you have $$\varphi=exp(ikx)$$ you se that the difference is not equal to zero. But if $$\varphi=0$$ at infinity, then $$\lambda=\lambda^*$$, giving a real eigen value.

23. Jun 28, 2008

### per.sundqvist

The last term in the equation tells you that a quantum current is comming out from the boundary. The simplest eigen value problem is for instance the following:

$$$V(r) = \left\{ \begin{array}{l l l} \infty & r<0\\ 0 & 0\leq r< L\\ V_0 & L\leq r< L+t\\ 0 & r\geq L+t\\ \end{array} \right.$$$

You could write the solution in terms of unknown coeffs and the k as:

$$$V(r) = \left\{ \begin{array}{l l l} \Psi_1 &=& A\sin(kx) ;\;0\leq r< L\\ \Psi_2 &=& B\exp(\kappa x)+C\exp(-\kappa x) ;\; L\leq r< L+t\\ \Psi_3 &=& t\exp(ikx) ;\; r\geq L+t\\ \end{array} \right.$$$

Matching functions and derivatives at the two intermediate points gives you a system like: $$M\vec{c}=0, det[M(E)]=0, k=\sqrt{\hbar^2E/2m}, \kappa=\sqrt{\hbar^2(V-E)/2m}$$. solving the determinant eqation in E numerically you will find an approxiamte solution like: $$E_n\approx\frac{\hbar^2}{2m}\left(\frac{n\pi}{L}\right)^2+i\varepsilon$$.

The imaginary part will contribute in the time-dependent solution as:
$$\mid\Psi\mid^2=\mid\psi\mid^2e^{i(i\varepsilon)t/\hbar}=\mid\psi\mid^2e^{-t/\tau} ;\;\tau=\hbar/\varepsilon$$