# Normalizing the wavefunction

1. Aug 4, 2011

### McLaren Rulez

Hi,

In Griffiths' Introduction to Quantum Mechanics, he proves an important result in the first chapter: If we normalize a wavefunction at t=0, it stays normalized at all later times. To do this, he considers the relation $$\frac{d}{dt}\int|\psi(x,t)|^{2}dx= \frac{i\hbar}{2m}[\psi^{*}\frac{\partial\psi}{\partial x} -\frac{\partial\psi^{*}}{\partial x}\psi]$$

Now the right hand side needs to be shown to be zero but this is only true if we make the assumption that $\psi$ is normalizable at all t. Griffiths says in a footnote that mathematically speaking, this is not necessary i.e. we can find functions which change from normalizable to non normalizable and still satisfy the Schrodinger Equation.

So my question is whether this sort of wavefunction has any physical meaning? Perhaps something like a particle disappearing over time?

Thank you.

2. Aug 4, 2011

### espen180

If there is absorption of particles in the system (for example by a detector), the norm of the wavefunction will not be conserved, but will decrease over time.

Unless there are non-normalizable states which are not scattering states I cannot imagine a normalizable state which changes to non-normalizable in a finite time.

3. Aug 4, 2011

### dextercioby

I don't know too much, so I can't think of another scenario but the interaction (pertubation) which would throw the initial system from a normalizable state (discrete spectrum) into a non-normalizable one.

4. Aug 4, 2011

### McLaren Rulez

I'm not familiar with the scattering and perturbation that both of you guys mentioned (think its part of my next QM course) but I understand what espen180 said about the absorption.
Isn't this a problem? If the norm of the wavefunction changes over time, the normalization factor is also a function of time. Let's call it A(t). But then, while we know that the original non normalized wavefunction $\psi$ satisfies the Schrodinger equation, what guarantee is there that $A(t)\psi$ also does?

5. Aug 5, 2011

### tom.stoer

It's all about the time evolution operator.

In general all unitary operators U acting on f as Uf conserve the norm |f|.

One can formally integrate the Schroedinger equation i.e. construct the time-evolution operator U(t). Formally this is just exp(-iHt), and b/c H is hermitean U is unitary which means it conserves the norm of the state.

6. Aug 5, 2011

### espen180

One way of modelling absorption is to choose a complex potential function. Then, H is no longer hermitian and so U(t) is no longer unitary.

7. Aug 5, 2011

### McLaren Rulez

Thank you for the replies.

As I understand it, the way we construct the time evolution operator is from the Schrodinger equation by integrating it with respect to time. So, we must first ensure that the Schrodinger equation is correct i.e. check that it is consistent with normalization. From what I see, this is not the case for the type of wavefunction that does not stay normalizable with time.

So, I don't see how we can answer this using the time evolution operator because that implicitly denies the problem of wavefunctions that don't normalize properly in the first place.

Thank you.

8. Aug 5, 2011

### Bill_K

You're given a 1-D finite square well potential V0. The spectrum of the Hamiltonian includes one or more (normalizable) bound states with E < 0, and a continuum of unnormalizable scattering states with E > 0. Suppose initially the system is in a bound state with energy E0 < 0.

At t=0, you turn on an additional time-varying potential V1 which oscillates at frequency ω. And E0 + ħω > 0, so this gives the system enough energy to escape the well to infinity. In fact since V1 includes a step function at t = 0, it contains a continuous range of frequencies, and as a result the system's energy will also have a continuous range. In terms of perturbation theory, V1 is a perturbation connecting a normalizable state with an unormalizable one, and the system has evolved from one to the other.

9. Aug 5, 2011

### Bill_K

Example 2. Let V(x,t) = kx2/2 for t < 0 and 0 for t > 0. In other words for t < 0 you have a harmonic oscillator potential. Suppose the system is initially in the ground state, so the wavefunction is a Gaussian. For t > 0 the particle finds itself in V = 0, and will evolve as a free particle. To describe this the initial Gaussian must be written as a superposition of plane waves.

Essential point: in some cases a superposition of unnormalizable functions is normalizable.

10. Aug 5, 2011

### McLaren Rulez

However, what I don't understand is this. When the wavefunction was normalizable, we could normalize it in the trivial way by putting a constant ahead of it. But now, when it changes to non-normalizable, we need to take a linear combination to make the wavefunction representing the system normalizable again, right?

How do we know that the constant we use now is the same as the one we used earlier? If it is not the same, then we have the problem of a normalization constant that is time dependent which is inconsistent with the Schrodinger equation.

Thank you very much for your help.

11. Aug 6, 2011

### McLaren Rulez

Anyone? I think I must be missing something simple since this is a basic question. Help me! Thank you.

12. Aug 6, 2011

### vanhees71

I'm not sure that I understand your problem. To make it fully clear: In the position representation the pure states of quantum mechanics are represented by a square-integrable (in the Lebesgue sense) function, i.e., the integral of the squared modulus of the wave function exists, and then by definition it's normalized to 1:

$$\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} |\psi(t,\vec{x})|^2=1.$$

There are no other functions that could represent the state of a quantum system!

What you have in the somewhat sloppy fasion of physicists handling operator theory in quantum mechanics are "eigenstates", which are in reality not true states but generalized ones, where "generalized" means "generalized functions" or distributions.

E.g. the momenum operator in the position representation (in the Schrödinger picture of time evolution) is given by ($\hbar=1$)

$$\hat{\vec{p}}=-\mathrm{i} \vec{\nabla}.$$

The generalized eigenfunctions of this operator are the plain waves,

$$u_{\vec{p}}(\vec{x})=\frac{1}{(2 \pi)^{3/2}} \exp(\mathrm{i} \vec{p} \cdot \vec{x}), \quad \vec{p} \in \mathbb{R}^3.$$

This is for sure not normalizable to 1 and thus cannot represent a state of a particle. A particle can never have a sharply determined momentum!

Now, from the theory of Fourier integrals you know that

$$\int_{\mathrm{R}^3} \mathrm{d}^3 \vec{p} u_{\vec{p}'}^*(\vec{x}) u_{\vec{p}}(\vec{x})=\delta^{(3)}(\vec{p}'-\vec{p}).$$

Now you can represent any state also in the momentum representation. This is given by

$$\tilde{\psi}(t,\vec{p})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} u_{\vec{p}}^*(\vec{x}) \psi(t,\vec{x}).$$

The inverse of this transformation is given by

$$\psi(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} u_{\vec{p}}(\vec{x}) \tilde{\psi}(t,\vec{x}).$$

With some caveats these integrals are all well defined, and the norm of the states is the same. More generally the scalar product between states is independent whether expressed in the position or the momentum representation.

All this leads to Dirac's bra-ket formalism, which has the great advantage that it formulates quantum theory independently from any representation. This makes the structure and physical content of quantum theory much more transparent than the old wave-mechanics approach, which is nothing else than quantum theory in position representation.

13. Aug 7, 2011

### McLaren Rulez

Okay my question is basically a problem raised in the very first chaper of Griffiths' Introduction to Quantum Mechanics.

If we normalize a wavefunction, how do we know that it stays normalized in the future? Renormalizing it repeatedly would imply that the normalization factor is a function of time, and that is inconsistent with the Schrodinger Equation.

To prove that once we normalize a wavefunction, it stays normalized, Griffiths uses a little bit of mathematical manipulation to arrive at the relation mentioned in the OP.

$$\frac{d}{dt}\int|\psi(x,t)|^{2}dx= (\frac{i\hbar}{2m}[\psi^{*}\frac{\partial\psi}{\partial x} -\frac{\partial\psi^{*}}{\partial x}\psi])|^{\infty}_{\infty}$$ The limits of that integral on the left are minus infinity and plus infinity.

In order for $$\int|\psi(x,t)|^{2}dx = 1$$ for all t, we must have $$\frac{d}{dt}\int|\psi(x,t)|^{2}dx= (\frac{i\hbar}{2m}[\psi^{*}\frac{\partial\psi}{\partial x} -\frac{\partial\psi^{*}}{\partial x}\psi])|^{\infty}_{\infty}=0$$ Griffiths remarks that this relation is indeed always zero since normalizable (not necessarily normalized but normalizable) wavefunctions always go to zero at infinity, meaning that the middle term is guaranteed to be zero.

When a wavefunction evolves from a normalizable one into a non-normalizable one (as in Bill_K's examples), how do we deal with it since the last condition is no longer true? This is my question.

Bill_K, you mentioned that we could take a superposition of these non-normalizable states and that superposition would still be normalizable, but I don't understand how we can do that. Let's say, I have states $\psi_{i}(x,0)$ which are all normalizable solutions to the initial conditions. Now let these states evolve to $\psi_{i}(x,t)$ and these are non normalizable. I would then take some superposition and call that the normalizable wavefunction for the system, say $\psi(x,t)=\sum c_{i}\psi_{i}(x,t)$ But then, there is no reason why this specific superposition is normalizable at t=0. That is

$\psi(x,0)=\sum c_{i}\psi_{i}(x,0)$

may not be normalizable. So I don't see how taking superpositions helps get around this problem unless you can prove that there is a set of $c_{i}$ that can do the job for both t=0 and any other t.

Thank you very much for the help.

14. Aug 9, 2011

### McLaren Rulez

Sorry to bump this up again, but does anyone know the answer?

15. Aug 9, 2011

### dextercioby

Well, it keeps its initial normalization, iff the Hamiltonian is time-independent in the Schrödinger picture (in which essentially the Hamiltonian is, if it wasn't, we could treat it as a perturbation). A time-independent Hamiltonian is generated by a unitary time-evolution which is known to conserve the scalar products, hence the norm of vectors in an ordinary Hilbert space.

There's no need to worry about the normalization, as long as the system is isolated, of course (an H-atom is an example). When interactions occur, then the system will have a different, usually time-dependent Hamiltonian which can have non-normalizable eigenstates. As the matter of fact, the new time-dependent Hamiltonian's eigenstates cannot be determined exactly and neither can we exactly determine how these functions evolve in time.
The purpose of perturbation theory is to give us approximations to transitions probabilities, but these transitions are actually between the energy levels of the unperturbed system, i.e. the one with the time-independent Hamiltonian.

16. Aug 10, 2011

### McLaren Rulez

Okay, thank you very much for all your help :)

17. Sep 1, 2011

### mmmboh

For this to be true, it must be assumed that the x derivative of the wave function is bounded near +\- infinity, or at least grows slower than the wave function goes to zero....how do we know this? Can we not have a wave function that goes to zero but oscillates quicker and quicker as x grows, or something of the sort?

18. Sep 2, 2011

### tom.stoer

This is a common missconception, and in general it is wrong!

You can construct normalizable wave functions with equidistant peaks, where the width of the peaks goes to zero whereas the height of the peaks increases (goes to infinity)! This clearly contradicts Griffiths' remark and proves that it's incorrect (afaik these counterexamples are physically irrelevant).

Last edited: Sep 2, 2011
19. Sep 2, 2011

### Galron

It generally goes to show that relying on mathematical axioms as hard science is fraught with danger. Ie axioms do not science make.

Normalising reality is akin to saying I do not believe in chaos... Or magic so any fairies must not exist by law.

Incidentally I started a thread in the philosophy section on this very question. In case you are wondering why the derail.

20. Sep 2, 2011

### BruceW

It was my understanding that the wavefunction going to zero at +/- infinity was part of the postulates of QM. So even though there are normalisable wavefunctions that do not go to zero at infinity, we are not allowed to choose them because the postulates of QM do not allow us.