# Normalization of Wave-Function

1. Jun 12, 2009

### Sparky_

In attempting to work through some basics of QM –

I have a question regarding a statement or a conclusion regarding “Normalizing the Wave Function”

After “turning the crank” authors show:

$$\frac {d}{dt} \int_ {-\infty}^{\infty}|\psi|^2 dx= \frac{ih}{2m}(\psi*\frac{d\psi}{dx} - \frac{\psi*}{dx}\psi )$$

I can mathematically get to this –

But I don’t “see” statements that follow this result.

Griffiths states, “But $$\psi$$ must go to zero as x goes to infinity."

A web search found this statement:
$$\frac {d}{dt} \int_ {-\infty}^{\infty}|\psi|^2 dx= \frac{ih}{2m}(\psi*\frac{d\psi}{dx} - \frac{\psi*}{dx}\psi )$$

“The above equation is satisfied provided $$|\psi|$$ goes to zero as $$|x|$$ goes to zero."
Can you provide clarifying information on these statements / conclusions.

I don’t understand the conclusion being drawn – it’s not intuitive to me where I could make the statement the authors make.

Last edited: Jun 12, 2009
2. Jun 12, 2009

### Cyosis

You know that $| \psi (x) |^2$ is the probability density function. If the wave function would not go to zero in +-infinity then the area below the graph of the probability density function would be infinite, instead of 1. We cannot allow the total probability to exceed 1 for physically realizable states.

Edit: something seems to be wrong with latex in general currently.

Last edited: Jun 12, 2009
3. Jun 12, 2009

### Nick89

For the first statement, if a wavefunction does not go to zero for $x = \pm \infty$, then it is not 'square integrable' and non-normalizable. This is easy to see if you try to normalize such a wavefunction. For example, suppose we have a wavefunction of the following form:
$$\psi (x) = Ax^2$$
Note that it does not go to zero for x to +/- infinity.

Now try normalizing it (which means it's absolute square, integrated over all space, should be 1, or in other words, the probability of finding your particle somewhere should be 100%):
$$\int_{-\infty}^{\infty} |\psi (x)|^2\, dx = \int_{-\infty}^{\infty} |Ax|^2 \, dx = |A|^2 \int_{-\infty}^{\infty} x^2 \, dx = |A|^2 \cdot \infty$$
This should be equal to one for physically realizable states, but it is impossible. You cannot choose any value for A that makes the integral 1.

In other words, the integral must converge, and for that to happen, the wavefunction must go to zero at +/- infinity.

We might have a different print of Griffiths, but I cannot find the second conclusion in bold in my book, nor does it make much sense to me.

4. Jun 12, 2009

### Bob_for_short

Think of psi as of a bound state: an oscillator or atomic wave function. Then psi vanishes at infinity. Now take a superposition of bound states. The integral will be the sum of integrals with different psis, each vanishing at infinity. Even in a large but finite box (plane waves) you have psi vanishing at infinity. So if there is no decay, the normalization of psi(t) is time independent.

Bob.

5. Jun 12, 2009

### muppet

There's one further point that can be made that's specific to this example, which is that the term in brackets is the probability current density (wiki it for precise details). In the formula above you have a definite integral (i.e. a number) equal to a function, so you need to evaluate the function at plus/minus infinity. For a wavefunction that vanishes at these boundaries- the physical significance of which has already been discussed- then this number is zero. What's more interesting is when you integrate over a finite interval (a,b), in which case you get a measure of how the probability of finding the particle within this region changes with time. In this case, the importance of the wavefunction vanishing at the boundaries is that it has a probabilistic interpretation at all, and it is assumed that psi is normalised.

6. Jun 15, 2009

### Sparky_

Nick89 -
The first quote (as you know) is from Griffiths)

Sorry - the second was found on a professor's web notes - University of Texas I think - where I was searching for further help on this topic.

Nick89, Cyosis, Bob_for_short, muppet and others -

follow-up question -

I agree that if you "know" that $$|\psi|^2$$ is a probability density (or behaves like a probability density - I know wording can be sticky) then the area under the curve has to be 1.0.

- If you didn't know how $$[|\psi|^2]$$behaves, could you make the statements:

“But $$|\psi|$$ must go to zero as x goes to infinity." - Griffiths

“The above equation is satisfied provided goes to zero as goes to zero." - some professor's notes

by seeing this alone: $$\frac {d}{dt} \int_ {-\infty}^{\infty}|\psi|^2 dx= \frac{ih}{2m}(\psi*\frac{d\psi}{dx} - \frac{\psi*}{dx}\psi )$$

Meaning by analysis alone - does the math tell me that “But $$|\psi|$$ must go to zero as x goes to infinity." (without knowing the probability density behavior)?

Does my question make sense? - I don't see how the statement can be made "by inspection of the resulting "derivation".

Thanks

7. Jun 15, 2009

### Nick89

I think the result (the fact that normalized wavefunctions stay normalized for all time) is only valid for wavefunctions that go to zero at +/- infinity. If your wavefunction does not have this property, it is not a physically realizable state, and it might behave differently.

So the fact that $\psi \to 0$ for $x=\pm \infty[/tex] is taken as 'known'. It does not follow from the math, but rather, the math follows from it. IF [itex]\psi$ goes to zero, THEN it stays normalized for all time, because the right hand side of your equation vanishes:
$$\frac{d}{dt}\int_{-\infty}^{\infty} |\Psi|^2\, dx = \frac{ih}{2m} \left[ \Psi^* \frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x} \Psi \right]_{-\infty}^{\infty} = \frac{ih}{2m} \left( 0 \times \frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x} \times 0 \right) = 0$$
So this tells you that the integral over all space of the absolute square of the wavefunction does not change in time, and hence, if it is normalized, it stays normalized.