# Griffiths 1.5: Normalization of a wave function

1. May 24, 2014

1. The problem statement, all variables and given/known data

Consider the wave function $$\Psi(x,t)=Ae^{-\lambda|x|}e^{-i\omega t}$$
Where $A$, $\lambda$, and $\omega$ are positive real constants.

(a)Normalize $\Psi$.
(b)Determine expectation values of $x$ and $x^2$.

2. Relevant equations
$$\int_{-\infty}^{+\infty}|\Psi(x,t)|^2dx=\int_{-\infty}^{+\infty}\Psi\Psi^{\ast}dx=1$$

3. The attempt at a solution

Correct me if I'm wrong, but in order to normalize the wave function you need to re-write A in terms of the other constants and $x$, and normalize it at t = 0. This is what I did:

$$\int_{-\infty}^{+\infty}|\Psi(x,0)|^2dx$$
$$|A|^2\int_{-\infty}^{+\infty}e^{-\lambda^2|x|^2}dx$$
$$|A|^2\frac{\sqrt{\pi}}{2\lambda}=1$$
$$A=\frac{\sqrt{2\lambda}}{(\pi^{1/4})}$$

Did I go wrong somewhere in (a) ?

2. May 24, 2014

### micromass

In this step, did you do

$$(e^x)^2 = e^{x^2}$$

because that would be incorrect.

3. May 24, 2014

I did, I should have payed more attention to the math;

Correcting my mistake:
$$\int_{-\infty}^{+\infty}|\Psi(x,0)|^2dx=|A|^2\int_{-\infty}^{+\infty}e^{-2\lambda|x|}dx$$
$$|A|^2=\frac{e^{-2\lambda|x|}}{-2\lambda}$$
From negative infinity to positive infinity.
$$|A|^2=\frac{1}{2\lambda}$$
$$|A|=\sqrt{\frac{1}{2\lambda}}$$

Is this correct? Or did I just formally prove the fact that I suck at integration?

4. May 24, 2014

### Saitama

I do not see how you get that. Notice that:

$$\int_{-\infty}^{+\infty}e^{-2\lambda|x|}dx=2\int_0^{\infty} e^{-2\lambda x}dx$$
Can you integrate this?

5. May 24, 2014

### micromass

This cannot be correct, the formula for $|A|$ can not contain an $x$.

You need to integrate

$$\int_0^{+\infty} e^{-2\lambda x}dx + \int_{-\infty}^0 e^{2\lambda x}dx$$

6. May 24, 2014

That was what I did, and I got:
$$-\frac{e^{-2\lambda x}}{2\lambda}|_0^{+\infty}+\frac{e^{2\lambda x}}{2\lambda}|_{-\infty}^0$$
and that is:
$$-\infty-(-\frac{1}{2\lambda})+\frac{1}{2\lambda}-\infty$$
Which means that
$$A=\sqrt{\frac{1}{\lambda}}$$
because the wave function disappears at plus or minus infinity.

7. May 24, 2014

### micromass

Not a fan of that notation. You shouldn't use infinite that way.

You know the integral you need to find is
$$\int e^{-2\lambda x}dx = -\frac{e^{2\lambda x}}{2\lambda} + C$$

So indeed, the integral I want you to find is
$$-\frac{e^{-2\lambda x}}{2\lambda}|_0^{+\infty}+\frac{e^{2\lambda x}}{2\lambda}|_{-\infty}^0$$

But this is equal to

$$\lim_{x\rightarrow +\infty} -\frac{e^{-2\lambda x}}{2\lambda} + \frac{e^{-2\lambda 0}}{2\lambda}+\frac{e^{2\lambda 0}}{2\lambda} - \lim_{x\rightarrow -\infty}\frac{e^{2\lambda x}}{2\lambda}$$

Both limits are $0$ if $\lambda>0$. Thus the above expression is $1/\lambda$

You have now calculated

$$\int_{-\infty}^{+\infty} |\Psi(x,0)|^2 dx = |A|^2 \int_{-\infty}^{+\infty} e^{-2\lambda |x|}dx = \frac{|A|^2}{\lambda}$$

You want this to be equal to $1$, so what should $|A|$ be?

8. May 24, 2014

Now I see where I was wrong:
$$\int_{-\infty}^{+\infty}e^{-2\lambda|x|}dx=2\int_0^{\infty} e^{-2\lambda x}dx=2|A|^2\int_{-\infty}^{+\infty}e^{-2\lambda|x|}dx=0-(-\frac{1}{2\lambda})$$
So:
$$|A|=\sqrt{\lambda}$$

9. May 24, 2014

### micromass

Right, this is the right result.

10. May 24, 2014

That was the dumbest mistake I made this year. I set A equal to 1 over lambda when I should have multiplied them and set them equal to one. Thanks for the help, though.

11. May 24, 2014

### micromass

12. May 24, 2014

Working them out now..

13. May 24, 2014

When solving for expectation values, is t set to zero as well? Is it so it is possible to integrate with respect to x?

14. May 24, 2014

### micromass

It doesn't matter whether you set $t$ equal to $0$ here. You need to take the modulus of $\Psi(x,t)$ anyway, and

$$|\Psi(x,t)| = |A| e^{-\lambda|x|} |e^{i\omega t}| = |A| e^{-\lambda |x|}$$

So the terms with $t$ drop anyway.

15. May 24, 2014

$$\langle x\rangle=\int_{-\infty}^{+\infty}x\lambda e^{-2\lambda|x|}dx$$
Thus:
$$\langle x \rangle=\left(-\frac{e^{-2\lambda x}}{4\lambda}|_0^{+\infty}\right)+\left(\frac{e^{2\lambda x}}{4\lambda}|_{-\infty}^0\right)$$
$$\langle x\rangle=\frac{1}{2\lambda}$$

Right?

16. May 24, 2014

### micromass

The function $x\lambda e^{-2\lambda |x|}$ is odd. So what does that mean about the integral? I think you must have made a sign error somewhere. You should get

$$\int_0^{+\infty} x\lambda e^{-2\lambda x}dx = \frac{1}{4\lambda}$$

and

$$\int_{-\infty}^0 x\lambda e^{-2\lambda x}dx = -\frac{1}{4\lambda}$$

17. May 24, 2014

I see my mistake. Also, how does the function being odd or even affect the integral? I'm a self learner so I'm not surprised at the fact that my knowledge is fragmented.
I'm guessing this means that the expectation value is 0. How can this be?

18. May 24, 2014

### micromass

It's a very easy test to compute integrals. It's always worth checking this before doing a long computation.

So, if a function is odd (meaning that $f(-x) = -f(x)$), then

$$\int_{-d}^d f(x)dx = 0$$

for all $d$ including $d=+\infty$. This can be seen by splitting up the integral into

$$\int_{-d}^0 f(x)dx + \int_0^d f(x)dx$$

and by applying substitution $x\rightarrow -x$ on the first integral.

You can also check the graph to see this easily.

If a function is even, then we can't compute the integral this way, but we can make some simplification:

$$\int_{-d}^d f(x)dx = 2\int_0^d f(x)dx$$

The proof is the same as in the case of odd functions.

19. May 24, 2014

### Saitama

@Radarithm: This is what I used when I wrote the definite integral. Can you relate?

20. May 24, 2014

Apparently that was an even function. Changing the sign of $x$ would not affect the entire function.