Wave Function: Normalization Constant

[teme92]

Homework Statement

Consider a free particle, initially with a well defined momentum $p_0$, whose wave function is well approximated by a plane wave. At $t=0$, the particle is localized in a region $-\frac{a}{2}\leq x \leq\frac{a}{2}$, so that its wave function is

$\psi(x)=\begin{cases} Ae^{-ip_0x/\hbar} & if -\frac{a}{2}\leq x \leq\frac{a}{2} \\0 & \text{otherwise} \end{cases}$

Find the normalization constant $A$ and sketch $Re(\psi(x))$, $Im(\psi(x))$ and $|\psi(x)|^2$.

Homework Equations

The Attempt at a Solution

So here's what I done:

$A^2\int_{-\frac{a}{2}}^\frac{a}{2} e^{-ip_0x/\hbar}dx=1$

$A^2.-\frac{\hbar}{ip_0}.e^{-ip_0x/\hbar}=1$

$A^2=-\frac{ip_0}{\hbar}.\frac{1}{e^{-ip_0a/2\hbar}-e^{-ip_0a/2\hbar}}$

Is this the correct method? Also I have no idea how to sketch the function asked. Any help would be greatly appreciated.

 

[George Jones]

$A^2\int_{-\frac{a}{2}}^\frac{a}{2} e^{-ip_0x/\hbar}dx=1$

How did you get this?

 

[teme92]

I took it from $P(t)=\int_{-\frac{a}{2}}^\frac{a}{2}|\psi(x,t)|^2=1$ subbed my wave function into that then. Is this method wrong?

 

[George Jones]

If $\psi \left( x \right) =Ae^{-ip_0x/\hbar}$, what is $\left| \psi \left( x \right) \right|^2$?

 

[teme92]

$|\psi(x)|^2=A^2e^{-2ip_0z/\hbar}$?

 

[George Jones]

$|\psi(x)|^2=A^2e^{-2ip_0z/\hbar}$?

Not quite. If $Z$ is a complex number, are $Z^2$ and $\left|Z\right|^2$ the same?

 

[teme92]

No, $|Z|^2$ is the modulus squared ie. $(\sqrt{a^2+b^2})^2$

 

[George Jones]

No, $|Z|^2$ is the modulus squared ie. $(\sqrt{a^2+b^2})^2$

Yes. What if the complex value $Z$ is expressed in polar notation?

 

[teme92]

Is it $cos\theta +isin\theta$?

 

[George Jones]

Well, $e^{i\theta} = \cos \theta + i \sin \theta$. What is $\left| e^{i\theta} \right|^2$ for real $\theta$?

 

[teme92]

$sin2\theta + 1$?

 

[George Jones]

$sin2\theta + 1$?

No. You're making this more difficult than it actually is. Forget about a and b and theta. Can you express $\left| Z \right|^2$ in terms of $Z$ and the complex conjugate of $Z$?

 

[kq6up]

Remember, also. The complex conjugate of $\psi$, and $|\psi|^2=\psi^{*}\psi$.

Chris

 

[teme92]

Oh $|\psi|^2=z\bar{z}$

 

[George Jones]

Yes! And if $z = e^{i\theta}$, then $\bar{z} =$?

 

[teme92]

Is it $\bar{z}=e^{-i\theta}$?

 

[George Jones]

When complex conjugating, what is done to every $i$?

 

[teme92]

Sorry I was supposed to put a minus in that I'll edit

 

[George Jones]

Is it $\bar{z}=e^{-i\theta}$?

And $z \bar{z} =$?

 

[teme92]

They $i\theta$ and $-i\theta$ cancel each other out so $e^0=1$

 

[George Jones]

Good. Now, go back to the original question.

 

[teme92]

So $|\psi(x)|^2 = 1$?

 

[George Jones]

What about the normalization constant $A$, which, without loss of generality, can be taken to be real and positive?

 

[teme92]

$A=\frac{1}{e^{-ip_0x/\hbar}}$?

 

[George Jones]

I took it from $P(t)=\int_{-\frac{a}{2}}^\frac{a}{2}|\psi(x,t)|^2=1$ subbed my wave function into that then. Is this method wrong?

Redo the above.

 

[teme92]

$A^2\int_{-\frac{a}{2}}^\frac{a}{2} e^{-2ip_0x/\hbar}dx=1$

$A^2.-\frac{\hbar}{ip_0}.e^{-i2p_0x/\hbar}=1$

$A^2=-\frac{ip_0}{\hbar}.\frac{1}{e^{-ip_0a/\hbar}-e^{-ip_0a/\hbar}}$

Is this correct and if so where do I go. Square rooting this seems wrong

 

[George Jones]

No. What is

$$\overline{e^{-ip_0x/\hbar}}?$$

 

[teme92]

$e^{ip_0x/\hbar}$?

 

[George Jones]

Yes.

 

[teme92]

I'm getting confused over what $(Ae^{-i{p_0}x/\hbar})^2$ is. And then the integrating of that.

 

[George Jones]

I'm getting confused over what $(Ae^{-i{p_0}x/\hbar})^2$ is. And then the integrating of that.

You are not supposed to integrate $(Ae^{-i{p_0}x/\hbar})^2$, you are supposed to integrate $|Ae^{-i{p_0}x/\hbar}|^2$.

 

[teme92]

So would that be $A^2(1)$ as $|e^{-i{p_0}x/\hbar}|^2=1$?

 

[George Jones]

Yes.

 

[teme92]

So $A=1$? Thanks for all the help George. How do I proceed to sketch the items asked? I assume $A$ is the real part and $e^{-i{p_0}x/\hbar}$ the imaginary part and $|\psi(x)|^2=1$. How do these look when sketched?

 

[George Jones]

So $A=1$?

No. Remember, you have to calculate

$$\int^{a/2}_{-a/2} \left| \psi \left(x\right) \right|^2 dx.$$