# Normalise wavefunction of hydrogen atom

1. Oct 17, 2011

### blueyellow

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

An electron in a hydrogen atom is described by the wavefunction:

psi(r) is proportional to (psi(subscript 100)+2psi(subscript 210)-3psi(subscript 32 -1) -4psi(subscript411))

where psi(nlm(subscript l)) are the eigenfunctions of the hydrogen atom with n, l, m(subscript l) the usual quantum numbers

a)normalise this wavefunction
b)evaluate the probability that the electron is measured to be in the ground state
c) evaluate the expectation value of the energy (you may assume that the energy levels of the hydrogen atom are given by E(subscript n)=(-1/(2(n^2)))a.u.)
d)Evaluate the expectation value of L(subscript z)
e)evaluate the probability that the electron is found to have orbital angular omentum l=1
f) evaluate the probability that, having obtained l=1, a subsequent measurement of L(subscript z) obtains the value m(subscript l)=0

3. The attempt at a solution

psi(r)=c(psi(100)+2psi(210)-3psi(32 -2)-4psi(411))
integral of (c^2) (psi(100)+2psi(210)-3psi(32 -2)-4psi(411))^2 =1
integral of (c^2) ((psi (100))^2 +psi(100)*2psi(210) +psi(100)*3psi(32 -2) - psi(100)*4psi(411) +2psi(210)psi(100)+4psi^2(210)+2psi(210)*3psi(32 -2) - 2psi(210)*4psi(411) -3psi(32 -2)*psi(100) -3psi(32 -2)*2psi(210) +9psi^2 (32 -2) +3psi(32-2)*4psi(411) -4psi(411)*psi(100) -4psi(411)*2psi(210) +12psi(411)*psi(32 -2) +16psi^2 (411))=1

2. Oct 17, 2011

### vela

Staff Emeritus
Now use the facts that the eigenfunctions are orthogonal to each other and that each eigenfunction is normalized.

3. Oct 19, 2011

### blueyellow

<psi|psi>=(c^2)((psi^2(100))+4(psi^2(210))+9(psi^2(32 -2))+16(psi^2(411)))=(1+4+9+16)(c^2)=1
c=1/(sqrt20)

??

b) ground state, n=1
probability = (1/(sqrt20))(psi(100))*(1/sqrt(20))(psi(100))=1/20

?
is this correct?

4. Oct 19, 2011

### vela

Staff Emeritus
1+4+9+16=30, but otherwise it looks fine.

5. Oct 22, 2011

### blueyellow

c) evaluate the expectation value of L(z)

psi=(1/(sqrt(30))(psi(100)+2psi(210)-3psi(32 -2) -4psi(100))

L(z)|psi(n, l, m>=m h-bar|psi(n, l, m>

<L(z)>=[h-bar/30](0+2*1-3*(-2)-4*1)=(h-bar)/15

I kind of did this by following an example I found on http://www.csun.edu/~jeloranta/CHEM352/ch2-examples.pdf

but I don't know whether I have done it right and why they added all of the values of l to get the expectation value,

and why even though psi=[1/(sqrt(30))]...
there expectation value would be h-bar/30... without the square root

6. Oct 22, 2011

### vela

Staff Emeritus
The expectation value $\langle \hat{L}_z \rangle$ is by definition $\langle \psi | \hat{L}_z | \psi \rangle$. Just take it step by step.

Suppose have the state $|\psi\rangle = \frac{1}{2}|2~1~1\rangle+ \frac{\sqrt{3}}{2}|3~2~{-2}\rangle$. If we apply $\hat{L}_z$ to it, we get
\begin{align*}
\hat{L}_z |\psi\rangle &= \hat{L}_z\left(\frac{1}{2}|2~1~1\rangle+ \frac{\sqrt{3}}{2}|3~2~{-2}\rangle\right) \\
&= \frac{1}{2}\hat{L}_z|2~1~1\rangle + \frac{\sqrt{3}}{2}\hat{L}_z|3~2~{-2}\rangle \\
&= \frac{1}{2}(1\hbar)|2~1~1\rangle + \frac{\sqrt{3}}{2}(-2\hbar)|3~2~{-2}\rangle \\
&= (1\hbar)\frac{1}{2}|2~1~1\rangle + (-2\hbar)\frac{\sqrt{3}}{2}|3~2~{-2}\rangle
\end{align*}so noting that the cross terms are going to be 0 because of the orthogonality of the eigenstates, we get
\begin{align*}
\langle \psi | \hat{L}_z |\psi\rangle &= \left(\frac{1}{2}\langle 2~1~1| + \frac{\sqrt{3}}{2}\langle 3~2~{-2} |\right) \left((1\hbar)\frac{1}{2}|2~1~1\rangle + (-2\hbar)\frac{\sqrt{3}}{2}|3~2~{-2}\rangle\right) \\
&= (1\hbar)\frac{1}{4}\langle 2~1~1|2~1~1\rangle + (-2\hbar)\frac{3}{4}\langle 3~2~{-2} |3~2~{-2}\rangle \\
&= (1\hbar)\frac{1}{4} + (-2\hbar)\frac{3}{4} \\
&= -\frac{5}{4}\hbar
\end{align*}If you see the pattern, you should be able to write the next-to-the-last line down immediately for a given a superposition.