Question

(a) Show that (@) = sin e- is an eigenfunction of both Î, and Î", where = -1 1 a 1 22 sin + sin 020 sin0 and derive the corresponding eigenvalues. You may use the identity 1 a 1 sin sin 2 sin sin 0 80 sino 31 (sin 00 (5 marks) (6) Consider the function $(,0,4)= A - 1/200 sin 6e-ip, 20 where A is a constant and an is the Bohr radius. This is a hydrogen atom energy eigenfunction with energy eigenvalue - Er/4, where Er is the Rydberg energy. Combining this information with the results of part (a), specify the n, I and m quantum numbers of this eigenfunction. Show that a suitable value of the normalization constant is A=1/(8V1032). Is this value uniquely defined? (8 marks) (c) Calculate the expectation value of 22 = p2 cos? O in the state defined by the energy eigenfunction (r, 0,6) of part (b). (4 marks) 3 You may find the following integrals useful in parts (b) and (c). 4 * sinº Odo 4 sino cosO do S*** p"e-/adr = n! "+1 for n= 1, 2, 3, ... 15

Answer 1

a)

A function(f) is called an eigen function of an operator(O) if:

Df = cf; where c= constant, called a sthe eigen value.

Thus, for the given function f:

For Lz;

$\hat{L_z} = \frac{\not{h}}{i}\frac{\partial }{\partial \phi}$

$\hat{L_z}f = \frac{\not{h}}{i}\frac{\partial }{\partial \phi}f = \frac{\not{h}}{i}\frac{\partial }{\partial \phi} (\sin \theta \exp(-i\phi))$

$\hat{L_z}f = \frac{\not{h \sin \theta}}{i}\frac{\partial \exp(-i\phi)}{\partial \phi} = \frac{\not{h \sin \theta}}{i}\exp(-i\phi) \times -i$

$\hat{L_z}f = -{\not{h}} f$

Thus, f is an eigen function of $\hat{L_z}$ , and its corresponding eigen value is = $(-\not{h})$

For $\hat{L}^2$ :

$\hat{L}^2 f= -\not{h}^2 [\frac{1}{\sin \theta} \frac{\partial }{\partial \theta}(\sin \theta \frac{\partial }{\partial \theta})+ \frac{1}{\sin^2 \theta }\frac{\partial^2 }{\partial \phi^2}] f$

$\hat{L}^2 f= -\not{h}^2 [\frac{1}{\sin \theta} \frac{\partial }{\partial \theta}(\sin \theta \frac{\partial }{\partial \theta})+ \frac{1}{\sin^2 \theta }\frac{\partial^2 }{\partial \phi^2}] (\sin \theta \exp(-i\phi))$

$\hat{L}^2 f= -\not{h}^2 [\frac{1}{\sin \theta} \frac{\partial }{\partial \theta}(\sin \theta \frac{\partial (\sin \theta \exp(-i\phi))}{\partial \theta})+ \frac{1}{\sin^2 \theta }\frac{\partial^2 (\sin \theta \exp(-i\phi))}{\partial \phi^2}]$

$\hat{L}^2 f= -\not{h}^2 [\frac{\exp(-i\phi)}{\sin \theta} \frac{\partial }{\partial \theta}(\sin \theta \frac{\partial (\sin \theta)}{\partial \theta})+ \frac{\sin \theta }{\sin^2 \theta }\frac{\partial^2 (\exp(-i\phi))}{\partial \phi^2}]$

Using the given identity to calculate the first term in the bracket:

$\hat{L}^2 f= -\not{h}^2 [\exp(-i\phi)(\frac{1}{\sin \theta} - 2 \sin \theta)+ \frac{1 }{\sin \theta }\times (-i)^2 \exp(-i\phi)]$

$\hat{L}^2 f= -\not{h}^2 [\frac{1}{\sin \theta} - 2 \sin \theta - \frac{1 }{\sin \theta }]\exp(-i\phi)$

$\hat{L}^2 f= \not{h}^2 (2 \sin \theta) \exp(-i\phi) =2 \not{h}^2 f$

Thus, f is an eigen function of $\hat{L}^2$ , and its corresponding eigen value is = $2 \not{h}^2$