# Question 1:

## Problem 1.10 from Townsend

The state $$\mid \psi \rangle = \frac{1}{2} \mid +z \rangle + \frac{i \sqrt{3}}{2} \mid -z \rangle$$

is a state with $S_n = + \hbar/2$ along a particular axis $\vec{n}$.

a) Determine $\langle S_x \rangle$ for this state.

*Hint: Let $\frac{\hbar}{2} = 1$* to help compare your results.

b) Determine $\langle S_y \rangle$ for this state.

*Tip: Once you've done the code for part a, you can copy and paste that code and make a few minor edits to speed things up*

c) Determine $\langle S_z \rangle$ for this state.

d) Interpret your results. What can you conclude about the state without further calculations (i.e. ignoring the results you got for $\vec{n}$ in the written homework).

e) Do these results match with what you got for the written homework?

# Question 2:

Question 1.11 in Townsend asks you to find $\langle S_x \rangle$, $\langle S_y \rangle$, and $\langle S_z \rangle$ for the following state vector:

$$\mid \psi \rangle = - \frac{i}{2} \mid +z \rangle + \frac{\sqrt{3}}{2} \mid -z \rangle$$

It then asks you to compare the results to question 1.10. The probabilities end up being the same for both states.

Call the state from this problem $\mid \psi*{11} \rangle $ and the state from problem 1.10 $\mid \psi*{10} \rangle$.

a) Calculate $\langle \psi*{11} \mid \psi*{10} \rangle$

b) Then calculate $\lvert \langle \psi*{11} \mid \psi*{10} \rangle \rvert^2$

c) Interpret your results.

# Question 3:

A quick and dirty way to find a perpendicular vector is to set each new coefficient to one over the value of the original coefficent and switch the sign on one of the terms. This vector won't be normalized but will be perpendicular to the original vector.

In problem 1.10 we are told the state vector given is for "spin-up" along some axis. A perpendicular vector would correspond to "spin-down" along that same axis.

a) Find a vector perpendicular to $\mid \psi*{10} \rangle$, show that it is in fact perpendicular to $\mid \psi*{10} \rangle$.

*Tip: If you have a single factor of $i$ (or in the case of Python 1j) in your equation, inverting that is the same as flipping the signs of one of your terms. If both terms have $i$ then you will still need to manually flip the sign of one of the terms.*

b) Calculate the inner product of this new vector with $\mid \psi_{11} \rangle$.

c) Interpret this result in part b. What can you tell about the state described by the new vector compared to the state described by $\mid \psi_{11} \rangle$?

d) Take the inner product of the perpendicular vector with itself. For a normalized vector this should be equal to 1 but here it isn't. Normalize the vector so the inner product with itself is equal to 1.

# Turn this in

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