The Theoretical Minimum: Probability/Spin Question

Question

Background

In The Theoretical Minimum, it states that any spin state can be represented by a linear combo of the basis vectors $|u\rangle$ and $|d\rangle$

It then goes on to show how this is done for $|r\rangle$ and $|l\rangle$ (spin prepped along the x-axis) by stating the following:

If you initially prepare apparatus A as $|r\rangle$ to measure $\delta_x$, then rotate it to measure $\delta_z$ …preparing apparatus either as $|u\rangle$ or $|d\rangle$…, there will be equal probabilities for $|u\rangle$ and $|d\rangle$. Therefore $a_u a_u^*$ and $a_d a_d^*$ must be equal to 1/2 and an appropriate function is

$$|r\rangle = \frac{1}{\sqrt{2}}|u\rangle + \frac{1}{\sqrt{2}}|d\rangle$$

Question

Why $\frac{1}{\sqrt{2}}|u\rangle$ instead of $\frac{1}{2}|u\rangle$? Does it have something to do with the fact that the components being complex numbers?

Answer 1

The probability is the square of the normalised wave-function.

In other words, the probability for every state is not $\frac{1}{\sqrt{2}}$ but rather the square of that, which is a half.

$$ P(\mid u\rangle) = P(\mid d\rangle) = (\frac{1}{\sqrt{2}})^2 = \frac{1}{2}$$

It is common practise to normalise wave-functions, such that the probability $$P_{total} = \left| \mid r\rangle\right|^2 = 1$$

This is a physical requirement. If we stuck with having $\frac{1}{2}$ rather than $\frac{1}{\sqrt{2}}$ then the probability for each would be $\frac{1}{4}$, and the total probability would be $\frac{1}{2}$, which is less than 1, this is simply unphysical. For example, if we are looking at the probability of finding a particle somewhere, and we look in all space, we would still only have a 50% of finding the particle. The normalisation is put in by hand to guarantee that this total probability on all space would sum up to 1 (i.e. 100%).