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

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


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?

  • $\begingroup$ Because of normalization of quantum states $\endgroup$
    – Slereah
    Mar 6, 2016 at 14:23

1 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%).

  • $\begingroup$ I forgot to add to my original question that I am a layperson when it comes to quantum mechanics in general and the mathematics of QM specifically (hence Susskind). Long story short, I'll need some additional explanation around your proposed answer; if you could start with explaining/defining normalisation that would help. $\endgroup$ Mar 6, 2016 at 20:10
  • $\begingroup$ MonaLisaOverdrive, see edits to the answer $\endgroup$
    – hsnee
    Mar 6, 2016 at 20:32
  • $\begingroup$ Give me a few to review and digest the edits...it's still not making sense but likely will once actually have time to think about it. $\endgroup$ Mar 9, 2016 at 18:30
  • $\begingroup$ This is called the Born rule $\endgroup$
    – innisfree
    Mar 13, 2016 at 11:38
  • $\begingroup$ Okay, I believe I've sorted this. I was stuck on a) the concept of normalization and b) other aspects of the equation. Normalization, if I'm understanding the answer correctly, is the necessary operation performed to make sure that the probability of a state (?) equals 1, which is an essential requirement of probability theory, and which is why the equation is written using one over the square root of two INSTEAD of one half. If that is correct, I'll address my other questions about the original equation in a different post. $\endgroup$ Mar 13, 2016 at 11:38

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