Background This question, from Quantum Mechanics: The Theoretical Minimum started with the following assumption(?)

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

I'm now attempting to derive(?) the above based on the initial information given, which I think, based on the paragraph immediately above, is as follows:

$$|r\rangle = x|u\rangle + y|d\rangle$$ and $$\langle r|u\rangle \langle u|r\rangle = \frac{1}{2}$$ and $$\langle r|d\rangle \langle d|r\rangle = \frac{1}{2}$$


  1. Is my assumption about the initial information correct?
  2. If #1 is 'yes' would solving the above be covered in basic linear algebra?

NOTE: Please do not provide a solution to the set of equations.

  • $\begingroup$ Yes and yes. This is exactly what you learn in linear algebra when you learn about basis decomposition. $\endgroup$
    – knzhou
    Apr 11 '16 at 21:20
  • $\begingroup$ @knzhou Thanks. And I was afraid you were going to say that. $\endgroup$ Apr 12 '16 at 13:21
  1. Yes, the state is a soverposition, with amplitudes x and y still to define, of the eigenstates up and down.
  2. You would just impose normalization for the state |r>, by performing a scalar product of |r> with itself. Also, you want the two probabilities to be the same ,i.e. the two projections of the state |r> over the 2 basis vectors up and down are equal.
  • $\begingroup$ Why a and b and not x and y? And, if possible to do so without making the answer incomplete, please remove the actual solution to the problem. $\endgroup$ Apr 10 '16 at 14:03
  • $\begingroup$ sorry i used a and b instead of x and y. but same thing $\endgroup$
    – Marco
    Apr 10 '16 at 14:13
  • $\begingroup$ No problem. Just remove the solution to the problem when you get a chance. Also, to clarify, the answer to #2 is "no", correct? $\endgroup$ Apr 10 '16 at 14:56
  • $\begingroup$ Also, are "normalization" and "scalar products" (I'm not familiar with either) covered in linear algebra? $\endgroup$ Apr 11 '16 at 13:17
  • $\begingroup$ yes. normalization means that you want the lenght of the vector to be 1. A scalar product is the sum of the products of the components of two vectors... you should check it out somewhere. $\endgroup$
    – Marco
    Apr 11 '16 at 16:20

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