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I solve for eigenvectors of Pauli X matrix. I get {1,-1} and {1, 1}. In exercises I solve, there is an answer $\frac{1}{\sqrt{2}} $ * {1,1} and $\frac{1}{\sqrt{2}} $ * {1,-1}. Where does that $\frac{1}{\sqrt{2}} $ come from?

EDIT:

Can you show me how to do it for a matrix [{1,0,0,0},{0,0,1,0},{0,1,0,0},{0,0,0,1}]? I get eigenvalues 1,1,1,-1. For eigenvector {a,b,c,d} I only get that b=c. Answers are as in the pdf, task 2:

http://web.mit.edu/2.111/www/2010/ps1Solutions.pdf

How do I know which a,b,c,d is 1 or 0?

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    $\begingroup$ Can you show us what have you tried when you got {1, -1}, {1, 1}? $\endgroup$ – G B May 18 '14 at 8:56
  • $\begingroup$ Calculated eigenvalues, plugged each eigenvalue into the matrix and solved for column vector that gives 0 vector after multiplying matrix by it. Got {x,x} and {x,-x}. $\endgroup$ – Ghostwriter May 18 '14 at 9:47
  • $\begingroup$ Looking up normalization in quantum mechanics will help you. $\endgroup$ – BMS May 18 '14 at 17:43
  • $\begingroup$ Take a look at this tutorial on dirac notation and note the steps for normalization youtube.com/watch?v=pBh7Xqbh5JQ $\endgroup$ – user6972 May 18 '14 at 19:24
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These values come from the normalization of the solutions you found.

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    $\begingroup$ Although correct, I feel that this answer is a little too short. You could perhaps explain why this normalization does not make a difference? $\endgroup$ – Danu May 18 '14 at 11:19
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Your solution is correct. But, as $\sigma_x$ can be viewed as a Hamiltonian of a spin subsystem, its eigenvectors are spin parts of the wavefunctions of the whole system.

We demand that the norm of eigenstates is 1, so that statistical interpretation of wavefunction makes sense. To achieve this, in this case you need to divide your solutions by $\sqrt2$. This multiplication doesn't change the fact that the vector is an eigenvector, of course, due to commutativity of multiplication of matrix by number.

To see that the factor has to be $\frac1{\sqrt2}$, just find the norm of current vectors:

$$||\vec v||=\sqrt{\vec v\cdot \vec v}=\sqrt{1^2+(-1)^2}=\sqrt2$$

Now solve for $a$: $$||a\vec v||=1.$$

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  • $\begingroup$ Ok, I know this, of course. But looking in mathematics only, how do I know if those factors are $\frac{1}{\sqrt{2}}$ instead of something else that sums to one? Can you help with my EDIT question as well? $\endgroup$ – Ghostwriter May 18 '14 at 12:33
  • $\begingroup$ See update about the factor. Your edit is another question, it should be asked as a separate one. Just a hint, though: eigenvectors are defined up to a multiplicative constant (and permutation of the vectors, of course), so you won't automatically get answer which is literally identical to expected by the teacher. You just have to prove that your solution is correct and satisfies all expected properties like normalization. $\endgroup$ – Ruslan May 18 '14 at 12:39
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Any multiple of an eigenvector is also an eigenvector. Eigenvectors of a matrix are also orthogonal to each other. It's conventional for eigenvectors to be normalized to unit length, because a set of orthogonal unit vectors make a good basis for a vector space, but normalization is not strictly required.

As a hint to your edited question, if you write out matrix $$ \left( \begin{array}{cccc} 1\\ &0&1\\ &1&0\\ &&&1 \end{array}\right) $$ you'll see that the bit in the middle looks like $\sigma_x$.

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  • $\begingroup$ I have already solved it "manually". So, is there any trick that allows to solve it faster provided that we have eigenvectors for σx? $\endgroup$ – Ghostwriter May 18 '14 at 16:08
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    $\begingroup$ @Makaveli Only that (in your notation) two of your four eigenvectors will have $b=\pm c$. $\endgroup$ – rob May 18 '14 at 18:59

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