athrun200
- 275
- 0
What's Wrong with My Eigenvector Calculation?
- Thread starter athrun200
- Start date
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 3K views
- 16,224
- 2,895
Your work is fine. Remember that an eigenvector is only unique up to a multiplicative constant. The eigenvector you found can be written
$$\left\lvert \frac{\hbar}{\sqrt{2}} \right\rangle = \begin{bmatrix} \frac{1-i}{2} \\ \frac{1}{\sqrt{2}} \end{bmatrix} = \frac{1}{\sqrt{2}}\begin{bmatrix}\frac{1-i}{\sqrt{2}} \\ 1 \end{bmatrix} = \frac{1}{\sqrt{2}} \begin{bmatrix}e^{-i\pi/4} \\ 1 \end{bmatrix}.$$ If you multiply that by ##e^{i\pi/8}## and ignore the normalization factor of ##1/\sqrt{2}##, you'll get the answer in the solution.
$$\left\lvert \frac{\hbar}{\sqrt{2}} \right\rangle = \begin{bmatrix} \frac{1-i}{2} \\ \frac{1}{\sqrt{2}} \end{bmatrix} = \frac{1}{\sqrt{2}}\begin{bmatrix}\frac{1-i}{\sqrt{2}} \\ 1 \end{bmatrix} = \frac{1}{\sqrt{2}} \begin{bmatrix}e^{-i\pi/4} \\ 1 \end{bmatrix}.$$ If you multiply that by ##e^{i\pi/8}## and ignore the normalization factor of ##1/\sqrt{2}##, you'll get the answer in the solution.
athrun200
- 275
- 0
Oh! Thank you very much.
But why do we bother to have such a complicated eigenvector?
The one with (1-i)/2 and 1/(sqrt2) is much easier to find. Why do we need to change it to the form of exp?
But why do we bother to have such a complicated eigenvector?
The one with (1-i)/2 and 1/(sqrt2) is much easier to find. Why do we need to change it to the form of exp?
- 16,224
- 2,895
You don't need to, but you have to admit there's certain symmetry there. And physicists like symmetry. 