- #1
What's Wrong with My Eigenvector Calculation?
- Thread starter athrun200
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Homework Help Overview
The discussion revolves around the calculation of eigenvectors in a quantum mechanics context. The original poster expresses confusion regarding their eigenvector result compared to a model answer.
Discussion Character
- Conceptual clarification, Assumption checking
Approaches and Questions Raised
- The original poster attempts to understand discrepancies between their eigenvector calculation and the model answer, questioning the necessity of transforming the eigenvector into a more complex form. Other participants discuss the uniqueness of eigenvectors and the relevance of symmetry in physics.
Discussion Status
Participants have engaged in clarifying the nature of eigenvectors and their representations. Some guidance has been offered regarding the multiplicative nature of eigenvectors, but questions about the complexity of the forms remain open for exploration.
Contextual Notes
There is an implied assumption that the eigenvector's complexity may relate to physical interpretations or conventions in quantum mechanics, though this is not explicitly resolved in the discussion.
- #2
vela
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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.
- #3
athrun200
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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?
- #4
vela
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You don't need to, but you have to admit there's certain symmetry there. And physicists like symmetry. 
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