- #1
What's Wrong with My Eigenvector Calculation?
- Thread starter athrun200
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In summary, the conversation discussed the process of finding an eigenvector and the importance of symmetry in physics. The model answer explained that an eigenvector is unique up to a multiplicative constant and showed how the eigenvector in the model answer could be simplified using exponential form. The person asking the question expressed gratitude and questioned why the simpler form of the eigenvector was not used. The model answer explained that while it is not necessary, the exponential form offers a certain symmetry that physicists appreciate.
- #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. 
- #5
paranormal
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As a scientist, it is important to carefully review your work and make sure all steps are correct. Have you checked your calculations and equations to ensure they are accurate? Additionally, have you considered the possibility of errors in your data or inputs? It may also be helpful to seek assistance from a colleague or professor to compare your work with the model answer and identify any discrepancies. Remember, in science, it is important to be thorough and precise in our methods and analyses. Keep practicing and don't get discouraged, as mistakes and challenges are a natural part of the scientific process.
FAQ: What's Wrong with My Eigenvector Calculation?
What is an eigenvector of a spin operator?
An eigenvector of a spin operator is a vector that, when acted upon by the spin operator, results in a scalar multiple of the original vector. This scalar multiple is known as the eigenvalue, and it represents the spin state of a quantum particle.
How is an eigenvector of a spin operator different from a regular vector?
An eigenvector of a spin operator is different from a regular vector in that it is a quantum state that represents the spin of a particle, rather than a physical direction or magnitude in space. It is also unique in that it is only defined up to a phase factor, rather than a specific value.
What is the importance of eigenvectors of spin operators in quantum mechanics?
Eigenvectors of spin operators are important in quantum mechanics because they represent the possible spin states of a particle, which is a fundamental property of matter. They also play a key role in the mathematical formulation of quantum mechanics, helping to describe and predict the behavior of quantum systems.
How are eigenvectors of spin operators used in experiments?
In experiments, eigenvectors of spin operators are used to determine the spin state of a particle. This is typically done by measuring the eigenvalue associated with a particular eigenvector, which can then be used to determine the spin state of the particle.
Can there be multiple eigenvectors of a spin operator for a single particle?
Yes, there can be multiple eigenvectors of a spin operator for a single particle. This is because the spin operator can have multiple eigenvalues, each associated with a unique eigenvector. These different eigenvectors represent different possible spin states of the particle.