2
$\begingroup$

I am solving the problem 5.6 from Zettili Quantum Mechanics book:

I have an Hamiltonian given by $\hat{H} = \epsilon \hat{\sigma} \cdot \hat{n}$ where $\hat{\sigma}$ represents the Pauli Matrices and $\hat{n}$ a arbitrary unit vector.

Using spherical coordinates I found the Eigenvalues, which are $E_{1} = \epsilon$ and $E_2 = -\epsilon$ but I can't find the Eigenvectors correctly. I find the following equation for the Eigenvectors equation:

$$ x \cos\theta + ye^{-i\phi} \sin\theta = E x$$ $$ x \sin{\theta}e^{i\phi} - y \cos\theta = E y$$

Where $E$ stands for the Eigenvalues in each case. For the first Eigenvalue I get the following relation between the x and y:

$$ \frac{x}{y} = \frac{e^{-i\phi}\sin\theta}{1-\cos\theta}$$ and from this point on I don't understand how the author proceeds with the solution. He writes that the previous relation is equal to this one:

$$\frac{x}{y} =\frac{\cos\frac{\theta}{2} e^{-i\phi/2}}{\sin\frac{\theta}{2} e^{i\phi/2} }$$

And that the numerator is the first element of the eigenvalue and the denominator is the second element of the eigenvector. But I don't understand why.

$\endgroup$
3
$\begingroup$

From your eigenvector equation we have: $$\frac{x}{y} = e^{-i\phi}\frac{ \sin \theta}{1-\cos\theta}=e^{-i\phi}\frac{2\sin(\theta/2)\cos(\theta/2)}{2\cos^2(\theta/2)}=\frac{\sin(\theta/2)e^{-i\phi/2}}{\cos(\theta/2)e^{i\phi/2}}$$ Where I've used the double angle formulas.

For your second question, keep in mind that for an eigenvalue-eigenvector pair $E,|v\rangle$, we have by definition: $$\hat H|v \rangle = E|v \rangle$$ Multiplying both sides by some constant $\alpha \in \mathbb C$ gives: $$\hat H(\alpha|v \rangle) = E(\alpha|v \rangle)$$ which means that $|v' \rangle = \alpha|v \rangle$ is also an eigenvector of $\hat H$ with the same eigenvalue. This means that you always have the freedom of multiplying the eigenvector by a constant when your calculating it, without worrying about it not being an eigenvector anymore. Another way you can look at this is that because of this freedom, any vector with components satisfying $x/y = {\sin(\theta/2)e^{-i\phi/2}}/{\cos(\theta/2)e^{i\phi/2}}$ is a valid eigenvector.

In conclusion, the eigenvector with components $x={\sin(\theta/2)e^{-i\phi/2}}$ and $y={\cos(\theta/2)e^{i\phi/2}}$ is one valid eigenvector, since it gives the right $x/y$ ratio. Another (completely valid) choice would be something like $x={\sin(\theta/2)e^{-i\phi/2}}/{\cos(\theta/2)e^{i\phi/2}}$ and $y=1$.

However, in Quantum Mechanics, people conventionally prefer to choose the normalized eigenstate (such that $\Vert v \Vert = 1$) between this infinite set of choices of eigenvectors; as this usually makes calculations (of probabilities, etc.) much simpler. Because of this, the author's choice is preferred by convention because $\Vert v \Vert^2 =|x|^2+|y|^2=\sin^2(\theta/2)+\cos^2(\theta/2)=1$.

$\endgroup$
  • $\begingroup$ Thank you very much!!! Your explanation was very instructive. I was with the same kind of problem in another question and I think you may have answered it as well. $\endgroup$ – Luh Nov 5 '18 at 21:32
  • $\begingroup$ You're more than welcome. $\endgroup$ – Sahand Tabatabaei Nov 5 '18 at 21:33
  • $\begingroup$ I just have one more question. When I get equations that lead to no information about the elements of the eigenvector I suppose we can set any value for the same reason as you have exposed? $\endgroup$ – Luh Nov 5 '18 at 21:35
  • $\begingroup$ Even though doing that is not wrong, I would try to get in the habit of sticking to the convention $\Vert v \Vert = 1$, since it will make your life easier in future courses. You can systematically find the eigenvector with this condition by writing $|x|^2+|y|^2=1$, which gives $|y|^2(1+|x/y|^2)=1$. Since you already know $x/y$, you can simply solve for $y$, and then get $x$ from the known $x/y$ value. $\endgroup$ – Sahand Tabatabaei Nov 5 '18 at 21:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.