# How to find the eigenvectors and eigenvalues of a hermitian operator?

While reading Theoretical Minimum by Leonard Susskind, I came across the exercise 3.4 where he asked to find the eigenvalues and the eigenvectors of the matrix that represents the $$\sigma_{n}$$ component of the operator $$\sigma$$ where: $$\sigma_{n}=\vec\sigma.\hat n$$ Which leads to: $$\sigma_n=\begin{pmatrix}n_{z}&(n_{x}-i.n_{y})\\(n_{x}+in_{y})&-n_{z}\end{pmatrix}$$ Given that: $$n_{z}=\cos(\theta)$$ $$n_x=\sin(\theta) \cdot \cos(\phi)$$ $$n_y=\sin(\theta) \cdot \sin(\phi)$$ How can I approach the question as I am a little bit confused, and I searched online for the answer and found that we need to find the determinant of the $$\sigma_n$$ where is equals: $$\begin{pmatrix}\cos(\theta)-\lambda&\sin(\theta)\cdot\cos(\phi)-i\cdot \sin(\theta)\cdot\sin(\phi)\\ \sin(\theta)\cdot \cos(\phi)+i \cdot \sin(\theta)\cdot \sin(\phi)&-\cos(\theta)-\lambda\end{pmatrix}$$ How does finding the determinant of $$\sigma_n$$ help solve the question and where did the $$\lambda$$ come from in the first place?

• Can you link the online answer you are talking about? And also, how familiar are you with linear algebra? So that people can answer your question accordingly. Dec 22, 2023 at 22:01
• Here is where I got the answer: quantum-abc.de/Exercises%20complete1.pdf. My mathematical knowledge is mostly from high school. Dec 22, 2023 at 22:18

From your post and the comments therein, it seems that you would benefit in reading a little bit more about linear algebra. There are plenty of resources for that, including some excellent posts on the Mathematics StackExchange.

In this post, I will only answer your question about why finding this determinant allows to compute the eigenvalues of a matrix. However, it's quite hard to answer such a question with high school knowledge only, so I'll assume just a tad bit of knowledge about matrices (hopefully not much, but feel free to ask questions in the comments).

We will need several facts:

1. If a matrix is not invertible, its determinant is nil. Reciprocally, it is invertible if its determinant isn't nil.
2. If there is a non-zero vector $$x$$ such that $$Mx$$ is the zero vector, then $$M$$ isn't invertible. Reciprocally, if the only $$x$$ such that $$Mx=0$$ is $$x=0$$, then $$M$$ is invertible.
3. If there is a non-zero vector $$x$$ such that $$Mx=\lambda x$$ for a certain $$\lambda$$, then $$\lambda$$ is an eigenvalue of $$M$$.

With that being said, let us consider an eigenvalue $$\lambda$$ of a matrix $$M$$. From Fact 3. we have: $$Mx=\lambda x$$ Which we can rewrite as: $$Mx=\lambda Ix$$ with $$I$$ being the identity matrix. This is equivalent to: $$(M-\lambda I)x=0$$ But now from Fact 2. we have that $$M-\lambda I$$ is not invertible, which means from Fact 1. that the determinant of $$M-\lambda I$$ is nil.

Thus, we have shown that if $$\lambda$$ is an eigenvalue of $$M$$, then the determinant of $$M-\lambda I$$ is nil.

Reciprocally, if $$\lambda$$ is not n eigenvalue of $$M$$, then we can show that $$M-\lambda I$$ is invertible. Suppose that: $$(M-\lambda I)x=0$$ Then it means that: $$Mx=\lambda x$$ But we have assumed that $$\lambda$$ is not an eigenvalue of $$M$$, so it necessarily means that $$x=0$$. Thus, from Fact 2. $$M-\lambda I$$ is invertible, which means from Fact 1. that its determinant is not nil.

All in all, we have shown that the eigenvalues of $$M$$ are exactly the solutions to the equation $$\text{det}(M-\lambda I)=0$$ Where we want to solve for $$\lambda$$.

In your case, you will find that the determinant is actually a degree 2 polynomial in $$\lambda$$. You should thus be able to quickly find the eigenvalues.

Once you know an eigenvalue $$\lambda$$, there are several ways to compute the eigenvectors. The simplest is simply to solve for $$x$$ in $$Mx=\lambda x$$. Such an $$x$$ isn't unique, but you're interested in linearly independent solutions here.

This might be a bit too much for an high-school level and I apologize for that. But to put it in a nutshell:

• The determinant of $$M-\lambda I$$ is called the characteristic polynomial of $$M$$, and its roots are exactly the eigenvalues of $$M$$.
• Using the definition of an eigenvalue, it is possible to find an eigenvector associated to a known eigenvalue.
• You should definitely get yourself an introductory course on linear algebra if you're interested in that topic, you'll learn a lot of things!
• As for learning linear algebra, I would highly recommend Greg Sanderson's Essence of Linear Algebra series to get an excellent intuitive understanding. If you are looking for something slightly advanced, then I would recommend Gilbert Strang's Linear Algebra Course at MIT. Dec 23, 2023 at 5:14
• Thanks a lot, but I why did you make a point about $x$ being 0 if $\lambda$ is not an eigenvalue of $M$? Dec 24, 2023 at 8:49
• @zizaaooo In this step, our goal is to show that if $\lambda$ is not an eigenvalue of $M$, then $M-\lambda I$ isn't invertible. In order to do this, we use the second part of Fact 2. : we show that the only $x$ satisfying $(M-\lambda I)x=0$ is $x=0$. Dec 24, 2023 at 9:27
• Do you mean that $M-\lambda$ is invertible because the only solution is $x=0$ if the $\lambda$ is not an eigenvalue of $M$? Because you wrote in the first line of your comment that $M-\lambda$ isn't invertible. I might be repeating what you just said but I want to make sure that I understood your answer. Dec 24, 2023 at 10:15
• @zizaaooo Oops, you're right! Its determinant isn't nil, so $M-\lambda I$ is invertible indeed, sorry for that! Dec 24, 2023 at 10:26