# Hermitian operators – a few theorems

References: edX online course MIT 8.05.1x Week 4.

Sheldon Axler (2015), Linear Algebra Done Right, 3rd edition, Springer. Chapter 7.

A hermitian operator ${T}$ satisfies ${T=T^{\dagger}}$. [Axler (and most mathematicians, probably) refers to a hermitian operator as self-adjoint and uses the notation ${T^*}$ for ${T^{\dagger}}$.]

As preparation for discussing hermitian operators, we need the following theorem.

Theorem 1 If ${T}$ is a linear operator in a complex vector space ${V}$, then if ${\left\langle v,Tv\right\rangle =0}$ for all ${v\in V}$, then ${T=0}$.

Proof: The idea is to show something even more general, namely that ${\left\langle u,Tv\right\rangle =0}$ for all ${u,v\in V}$. If we can do this, then setting ${u=Tv}$ means that ${\left\langle Tv,Tv\right\rangle =0}$ for all ${v\in V}$, which in turn implies that ${Tv=0}$ for all ${v\in V}$, implying further that ${T=0}$.

Zwiebach goes through a few stages in developing the proof, but the end result is that we can write

 $\displaystyle \left\langle u,Tv\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{4}\left[\left\langle u+v,T\left(u+v\right)\right\rangle -\left\langle u-v,T\left(u-v\right)\right\rangle \right]+\ \ \ \ \ (1)$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{1}{4i}\left[\left\langle u+iv,T\left(u+iv\right)\right\rangle -\left\langle u-iv,T\left(u-iv\right)\right\rangle \right] \ \ \ \ \ (2)$

Note that all the terms on the RHS are of the form ${\left\langle x,Tx\right\rangle }$ for some ${x}$. Thus if we require ${\left\langle x,Tx\right\rangle =0}$ for all ${x\in V}$, then all four terms are separately 0, meaning that ${\left\langle u,Tv\right\rangle =0}$ as desired, completing the proof. $\Box$

Although we’ve used the imaginary number ${i}$ in this proof, we might wonder if it really does restrict the result to complex vector spaces. That is, is there some other decomposition of ${\left\langle u,Tv\right\rangle }$ that doesn’t required complex numbers that would still work?

In fact, we don’t need to worry about this, since there is a simple counter-example to the theorem if we consider a real vector space. In 2-d or 3-d space, an operator ${T}$ that rotates a vector through ${\frac{\pi}{2}}$ always produces a vector orthogonal to the original, resulting in ${\left\langle v,Tv\right\rangle =0}$ for all ${v}$. In this case, ${T\ne0}$ so the theorem is definitely not true for real vector spaces.

Now we can turn to a few theorems about hermitian operators. First, since every operator on a finite-dimensional complex vector space has at least one eigenvalue, we know that every hermitian operator has at least one eigenvalue. This leads to the first theorem on hermitian operators.

Theorem 2 All eigenvalues of hermitian operators are real.

Proof: Since at least one eigenvalue ${\lambda}$ exists, let ${v}$ be the corresponding non-zero eigenvector, so that ${Tv=\lambda v}$. We have

$\displaystyle \left\langle v,Tv\right\rangle =\left\langle v,\lambda v\right\rangle =\lambda\left\langle v,v\right\rangle \ \ \ \ \ (3)$

Since ${T=T^{\dagger}}$ we also have

$\displaystyle \left\langle v,Tv\right\rangle =\left\langle T^{\dagger}v,v\right\rangle =\left\langle Tv,v\right\rangle =\left\langle \lambda v,v\right\rangle =\lambda^*\left\langle v,v\right\rangle \ \ \ \ \ (4)$

Equating the last two equations, and remembering that ${\left\langle v,v\right\rangle \ne0}$, we have ${\lambda=\lambda^*}$, so ${\lambda}$ is real. $\Box$

Next, a theorem on the eigenvectors of distinct eigenvalues.

Theorem 3 Eigenvectors associated with different eigenvalues of a hermitian operator are orthogonal.

Proof: Suppose ${\lambda_{1}\ne\lambda_{2}}$ are two eigenvalues of ${T}$, and ${v_{1}}$ and ${v_{2}}$ are the corresponding eigenvectors. Then ${Tv_{1}=\lambda_{1}v_{1}}$ and ${Tv_{2}=\lambda_{2}v_{2}}$. Taking an inner product, we have

 $\displaystyle \left\langle v_{2},Tv_{1}\right\rangle$ $\displaystyle =$ $\displaystyle \lambda_{1}\left\langle v_{2},v_{1}\right\rangle \ \ \ \ \ (5)$ $\displaystyle \left\langle v_{2},Tv_{1}\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle Tv_{2},v_{1}\right\rangle \ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \lambda_{2}\left\langle v_{2},v_{1}\right\rangle \ \ \ \ \ (7)$

where in the last line we used the fact that ${\lambda_{2}}$ is real when taking it outside the inner product. Equating the first and last lines and using ${\lambda_{1}\ne\lambda_{2}}$, we see that ${\left\langle v_{2},v_{1}\right\rangle =0}$ as required.$\Box$

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