Is it Hermitian or not?

Question

I have a question, since I'm realy consfused.

I am doing quantum information theorem, and there's a theorem that says, that for Hermitian $A\in L(H)$ ($H$ some finite dimensional Hilbertspace) then it has a spectral decomposition:

$$A = \sum_{i=1}^d \lambda_i \left|i\right\rangle \left\langle i\right| $$

But is for example $A=\left|0\right\rangle \left\langle 1\right| + \left|1\right\rangle \left\langle 0\right|$ not also hermitian, since i thought ($\left|0\right\rangle \left\langle 1\right|)^* = \left|1\right\rangle \left\langle 0\right|)$.

But this operator doesnt have such spectral decomposition. So I guess i'm making a wrong assumption here?

Answer 1

Yes, the operator $A$ does have such a spectral decomposition. However note that the vectors that you have written as $| i \rangle$ in this decomposition do not have to be in the computational basis. That is, one has that $$ A = \sum_i \lambda_i | \psi_i \rangle \langle \psi_i |, $$ for some orthonormal vectors $| \psi_i \rangle$. But usually it is not the case that one can take $| \psi_i \rangle = | i \rangle$, the basis in which you were given your original description of the operator $A$.

In your example, let $| + \rangle = \frac{1}{\sqrt{2}} (| 0 \rangle + | 1 \rangle)$ and $| - \rangle = \frac{1}{\sqrt{2}} (| 0 \rangle - | 1 \rangle)$.

You can check that $A = | + \rangle \langle + |-| - \rangle \langle - |$.

If you want to learn more about this, look for information about "diagonalizable matrices". All hermitian matrices are diagonalizable by unitaries: $A = U D U^\dagger$, where $U$ is a unitary and $D$ is a diagonal matrix. Nielsen and Chuang's book on quantum information theory has a short review of linear algebra in Section 2.1.

Answer 2

The operator $A$ you gave is Hermitian, and it does have a spectral decomposition. For ease of viewing, let's write $A = uv^{\ast} + vu^{\ast}$, where $u = |0 \rangle$ and $v = | 1 \rangle$ are unit vectors. Then you can rewrite this as

$$ A = \left( \frac{u+v}{\sqrt{2}} \right) \left( \frac{u+v}{\sqrt{2}} \right)^{\ast} - \left( \frac{u-v}{\sqrt{2}} \right) \left( \frac{u-v}{\sqrt{2}} \right)^{\ast}$$