Given a three dimensional Hilbert space with the three basis vectors $|1\rangle, |2\rangle, |3\rangle,$ and two state vectors, $|\psi_1 \rangle = a|1\rangle -b |2\rangle +c|3\rangle, |\psi_2 \rangle = b|1\rangle + a |2\rangle,$ with $a,b,c \in \mathbb{C},$ I need to find if the operator $A = |\psi_1' \rangle \langle \psi_1'| + |\psi_2' \rangle \langle \psi_2'|$ is a hermitian operator, whereby $|\psi_1' \rangle, |\psi_2' \rangle,$ are the corresponding normed vectors. In case the scalars $a,b,c$ were reals, the matrix of $A$ would be symmetric as sum of two symmetric matrices and thus hermitian, otherwise I do not see if $A$ can be hermitian. Can somebody provide some insight?
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$\begingroup$ To start, make e.g. use of the fact that the adjoint of a sum is the sum of the adjoints. Then you have reduced the problem to find the adjoint of an operator of the form $|\psi\rangle\langle \psi|$, which should be easy to find given the definition of the adjoint. More generally, you can compute the adjoint of $|\phi_1\rangle\langle \phi_2|$ for any two vectors $\phi_1,\phi_2$. Your cases then follow as a special case where $\phi_1=\phi_2$. $\endgroup$– Tobias FünkeCommented Jul 3, 2023 at 17:21
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Your operator $A$ is Hermitian if, for any two vectors $\phi_1$ and $\phi_2$, you have that
$$\langle \phi_1,A\phi_2\rangle = \langle A\phi_1,\phi_2\rangle$$ In bra-ket notation, this is $$\langle \phi_1|A|\phi_2\rangle = \overline{\langle \phi_2|A|\phi_1\rangle}$$
Computing the left-hand side in your example yields $$\langle \phi_1|\psi_1\rangle\langle\psi_1|\phi_2\rangle + \langle \phi_1|\psi_2\rangle\langle \psi_2|\phi_2\rangle$$
All you have to do is compute the right-hand side and check to see whether they're equal. The result does not depend on the basis you choose, what $\psi_1$ or $\psi_2$ are, and in particular whether they are normalized or whether their components in your chosen basis happen to be real-valued.
answered Jul 3, 2023 at 17:39
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$\begingroup$ Thanks. The right-hand side equals the expression you showed. I have an additional question. Why would zero be an eigenvalue of such an operator ? Again, in case the components are real, the matrix of the operator would have rows that are proportionate to each other, meaning that the determinant would vanish, and thus at least one eigenvalue would be zero. $\endgroup$ Commented Jul 3, 2023 at 18:21
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1$\begingroup$ @user996159 If the Hilbert space is 3-dimensional, then there must exist some vector (in fact, an entire subspace of them) which is orthogonal to both $\psi_1$ and $\psi_2$, and so the action of $A$ on that vector yields zero. For intuition, you can think about $\mathbb R^3$ rather than $\mathbb C^3$. At most, any two nonzero vectors in $\mathbb R^3$ span a plane (they span a line if they happen to be linearly dependent). As a result, you can always find vectors which are orthogonal to both of them. The same is true in $\mathbb C^3$. $\endgroup$ Commented Jul 3, 2023 at 18:24
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$\begingroup$ Why is this restricted to a 3-dimensional Hilbert space ? $\endgroup$ Commented Jul 3, 2023 at 18:33
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1$\begingroup$ @user996159 Because of the first six words in your question. If the dimension of your space is greater than three, then the same argument holds, but the dimension of the eigenspace which is annihilated by $A$ is larger. $\endgroup$ Commented Jul 3, 2023 at 18:35
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$\begingroup$ It is unclear to me if this argument applies to the specific operator $A$ above or to all hermitian operators in QM. In other words, by means of the explanation of annihilation of an eigenspace by some hermitian operator $\hat A,$ does that mean that each hermitian operator in QM must have at least one eigenvalue zero ? $\endgroup$ Commented Jul 3, 2023 at 18:43