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I'd like to ask the community to please verify the first two proofs below and help me get through the last one since I seem to be stuck. Thank you in advance.

Proof 1: Given two noncommutting operators $A$ and $B$, that is, $[A,B] = C$, show that such operators can't have the same eigenvectors.

Attempt:

Let's try to prove it by examining the opposite scenario, that is, the case of them sharing the same eigenvector $|\phi \rangle$:

$$\begin{cases} A |\phi \rangle = a | \phi \rangle = |\psi_1 \rangle \\ B |\phi \rangle = b | \phi \rangle = |\psi_2 \rangle \end{cases}$$

$$B\left( A |\phi \rangle\right) = B |\psi_1 \rangle = b |\psi_1 \rangle = ba |\phi \rangle $$

$$A \left( B |\phi \rangle \right) = A |\psi_2 \rangle = a |\psi_2 \rangle = ab |\phi \rangle $$

$$[A,B]|\phi \rangle = \left( AB - BA\right)| \phi \rangle = \left( ab - ba \right)|\phi \rangle = C|\phi \rangle = c |\phi \rangle $$

$$(ab - ba) = c \neq 0$$

We see that the above equation should equal zero instead, thus they can't share the same eigenvector $|\phi \rangle$ if they don't commute.

Proof 2: Show that if the eigenvalue of an operator is a real number then said operator must be hermitian.

Attempt:

Let's use $A$ as our operator, acting on its eigenvector $|\psi \rangle$, then

$$\begin{cases} A|\psi \rangle = a|\psi \rangle \\ \left( A |\psi \rangle \right)^{\dagger} = \langle \psi| A^{\dagger} = a^*\langle \psi| \end{cases} $$

$$\langle \psi| \left( A |\psi \rangle \right) = \langle \psi| a | \psi \rangle = a\langle \psi | \psi \rangle$$

$$\left( \langle \psi|A^{\dagger} \right) |\psi \rangle = \langle \psi|a^*|\psi \rangle = a^* \langle \psi| \psi \rangle $$

$$\langle \psi| A |\psi \rangle - \langle \psi|A^{\dagger} |\psi \rangle = a\langle \psi | \psi \rangle - a^* \langle \psi| \psi \rangle $$

According to definition, a hermitian operator is any self-adjoint operator, i.e. $A = A^{\dagger}$, also from Hilbert space, one knows that the inner product of a vector by itself is always zero or positive, i.e. $\langle \psi|\psi \rangle \geq 0$, being null if and only if $|\psi \rangle = 0$, and this is not the case here, since it's a trivial one, so

$$\langle \psi| A |\psi \rangle - \langle \psi|A^{\dagger} |\psi \rangle = \left(A - A^{\dagger} \right)\langle \psi|\psi \rangle = a\langle \psi | \psi \rangle - a^* \langle \psi| \psi \rangle = \left( a - a^*\right) \langle \psi|\psi \rangle = 0$$

$$a=a^*$$

Only real numbers are their own conjugate.

Proof 3: Given an eigenvalue-eigenvector equation, suppose that the vectorstate depends on an external parameter, e.g. time, and that over it acts an operator that is the fourth derivative w.r.t. time. If this operator is hermitian, find the most general operator possible that satisfies these conditions and what are the boundary conditions on the eigenfunctions that are needed.

Attempt:

Using $A$ as our hermitian operator and $|\psi (t) \rangle$ for its time dependent eigenvector, and $a(t)$ for its eigenvalue, I suppose the simplest eigenvalue-eigenvector equation one can write would be

$$A|\psi (t) \rangle = a(t) |\psi (t) \rangle$$

As for the operator, I'd say it follows directly that

$$A \sim \frac{\partial^4}{\partial t^4} $$

$$a(t) = a^*(t)$$

Now, based on the assumption I've been correct so far, comes the part where I'm stuck. I understand the general concept of eigenvalues and eigenfunctions, i.e. $a(t)$ would generate a set of eigenfunctions $\psi(t)$, however I'm not fully aware which boundary conditions would be necessary from the provided information aside from the fact that $a(t)$ is a real number. Any hints?

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closed as off-topic by ACuriousMind, John Rennie, Danu, Martin, yuggib May 31 '15 at 9:32

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  • $\begingroup$ For your second proof, well it is just wrong because there exists operators, which are not hermitian but have real eigenvalues. eg. $\begin{pmatrix} 1 &2 \\1 &1 \end{pmatrix}$. What you have shown is if the operator is hermitian then the eigenvalues are real not the other way around. The first one seems ok. $\endgroup$ – Gonenc May 30 '15 at 21:39
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    $\begingroup$ BTW:Please note that this is not a homework help site. See How do I ask homework questions on Physics Stack Exchange? and Should any check my work questions be made on topic? posts on meta for more information. $\endgroup$ – Gonenc May 30 '15 at 21:41
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Statement 1: did you invent it yourself? the rough idea is correct nevertheless the statement as it is right now is false. Consider the following matrices $ \begin{pmatrix} 1 & 0 \\ 0 & A\end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 0 & B\end{pmatrix}$ they don't commute if $A$ and $B$ don't commute, yet they have a common eigenvector. One can certainly make a correct statement with some "for all" or "exists".

Note also that you assumed $C|\phi\rangle =c|\phi\rangle$, i.e. $|\phi\rangle$ eigenvector of C. It is not useful and doesn't modify the argument.

Statement 2: did you invent it yourself again? I guess all operators have at least an eigenvalue but that is not so obvious to begin with. Then I'm sure you implicitly assumed that your operator is normal, like in the spectral theorem. Then that may be true.

Otherwise the operator is not even "diagonalizable" (or the equivalent in possibly infinite dim. spaces) a fortiori not hermitian.

Statement 3: I'm sure you got it from some old russian book. I'm not familiar enough to be sure of what I'll say but by comparing with the eigenvector of the Hamiltonian operator I would say that the dependence of the vector can be entirely factorized in the following sense $$ |\psi(t)\rangle = a(t) |\psi(0)\rangle $$ In which case one just has to solve $\frac{d^4 a(t)}{dt^4} = a(t)$ and calling upon a vague souvenir I guess the general solutions are a constant factor times the exponential, so the function is completely determine by a single initial condition ( or here boundary condition)

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