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For simplicity, I consider a 2 dimensional Hilbert space.

In Q.M we have a freedom of choice for a global phase of a wavefunction. But here I am interested in the freedom of choice we could have for an operator.

Consider an operator $A$.

I write it as : $$A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$$

I have a memory that during a class a teacher said that we have a freedom to chose the absolute phase of this matrix. So, I can change $A$ to $A e^{i \phi}$ and all the prediction I will make will be unchanged.

However I'm not sure if my memory is good or not because if I take the example of $A$ being an observable, the elements on the diagonal will then be complex numbers.

I would like to know if indeed we don't have such degree of freedom on operators in Q.M : can we multiply the operators we consider by a given phase ? Because for me it seems to be wrong.

The only way to have a compatibility for me would be to multiply all the operators used in the problem by the same $e^{i \phi}$ because then when applied on the wave function, it would be equivalent as adding a global phase on the wavefunction.

But I would like to check with you if it is the only possible operation we can do ?

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I suspect your memory (or your teacher? surely not!) is at fault. If an operator represents an observable then it must be Hermitian: $A^\dagger=A$. So $a$ and $d$ must be real, and $c=b^*$.

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You can choose the phase of the state vectors freely. But a change of phase by $\mathrm{e}^{\mathrm{i}\phi}$ on the vectors acts on operators as $A\mapsto \mathrm{e}^{\mathrm{i}\phi}A\mathrm{e}^{-\mathrm{i}\phi} = A$, leaving them unchanged.

There is another sense in which the "irrelevant phase" can act on operators, in that symmetry groups should not be represented as unitary operators, but projective unitary operators. However, in this case it only acts on unitary operators $U$ as $U\mapsto U\mathrm{e}^{\mathrm{i}\phi}$, which shifts the Hermitian observable $H$ with $U=\mathrm{e}^{\mathrm{i}Ht}$ (for some $t\in\mathbb{R}$) by the phase as $H\mapsto H+\phi$. Shifting all values of an observable by a constant doesn't change anything about the physics, either - it's just choosing the zero value for the observable.

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It's not that hard to give a rule-based answer to this question, which is that observables in quantum mechanics are supposed to be self-adjoint, so that if $A$ is an observable, then $e^{i\phi}A$ will in general not be an observable. This is equivalent to the condition that all eigenvalues of $A$ be real.

But I think it's more helpful to understand why the rule should be that way. We do, after all, have measurements in the real world that are complex. For example, I can measure the impedance of a circuit element, and it can be imaginary. All we really need is for our operator to have the property that its eigenvectors are orthogonal to one another, so that if we measure the same observable twice in a row on the same system, we get the same answer the second time. For this to happen, it turns out that it's enough for the operator to be normal, which is defined by $A^*A=AA^*$. So this may have been what your instructor had in mind. If $A$ is normal, making it a possible observable, then so is $iA$, for example.

But we should consider what the complex phase of a measurement in quantum mechanics would really mean. Whenever we do an application of the complex number system, we have to deal with the fact that there are two square roots of $-1$, which are $i$ and $-i$. Since the mathematical properties of these two roots are identical, there is always some arbitrary choice involved. For example, capacitors have negative imaginary impedance and inductors a positive imaginary impedance, but this is an arbitrary convention. We could flip it around.

In the application to quantum mechanics, we do have some phase conventions, such as that for a state with positive energy, the complex phase rotates clockwise, like $e^{-i\omega t}$, not counterclockwise like $e^{i\omega t}$. These phase conventions are purely conventions. We can never test which convention is right, because the structure of quantum mechanics keeps us from ever measuring a phase except relative to another phase. This disconnects the phase-conventions of wavefunctions from any other set of phase-conventions that we might establish, such as the one for impedance. This disconnection means that if $A$ is hermitian, then there is never any way to assign a different physical significance to $iA$ or $-iA$. The non-hermitian versions $e^{i\phi}A$ are redundant, so we don't lose anything by just ruling them out and insisting on the hermitian one.

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I believe what you actually want is the following:

For $A$ being a $2\times 2$ Hermitian matrix (so that it's an observable), you are writing it is in a specific complete basis

$\phi_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$

$\phi_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$

In principle you can choose any orthonormal complete basis you like, which corresponds to a unitary transformation $U$ on your basis $\phi_1 \rightarrow \phi_1' = U \phi_1$, $\phi_2 \rightarrow \phi_2' = U \phi_2$. In the new basis, $A \rightarrow A'=U A U^{-1}$.

Now if you don't bother to change your basis so generally, but only want to change the phase of each basis, then you have

$U = \begin{bmatrix} e^{i \theta_1} & 0 \\ 0 & e^{i \theta_2} \end{bmatrix}$

which is just a special case, so that you still have $A \rightarrow A'=U A U^{-1}$, which is still Hermitian because $U$ is unitary.

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