Let's say I have a quantum system with 2 observables $\hat{O}_1$ and $\hat{O}_2$. Those are supposed to be "functions"of $\hat{X}$ and $\hat{P}$. Let's say I want to Transform $\hat{O}_1$ using an unitary transformation: $$\hat{O}_1' = \hat{U}^{-1} \hat{O}_1 \hat{U}$$ By doing that, do I have to transform the second operator in the same way?

Here is why I think I have to:

Instead of transforming the operator, transforming the states according to $| \Psi \rangle ' = \hat{U}| \Psi \rangle$ will yield exactly the same matrix elements for $\hat{O}_1$. But this operation will automatically also chanage the matrix elements of Operator $\hat{O}_2$ in the same way that transforming operator $\hat{O}_2$ would have yielded.

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    $\begingroup$ Let's say I have two numbers, $x$ and $y$. Let's say I want to add $3$ to the first of these numbers: $z=x+3$. By doing that, do I have to add $3$ to $y$ at the same time? (Answer: It depends entirely, of course, on what I'm trying to accomplish.) $\endgroup$ – WillO Jul 20 '17 at 13:25

Yes. A change of basis has to be global or otherwise your system will not give you the same physics. As an analogy imagine you want to rotate your basis vectors $\{\hat x,\hat y,\hat z\}$. To get another equivalent set of basis vectors you would rotate all basis vectors, not just some of them.

In the specific example you have, if you want to work with $\vert\Psi_k'\rangle$ and compute overlaps like $$ \langle \Psi'_m\vert {\hat O}_\alpha \vert\Psi'_k\rangle $$ it is quite easy to expand and see that $$ \langle \Psi_m\vert U^{-1} {\hat O}_\alpha U \vert\Psi_k\rangle $$ so that working with ${\hat O}'_\alpha = U^{-1} \hat O_\alpha U$ in the basis $\{\vert\Psi_k\rangle\}$ is the same as working with $\hat O_\alpha$ in the basis $\{\vert\Psi'_k\rangle\}$


Yes, your are right. You have to transform both. I try to explain it by analogy. If you have two matrices $A,B\in GL(n,\mathbb{R})$. By picking a basis you also determine the explicit form of these two matrices. What you do by transforming the matrices is that you transform the basis, thus both matrices look different now. The same applies for operators, which in quantum mechanics are kind of the infinite dimensional analogues of matrices.

As an example take the Pauli matrices, which represent the operators for measuring spin in $x,y$ and $z$ direction $$\sigma_x=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},\sigma_y=\begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix},\sigma_z=\begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}$$ With respect to the canonical basis $e_1=(1,0)$ and $e_2=(0,1)$ we find that $\sigma_z$ is diagonal with eigenvalue $\pm 1$ (corresponding to spin up or down). If I now choose another basis, let's say $\tilde e_1=(1,1)$ and $\tilde e_2=(1,-1)$ we find that $\sigma_x$ is diagonal with eigenvalues $\pm 1$, but $\sigma_z$ is not diagonal anymore.


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