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In J.J. Sakurai's Modern Quantum Mechanics, he introduces the concept of 'Unitary Equivalent Observables'.

If $|a^{'}\rangle$ and $|b^{'}\rangle$ are the orthonormal bases eigenkets of two non-commuting hermitian operators connected by the unitary operator $\hat{U}$ ($|b^{'}\rangle = \hat{U}|a^{'}\rangle$), we can construct a unitary transform of $\hat{A}$ as $\hat{U}\hat{A}\hat{U^{\dagger}}$ such that $$\hat{A}|a^{'}\rangle=a^{'}|a^{'}\rangle$$ and $$\hat{U}\hat{A}\hat{U^{\dagger}}|b^{'}\rangle=a^{'}|b^{'}\rangle.$$

However, the $|b^{'}\rangle$ satisfy: $$\hat{B}|b^{'}\rangle=b^{'}|b^{'}\rangle.$$

So, $\hat{B}$ and $\hat{U}\hat{A}\hat{U^{\dagger}}$ are simultaneously diagonalizable.

My question is this. Are $\hat{B}$ and $\hat{U}\hat{A}\hat{U^{\dagger}}$ the same operators? Sakurai's own example using the spin operators $\hat{S_{x}}$ and $\hat{S_{z}}$ suggests so. Is there an example where they aren't?

Also, what exactly does $\hat{U}\hat{A}\hat{U^{\dagger}}$ signify? As in, $\hat{U}\hat{A}\hat{U^{\dagger}}$ is the expression we'd get if we knew $\hat{A}$ and $\hat{U}$ in the $|b^{'}\rangle$ basis and wanted to know its matrix elements in the $|a^{'}\rangle$ basis.

$$\langle a^{'}|\hat{A}|a^{''}\rangle = \langle a^{'}|\hat{U^{\dagger}}\hat{U}\hat{A}\hat{U^{\dagger}}\hat{U}|a^{''}\rangle = \langle b^{'}|\hat{U}\hat{A}\hat{U^{\dagger}}|b^{''}\rangle$$

How does that expression then give us an eigenvalue equation with $|b^{'}\rangle$? I think I'm missing something obvious here.

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If you look at the passive picture of the transformation, it will be easier to understand. For an Active picture, you look at here.

Suppose an active transformation under a unitary transformation $U$ such that for all vector $$|V\rangle\rightarrow U|V\rangle$$

Under this transformation, the matrix elements of any operator $\Omega$ are modified $$\langle V'|\Omega|V\rangle\rightarrow \langle UV'|\Omega|UV\rangle=\langle V'|U^\dagger\Omega U|V\rangle$$ It's clear that the same change would be affected if we left the vectors alone and subjected all operators to the change $$\Omega \rightarrow U^\dagger \Omega U$$

which is called passive transformation.


Back to original problem: $$A|a'\rangle=a'|a'\rangle$$ which is an eigen equation.

Under an active transformation : $$|b'\rangle=U|a'\rangle$$ which has the same effect if we change the operator

$$A\rightarrow U^\dagger A U$$ so that the Eigen equation will become $$(U^\dagger A U) |a'\rangle=a'|a'\rangle$$

Now let's see the following : $$B|b'\rangle=b'|b'\rangle$$ under the active transformation $|a'\rangle=U^\dagger|b'\rangle$ : $$B\rightarrow U BU^\dagger$$ so that in passive picture:

$$U BU^\dagger |b'\rangle = b'|b'\rangle$$

putting $|b'\rangle=U|a'\rangle$ : $$B|a'\rangle=b'|a'\rangle$$ comparing two equation : $B=U^\dagger AU$ and $b'=a'$.

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  • $\begingroup$ Why exactly are you able to claim that the original eigenvalue equation is still valid after an active transformation? Clearly $U^\dagger A U |b'\rangle = a' |b'\rangle$ but I don't see why $U^\dagger A U |a'\rangle = a' |a'\rangle$ should be true in general. Is there an extra assumption that I am missing here? $\endgroup$ Oct 6 at 14:45

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