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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.

Urb
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DentPanic42
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2 Answers2

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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 $b'=a'$.

Young Kindaichi
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  • 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? – By Symmetry Oct 06 '21 at 14:45
  • Was this ever answered^? – EE18 May 31 '22 at 15:58
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They have the same eigenvectors (modulo some degeneracy) but they can have different eigenvalues. This can be obvious by making the trivial choice $U=I$ : for example $U^{\dagger}AU=\mathrm{diag}(1, -1)$ and $B=\mathrm{diag}(1, 0)$ corresponds to a case where $a'$ is not always equal to $b'$.

The physical interpretation of this case would correspond to observable which have the same eigenstates, so they can be measured simultaneously, but they do not necessary have the same eigenvalues. They are compatible, but do not measure the same physical quantity

Frédéric Grosshans
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