Undergrad Symmetries in quantum mechanics and the change of operators

[Lebnm]

When we make a symmetrie transformation in a quantum system, the state $|\psi \rangle$ change to $|\psi' \rangle = U|\psi \rangle$, where $U$ is a unitary or antiunitary operator, and the operator $A$ change to $A'$. If we require that the expections values of operators don't change, we have $$\langle \psi '| A' | \psi ' \rangle= \langle \psi | U^{\dagger}A'U | \psi \rangle =\langle \psi | A | \psi \rangle,$$ what suggest that $A' = UAU^{\dagger}$. However, in some textbooks, like Sakurai, the change in the operator is $A' = U^{\dagger}AU$. Why this difference appear? They are equivalent?

 

[stevendaryl]

Check to make sure of the definition of $U$. Are they saying $|\psi'\rangle = U |\psi\rangle$ or $|\psi\rangle = U |\psi'\rangle$?

In the first case, $A' = U A U^\dagger$, in the second case, $A' = U^\dagger A U$

 

[Lebnm]

Thanks for answering. I think it's the first case. In section 4.3, Sakurai talk about discrete symmetries. The first example is the lattice translation; he considerer a periodic potential $V$ and define the translation operator $\tau$ such that $\tau (l) |x \rangle = |x + l \rangle$ ($| x \rangle$ is a eigenstate of the position operator). So he write $$\tau ^{\dagger}(a) V(x) \tau (a) = V(x + a).$$ But shouldn't it be: $\tau(a) V(x) \tau ^{\dagger}(a) = V(x + a)$?

 

[stevendaryl]

I think it's confusing to mix up the operator $\hat{V}$ with its value $V(x)$, which is just a number. So here's the way I would say it:

$\hat{V}$ is an operator which acts on position eigenstates as follows:

$\hat{V} |x\rangle = V(x) |x\rangle$

The $V(x)$ on the right side of the equals is just a number, not an operator, so acting on it with $\tau(a)$ or $\tau^\dagger(a)$ would do nothing. Similarly,

$\hat{V} |x+a\rangle = V(x+a) |x+a\rangle$

which is equivalent to:

$\hat{V} \tau(a) |x\rangle = V(x+a) \tau(a) |x\rangle$

Now, act on both sides with $\tau^\dagger(a)$:

$\tau^\dagger(a) \hat{V} \tau(a) |x\rangle = \tau^\dagger(a) V(x+a) \tau(a) |x\rangle$

But remember, on the right-hand side, $V(x+a)$ is just a number, not an operator. So $\tau^\dagger(a)$ commutes with it, to give:

$\tau^\dagger(a) V(x+a) \tau(a) |x\rangle = V(x+a) \tau^\dagger(a) \tau(a) |x\rangle = V(x) |x\rangle$ (since $\tau^\dagger(a) = (\tau(a))^{-1}$).

So we have:
$\tau^\dagger(a) \hat{V} \tau(a) |x\rangle = V(x+a) |x\rangle$

So $\tau^\dagger(a) \hat{V} \tau(a)$ is the operator which acts on $|x\rangle$ and returns $V(x+a) |x\rangle$