I Symmetries in quantum mechanics and the change of operators

  • Thread starter Lebnm
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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

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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##
 
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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

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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##
 

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