# Transformations of Operators and Symmetry

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## Transformations of Operators

In the previous section, we discussed operators as transformations of vectors. In many cases, however, we will be interested in how operators, observables in particular, will transform under the action of another operator. Given an operator $\hat{A}$ and a transformation $\hat{T},$ we define the transformed operator $\hat{A}'$ as follows. Given the relation,

$\hat{A}|\psi\rangle=|\phi\rangle,$

between two vectors $|\psi\rangle$ and $|\phi\rangle$, the operator $\hat{A}'$ is the operator giving the relation between $|\psi'\rangle=\hat{T}|\psi\rangle$ and $|\phi'\rangle=\hat{U}|\phi\rangle;$ i.e.,

$\hat{A}'|\psi'\rangle=|\phi'\rangle.$

To find $\hat{A}',$ let us first act on both sides of the original relation with $\hat{T}:$

$\hat{T}\hat{A}|\psi\rangle=\hat{T}|\phi\rangle$

We now introduce the identity between $\hat{A}$ and $|\psi\rangle$ in the form, $\hat{T}^{-1}\hat{T}:$

$\hat{T}\hat{A}\hat{T}^{-1}\hat{T}|\psi\rangle=\hat{T}|\phi\rangle$

Using the above definitions of $|\psi'\rangle$ and $|\phi'\rangle,$ we may write this as

$\hat{T}\hat{A}\hat{T}^{-1}|\psi'\rangle=|\phi'\rangle$

We see then that the transformed operator $\hat{A}'=\hat{T}\hat{A}\hat{T}^{-1}.$ In matrix form, this would simply correspond to a similarity transformation of $\hat{A}.$

Of particular importance is the case in which $\hat{T}$ is unitary and $\hat{A}$ is an observable. This is because, in addition to preserving the normalization of the state vectors, as mentioned in the previous section, unitary transformations also preserve the Hermitian nature of $\hat{A}:$

$\hat{A}'^{\dagger}=(\hat{T}\hat{A}\hat{T}^\dagger)^\dagger=\hat{T}\hat{A}^\dagger\hat{T}^\dagger=\hat{T}\hat{A}\hat{T}^\dagger=\hat{A}'$

## Symmetry and its Role in Quantum Mechanics

Having discussed the transformation of operators, we will now apply our results to discuss symmetries of the Hamiltonian, a very important topic. As alluded to in the previous section, identifying the symmetries of the Hamiltonian will allow us to greatly simplify the problem at hand. In addition, in both classical and quantum mechanics, symmetry transformations become important due to their relation to conserved quantities via Noether's Theorem. In quantum mechanics, the importance of symmetries is further enhanced by the fact that measurements of conserved quantities can be exact in spite of the probabilistic nature of quantum predictions.

Given a unitary transformation $\hat{U},$ we say that it is a symmetry of the Hamiltonian if it leaves the Hamiltonian invariant; i.e., if $\hat{H}'=\hat{U}\hat{H}\hat{U}^\dagger=\hat{H}.$ We will now show that, if a transformation is a symmetry of the Hamiltonian, then it commutes with the Hamiltonian. To see this, let us take the relation,

$\hat{H}|\psi\rangle=|\phi\rangle,$

and act on both sides with $\hat{U}:$

$\hat{U}\hat{H}|\psi\rangle=\hat{U}|\phi\rangle$

Now, if $\hat{U}$ is a symmetry of the Hamiltonian, then it must also be true that

$\hat{H}\hat{U}|\psi\rangle=\hat{U}|\phi\rangle.$

Subtracting these two equations, we see that, because $|\psi\rangle$ is arbitrary, the Hamiltonian commutes with the transformation operator; i.e., $[\hat{H},\hat{U}]=0.$

This is a very important result; we know that, if two operators commute, then it is possible to simultaneously diagonalize them. This implies that every symmetry of the Hamiltonian has a "good quantum number" associated with it that we may use to describe the eigenstates of the Hamiltonian.

To help illustrate this fact, let us consider the parity, or inversion, operator, $\hat{P}:$

$\hat{P}f(x)=f(-x).$

The parity operator commutes with the Hamiltonian $\hat{H}$ if the potential is symmetric; i.e., $\hat{V}(x)=\hat{V}(-x)$. Since the two commute, the eigenfunctions of the Hamiltonian can be chosen to be eigenfunctions of the parity operator. This means that, if the potential is symmetric, then the eigenstates of the Hamiltonian can be chosen to have definite parity (even or odd).

## Problem

(From a Quantum Mechanics assignment in the Department of Physics, UF)

Consider an N state system, with N even and the states labeled as $|1\rangle, |2\rangle, \ldots, |N\rangle,$ described by the Hamiltonian,

$\hat{H}=\sum_{n=1}^{N} (|n\rangle \langle n+1| + |n+1\rangle \langle n|).$

Notice that the Hamiltonian, in this form, is manifestly Hermitian. Assume periodic boundary conditions; i.e, $|N+1\rangle = |1\rangle.$ One may therefore think of this Hamiltonian as describing a particle on a circle.

(a) Define the translation operator, $\hat{T},$ as taking $|1\rangle \to |2\rangle, |2\rangle \to |3\rangle ,...,|N\rangle \to |1\rangle.$ Write $\hat{T}$ in a form like $\hat{H}$ in the first equation and show that $\hat{T}$ is both unitary and commutes with $\hat{H},$ thus showing that $\hat{T}$ is a symmetry of the Hamiltonian.

(b) Find the eigenstates of $\hat{T}$ by using wavefunctions of the form,

$|\psi(k)\rangle = \sum_{n=1}^{N} e^{ikn}|n\rangle.$

What are the eigenvalues associated with these eigenstates? Do all these eigenstates have to be eigenstates of $\hat{H}$ as well? If not, do any of these eigenstates have to be eigenstates of $\hat{H}?$ Explain your answer.

(c) Next, consider the operator $\hat{F}$ which takes $|n\rangle \to |N+1-n\rangle.$ Write $\hat{F}$ in a form like $\hat{H}$ in the first equation and show that $\hat{F}$ is both unitary and commutes with $\hat{H},$ thus showing that $\hat{F}$ is also a symmetry of the Hamiltonian.

(d) Find a complete set of eigenstates of $\hat{F}$ and their associated eigenvalues. Do all these eigenstates have to be eigenstates of $\hat{H}$ as well? If not, do any of these eigenstates have to be eigenstates of $\hat{H}?$ Explain your answer.