Transformations of Operators and Symmetry
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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{A}} and a transformation Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{T},} we define the transformed operator Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{A}'} as follows. Given the relation,
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{A}|\psi\rangle=|\phi\rangle,}
between two vectors Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\psi\rangle} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\phi\rangle} , the operator Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{A}'} is the operator giving the relation between Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\psi'\rangle=\hat{T}|\psi\rangle} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\phi'\rangle=\hat{U}|\phi\rangle;} i.e.,
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{A}'|\psi'\rangle=|\phi'\rangle.}
To find Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{A}',} let us first act on both sides of the original relation with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{T}:}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{T}\hat{A}|\psi\rangle=\hat{T}|\phi\rangle}
We now introduce the identity between Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{A}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\psi\rangle} in the form, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{T}^{-1}\hat{T}:}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{T}\hat{A}\hat{T}^{-1}\hat{T}|\psi\rangle=\hat{T}|\phi\rangle}
Using the above definitions of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\psi'\rangle} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\phi'\rangle,} we may write this as
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{T}\hat{A}\hat{T}^{-1}|\psi'\rangle=|\phi'\rangle}
We see then that the transformed operator Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{A}'=\hat{T}\hat{A}\hat{T}^{-1}.} In matrix form, this would simply correspond to a similarity transformation of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{A}.}
Of particular importance is the case in which Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{T}} is unitary and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{A}:}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{U},} we say that it is a symmetry of the Hamiltonian if it leaves the Hamiltonian invariant; i.e., if Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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,
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{H}|\psi\rangle=|\phi\rangle,}
and act on both sides with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{U}:}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{U}\hat{H}|\psi\rangle=\hat{U}|\phi\rangle}
Now, if Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{U}} is a symmetry of the Hamiltonian, then it must also be true that
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{H}\hat{U}|\psi\rangle=\hat{U}|\phi\rangle.}
Subtracting these two equations, we see that, because Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\psi\rangle} is arbitrary, the Hamiltonian commutes with the transformation operator; i.e., Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [\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, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{P}:}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{P}f(x)=f(-x).}
The parity operator commutes with the Hamiltonian Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{H}} if the potential is symmetric; i.e., Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle N} state system, with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle N} even and the states labeled as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |1\rangle, |2\rangle, \ldots, |N\rangle,} described by the Hamiltonian,
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |N+1\rangle = |1\rangle.} One may therefore think of this Hamiltonian as describing a particle on a circle.
(a) Define the translation operator, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{T},} as taking Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |1\rangle \to |2\rangle, |2\rangle \to |3\rangle ,...,|N\rangle \to |1\rangle.} Write Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{T}} in a form like Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{H}} in the first equation and show that is both unitary and commutes with thus showing that is a symmetry of the Hamiltonian.
(b) Find the eigenstates of by using wavefunctions of the form,
What are the eigenvalues associated with these eigenstates? Do all these eigenstates have to be eigenstates of as well? If not, do any of these eigenstates have to be eigenstates of Explain your answer.
(c) Next, consider the operator which takes Write in a form like in the first equation and show that is both unitary and commutes with thus showing that is also a symmetry of the Hamiltonian.
(d) Find a complete set of eigenstates of and their associated eigenvalues. Do all these eigenstates have to be eigenstates of as well? If not, do any of these eigenstates have to be eigenstates of Explain your answer.