Transformations of Operators and Symmetry

Symmetry of any quantum mechanical state is determined by how the state transforms under certain mathematical transformations, examples being translation and rotation. A symmetry transformation is a transformation that keeps the physical characteristics of the system unchanged (for example, a rotation of a spherical object). Of special importance are problems for which the Hamiltonian is left invariant under a symmetry transformation.

In addition, in both classical and quantum mechanics, symmetry transformations become important due to their relation to conserved quantities. Moreover in quantum mechanics the importance of symmetries is further enhanced by the fact that observation of conserved quantities can be exactly predictable in spite of the probabilistic nature of quantum predictions.

Let us consider an arbitrary transformation of an arbitrary state | n > to be given by the operator U such that the transformation gives | n > $\rightarrow$ U | n >

If U produce a symmetry transformation, the following theorems hold.

If the operator U produces a symmetry transformation on all ket vectors, then it must commute with the hamiltonian.

Proof: By definition of a symmetry transformation, the operator U could transform an energy eigenstate either to itself or another eigenstate degenerate to it. Hence, if | E_i > is an eigenstate of H with eigenvalue Ei then

$HU|E_{i}>=HU|E_{i}^{'}>=E_{i}|E_{i}^{'}>=E_{i}U|E_{i}>=UE_{i}|E_{i}>=UH|E_{i}>$

$|E_{i}>$ and $|E_{i}^{'}>$

Therefore we can write,

$\left[H,U\right]|E_{i}>=0$

This is valid for all energy eigenstates $|E_{i}>.$

Now From the completeness theorem any arbitrary state $|n>$

can be written as a linear combination of the eigenstates $|E_{i}>$ .

Hence, we can write,

$\left[H,U\right]|n>=0$

Since $|n>$

is an arbitrary ket vector, we can conclude that

$\left[H,U\right]=0$

Commutators & symmetry

We can define an operator called the parity operator, ${\hat {P}}$ which does the following:

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

The parity operator commutes with the Hamiltonian ${\hat {H}}$ if the potential is symmetric, ${\hat {V}}(r)={\hat {V}}(-r)$. 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, the solutions can be chosen to have definite parity (even and odd functions).