# Definitions of parity conservation

1. Jan 3, 2016

### Happiness

Definition 1: The expectation value of the observable related to the parity operator $\hat{P}$ is constant over time. That is,

$$\frac{d}{dt}\langle P\rangle=0$$
$$\int\Psi^*(r)\ \hat{P}\ \Psi(r)\ dr=constant$$
\begin{align}\int\Psi^*(r)\ \Psi(-r)\ dr=constant\end{align}

Definition 2: If the physical process proceeds in exactly the same way when referred to an inverted coordinate system, then parity is said to be conserved. If, on the contrary, the process has a definite handedness, then parity is not conserved in that physical process.

In particular, the expectation values of all observables $A$'s are invariant under the parity transformation. That is,

\begin{align}\int\Psi^*(r)\ \hat{A}\ \Psi(r)\ dr=\int\Psi^*(-r)\ \hat{A}\ \Psi(-r)\ dr\end{align}

I suppose both definitions are equivalent. How, then, do we prove (1) implies (2) and vice versa?

2. Jan 3, 2016

### Ravi Mohan

The expectation values of the observables may not be invariant under the parity transformation. Only those observables which commute with $\hat{P}$ will have invariant expectation values. This can be shown as follows
$$\langle \Psi|\hat{A}|\Psi\rangle = \langle \Psi|\hat{A}\hat{P}^\dagger\hat{P}|\Psi\rangle$$
since parity operator is unitary operator. And now if $[\hat{A},\hat{P}^\dagger]=0$, then the above expression becomes
$$\langle \Psi|\hat{P}^\dagger\hat{A}\hat{P}|\Psi\rangle$$
which proves the equality of equation 2 in OP.

Now if the time evolution of a physical process is invariant under parity that means $[\hat{P},\hat{H}]=0$ which further implies $\frac{\partial \hat{P}}{\partial t}=0$.

3. Jan 3, 2016

### Happiness

Is this a typo? $\frac{\partial \hat{P}}{\partial t}=0$ should be $\frac{\partial}{\partial t}\langle P\rangle=0$?

In other words, can I say that only for those observables whose expectation values are constant over time, ie., $\frac{\partial}{\partial t}\langle A\rangle=0$, are their expectation values invariant under the parity transformation?

By definition 1, $\frac{\partial}{\partial t}\langle P\rangle=0$, which implies $[\hat{P}, \hat{H}]=0$. Together with $\frac{\partial}{\partial t}\langle A\rangle=0$, which implies $[\hat{A}, \hat{H}]=0$, we have $[\hat{A}, \hat{P}]=0$, which as shown by you, implies its expectation value is invariant under the parity transformation.

How does this imply definition 2? Restricting the class of observables only to those where $\frac{\partial}{\partial t}\langle A\rangle=0$ seems to make definition 2 false.

4. Jan 3, 2016

### Ravi Mohan

Actually I abused the notation. $\frac{\partial \hat{P}_S}{\partial t}=0$ because parity has no intrinsic time dependence. The Heisenberg equation implies $\frac{\partial\hat{P}_H}{\partial t}=0$.

No.
If you have three operators $\hat{A}, \hat{B}\text{ and }\hat{C}$ such that $[\hat{A},\hat{B}]=[\hat{B},\hat{C}]=0$. Then this does not imply that $[\hat{A},\hat{C}]=0$.
The condition/restriction for the expectation values to be parity invariant is that the parity operator should commute with the observable (does not matter if the expectation value of the observable is time varying).

Now definition #2 "If the physical process proceeds in exactly the same way when referred to an inverted coordinate system ..." essentially means that parity operator commutes with the Hamiltonian. My earlier response was asserting that definition #2 implies definition #1.