Chapter 6. Symmetries and Conservation Laws - (4)

6.4 Parity

6.4.1 Parity in 1D

A spatial inversion is implemented by the parity operator \[\hat{\Pi}\]

In 1D, \[\hat{\Pi}\psi(x)=\psi'(x)=\psi(-x)\]

Now we talk about properties of Parity operator.

\[\hat{Pi^{i}=\hat{\Pi}\] 

\[\hat{\Pi}^{-1}=\hat{\Pi}\]

 

And \[\hat{\Pi}\] is a hermitian operator, so we get \[\hat{\Pi}^{\dagger}=\hat{\Pi}\].

Here's the reason. 

\[<\psi|\hat{\Pi}\psi>=<\hat{\Pi}^{\dagger}\psi|\psi>=<\hat{\Pi}\psi|\psi>=\int_{\infty }^{\infty} \psi^*(-x)\psi(x)dx=\int_{\infty }^{\infty} \hat{\Pi}^*\psi(x)\psi(x)dx=<\hat{\Pi}\phi|\phi>\]

Replace \[x=-x'\] 

 

Since \[\hat{\Pi}\] is unitary.

\[\hat{\Pi}^* =\hat{\Pi}=\hat{\Pi}\]

 

For operator Q, \[\hat{\Pi}|\phi_{n}>=\lambda|\phi_{n}>\]

\[|\lambda_{n}|^2=1\]

 

Operators transform under a spatial inversion as \[\hat{Q}'=\hat{\Pi}^{\dagger}\Q\hat{\Pi} \]

\[\hat{x}'=\hat{\Pi}^{\dagger}\hat{x}\hat{\Pi}=-\hat{x}\]

\[\hat{p}'=\hat{\Pi}^{\dagger}\hat{p}\hat{\Pi}=-\hat{p}\]

 

\[\hat{p'}\phi(x)=\hat{\Pi}^{\dagger}\hat{Pi}\phi(x)=\Pi^{\dagger}\hat{p}\phi(-x) = \hat{\Pi}^{\dagger} [-i\hbar \frac{\partial \phi(-x)}{\partial x}]=\hat{\Pi}[(-i\hbar\frac{\partial \phi(-x)}{\partial x})]=-i\hbar \frac{\partial \phi(x) }{\partial -x}] = -\hat{p}\phi(x) \]

The system has inversion symmetry if the hamiltonian is unchanged by a parity transformation \[\hat{H}=\hat{\Pi}^{\dagger}\hat{H}\hat{\Pi}=\hat{H}\]

(which means \[\hat{\Pi}\] is an inversion operator. and it is a commute with \[H\].

Which means a system has inversion symmetry(Hamiltonian is unchanged by a parity transformation)

 

Now let's consider a particle of mass m in a 1D potential V(x). The inversion symmetry means, again V(-x)=V(x), potential function is even.

 

The implications of inversion symmetry.

1. We can find a complete set of simultaneous eigenstates of \[\hat{\Pi}] and \[\hat{H}]
   \[\hat{\Pi}\psi_{n} (x)=\pm\psi_{n}(x)\]
2. According to Ehrenfest's theorem, if the hamiltonian has an inversion symmetry, parity is conserved.
   \[\frac{\mathrm{d <\Pi>} }{\mathrm{d} x}=\frac{i}{\hbar}<[\hat{\Pi},\hat{\Pi}]>=0\]

6.4.2 Parity in 3D

\[\hat{\Pi}\psi(\vec{r})=\psi(-\vec{r})\]

\[\hat{r}'=\hat{\Pi}^{\dagger}\hat{r}\hat{\Pi}=-\hat{r}\]

\[\hat{p}'=\hat{\Pi}^{\dagger}\hat{p}\hat{\Pi}=-\hat{p}\]

\[\hat{Q}'(\hat{r},\hat{p})=\hat{\Pi}^{\dagger}\hat{Q}(\hat{r},\hat{p})\hat{\Pi}=\hat{Q}'(\hat{-r},\hat{-p})\]

 

We solved Example 6.2. and now we know the difference between scalar and vector, they show different act about rotation operator.

 

6.4.3 Parity Selection Rules

Selection rules tell us when the matrix element is zero based on the symmetry of the situation. 

Matrix element 

\[<b|\hat{Q}|a>\] -------------------------(1)

 

\[\hat{H}, \hat{L}^{2},\hat{L}_{z}\] are all commute, which means they have common eigenstates.

 

Consider the matrix element of electric dipole operator : 

\[\hat{p}_{e}=q\hat{r}\], 

which is odd under parity.

\[\hat{\Pi}^{\dagger}\hat{p}_{e}\hat{\Pi}=-\hat{p}_{e}\]

The selection rules for the electric dipole operator determine which atomic transitions are allowed and which are forbidden.

 

Now consider the matrix elements of the electric dipole operator, by using same method(?) we used in (1)

\[<n'l'm'|\hat{p}_{e}|nlm>=<n'l'm'|\hat{\Pi}^{\dagger}\hat{p}_{e}\hat{\Pi}|nlm>=<n'l'm'|\(-1)^{l'}\hat{p}_{e}(-1)^{l}|nlm>=(-1)^{l+l'+1}<n'l'm'|\hat{p}_{e}|nlm>\]

 

So we see \[<n'l'm'|\hat{p}_{e}|nlm>=0, if/l+l'/is/even\]

This is called Laporte's rule

Matrix elements of the electric dipole moment operator vanish between states with the same parity.

 

Reference: Introduction to quantum mechanics  3rd ed. (by David J.Griffith)

 

그나저나 Bra-Ket notation을 여기서 치려니 잘 안 쳐지네요.

    \documentclass[12pt,twoside]{article}
    \usepackage{amsmath}
    \usepackage{amssymb}
    \usepackage{amsthm}
    \usepackage{amsfonts}
    \usepackage{braket}

    \begin{document}
    \begin{equation}
    \ket \psi{(t)}
    \end{equation}
    \end{document}

 

뭐 이런것 같은데, 해도 안 되고;;;