Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 11, Exercises 11.4.1 – 11.4.4.
A parity transformation reflects all the coordinate axes through the origin, so that, in one dimension and in three dimensions the position vector . In one dimension, a parity transformation is the same as reflection in a point-sized mirror placed at the origin. It might seem that in three dimensions, parity is more than just a reflection in a plane mirror, but in fact it can be shown that it is equivalent to such a reflection followed by a rotation. To see this, suppose we place a mirror in the plane, so that the axis gets reflected into . This converts a right-handed rectangular coordinate system (where the direction of the axis is determined by the direction of your thumb on your right hand when you curl your fingers through the right angle between the positive and axes) into a left-handed coordinate system (the direction of the new axis is found by doing the finger-curling maneuver with your left hand). However, merely reflecting the axis in the plane leaves the and axes unchanged. Now if we rotate the plane by an angle (or ) about the axis, then the axis gets rotated into the axis, and the axis gets rotated into the axis. In this sense, the 3-d parity transformation is equivalent to a reflection (since pretty well every physical phenomenon is invariant under a rotation).
To apply parity to quantum state vectors, we define a parity operator to have the following action on the basis:
From this definition we can see the effect on an arbitrary state by inserting a complete set of states:
In the third line we made the substitution , so that and the limits of integration get swapped. As a result of this, the effect of parity in the basis representation of a state vector is
Parity therefore simply converts wherever it occurs in the function .
One special case of this is the momentum eigenstate which has the form in the basis of
The parity transformation gives
Another way of looking at this is that parity changes to and leaves the alone, so that
[You might think that if parity transforms and then the effect on should be to switch the signs of both and and thus leave the state unchanged. However, this isn’t correct, as we can express a state vector in either the basis (in which ) or in the basis (in which ) but not both at the same time.]
A few properties of can be derived fairly easily. First, since applying twice in succession to the same state swaps and back again, it leaves that state unchanged. Since this is true for all states, we must have
from which we see that is its own inverse, so
We can also see that is Hermitian by considering
In the second line we used the same trick as in the derivation of 5 to substitute . Thus we see that
The condition shows that is Hermitian, and the condition shows that is unitary.
Finally, any operator whose square is the identity operator has eigenvalues , as we can see as follows. Suppose is an eigenvector of with eigenvalue . Then
Therefore , so .
We can also define by examining its effect on operators, rather than states. Consider
However, this is equivalent to
Thus we can write
and similarly for the momentum
Eigenstates of parity are said to be even if the eigenvalue is and odd if the eigenvalue is . Mathematically, the basis representation of such eigenstates are even or odd functions of , respectively.
The Hamiltonian is parity invariant if a parity transformation leaves it unchanged, so that
Since , this condition is equivalent to
Using the same argument as with conservation of momentum, if this commutator is valid at all times (if is time-independent this is automatic; if is time-dependent, then we must impose the commutator at all times), then must also commute with the propagator , since depends only on . In this case, if we start with a system in a definite parity state (even or odd), then the parity of the state doesn’t change with time. This follows because if then if (where ), then we can let the state evolve in time by applying the propagator to it, so that we have
Applying the parity operator to this and using the commutator, we have
Thus the parity of the evolved state is the same as the parity of the initial state.
Parity is not always conserved in physics. A notable parity-violating reaction is a decay involving the weak nuclear force. Shankar describes one such case with the decay of an isotope of cobalt: . Another example is in Shankar’s exercise 11.4.3.
Suppose that in one particular reaction which emits an electron, the electron’s spin is observed to be always parallel to its momentum. For the purposes of this argument, we can regard an electron’s spin as being caused by some physical rotation of the electron. Suppose in one such reaction, the electron’s spin is in the direction (using the right-hand rule for calculating the direction of angular momentum, so that viewed from above, the electron is rotating counterclockwise) and therefore its momentum is also in the direction. Now reflect this reaction in a mirror lying in the plane. This reflection will invert the direction of rotation (think of viewing a spinning top in a mirror) so that the spin direction will now point in the direction, but since the momentum vector is parallel to the plane of the mirror, it will not be inverted. Thus the spin and momentum are now anti-parallel after a parity transformation, showing that parity in this case is not conserved.
Finally, Shankar includes a curious problem (11.4.2) which, as far as I can tell, doesn’t have anything to do with parity, but I’ll include it here for completeness. Suppose we have a particle that moves in a potential
This potential is periodic with a period of , so if we translate the system according to for some integer , the potential is unchanged. The problem is to show that momentum is not conserved in this case. The conservation of momentum argument, valid for infinitesimal translations, relied on Ehrenfest’s theorem, which states that
If the momentum commutes with the Hamiltonian, then, on average, the momentum is conserved. Now in this case we can calculate the commutator using the result
We can write the potential as a series:
The commutator is therefore
Therefore, Ehrenfest’s theorem gives us (since presumably is of the form with the kinetic energy depending only on , so it commutes with ):
Since the cosine is periodic, we can’t actually calculate a unique value for its average, although if we do the average over an exact number of periods, the average is still zero. I have a feeling that I’m missing something obvious here, so any suggestions are welcome.