Solving Parity in B.R. Martin's Nuclear and Particle Physics

[Bill Foster]

I'm reading a book: B.R. Martin's Nuclear and Particle Physics - An Introduction.

In section 1.3.1 on Parity, it states the following:

If the particle in an eigenfunction of linear momentum p, i.e.

$$\psi($$x$$,t)=\psi_p($$x$$,t)=$$exp$$[i($$p·x$$-Et)]$$

When dealing with p·x in the exponent, which should p and x be treated as - vectors or operators?

Suppose I work it out.

If exp$$[i($$p·x$$-Et)]$$ is an eigenfunction of momentum, then I assume that means that:

$$\hat{p}\psi($$x,t)=$$-i\hbar\frac{\partial}{\partial x}$$exp$$[i($$p·x$$-Et)]$$

Now if p and x are treated as vectors, then I will have to rewrite p·x as pxcosθ, which would lead to $$\hbar p \cos{\theta}$$exp$$[i($$p·x$$-Et)]$$, right?

Then the author goes on to say

and so a particle at rest, with p = 0, is an eigenstate of parity.

I don't really understand that. Because if I did the above math correctly, then if p = 0, then $$\hbar p \cos{\theta}$$exp$$[i($$p·x$$-Et)]$$ would also be 0.

Can someone clear this up for me? Thanks.

 

[CompuChip]

I'm reading a book: B.R. Martin's Nuclear and Particle Physics - An Introduction.

Sorry, don't know that one.

When dealing with p·x in the exponent, which should p and x be treated as - vectors or operators?

They should be vectors, because you are taking the inner product.

If exp$$[i($$p·x$$-Et)]$$ is an eigenfunction of momentum, then I assume that means that:

$$\hat{p}\psi($$x,t)=$$-i\hbar\frac{\partial}{\partial x}$$exp$$[i($$p·x$$-Et)]$$

Actually that doesn't say much. Basically you just wrote out what $\hat p$ looks like in "real"-space. Being an eigenfunction means that
$$\hat p \psi(x, t) = p \psi(x, t)$$
where the p on the right hand side is just a number. If you want to write it in real space, then you get
$$-i \hbar \frac{\partial}{\partial x} e^{i \vec p \cdot \vec x - E t} = P e^{i \vec p \cdot \vec x - E t}$$
I have called the eigenvalue P here, because from a mathematical point of view, it is just a number which needn't have anything to do with the momentum vector $\vec p$. In linear algebra, you would conventionally use $\lambda$ instead of P.

Now if p and x are treated as vectors, then I will have to rewrite p·x as pxcosθ, which would lead to $$\hbar p \cos{\theta}$$exp$$[i($$p·x$$-Et)]$$, right?

That is a possibility, where you go to polar coordinates in which you express a vector $\vec p$ in terms of a length $p$ and two angles $\theta, \phi$ and you choose your axes such that one of the variables $\theta$ is precisely the angle between $\vec p$ and $\vec x$ (that is, you make $\vec p$ lie along the z-axis).

Then the author goes on to say
"and so a particle at rest, with p = 0, is an eigenstate of parity."
I don't really understand that. Because if I did the above math correctly, then if p = 0, then $$\hbar p \cos{\theta}$$exp$$[i($$p·x$$-Et)]$$ would also be 0.

You did the math correctly. Note that if $\vec p = \vec 0$, then in your polar coordinates in which p is the radial coordinate, also $p = 0$. So even after multiplying by $\hbar \cos\theta e^\cdots$ it will be zero. That entire exercise was just to show that indeed the exponential is an eigenfunction of the momentum operator, and you can read of the eigenvalue (it's the lambda in $\hat p \psi = \lambda(p) \psi$).

I think the point the author is trying to make is, that if the momentum vector is zero, there is no dependence on $\vec x$ anymore. So if you flip all the spatial coordinates $\vec x \to -\vec x$ (meaning x to -x, y to -y, z to -z, t to t) then the eigenfunction won't change.

Hope that lifts some confusion?