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Bill Foster
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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:
Suppose I work it out.
If exp[tex][i([/tex]p·x[tex]-Et)][/tex] is an eigenfunction of momentum, then I assume that means that:
[tex]\hat{p}\psi([/tex]x,t)=[tex]-i\hbar\frac{\partial}{\partial x}[/tex]exp[tex][i([/tex]p·x[tex]-Et)][/tex]
Now if p and x are treated as vectors, then I will have to rewrite p·x as pxcosθ, which would lead to [tex]\hbar p \cos{\theta}[/tex]exp[tex][i([/tex]p·x[tex]-Et)][/tex], right?
Then the author goes on to say
I don't really understand that. Because if I did the above math correctly, then if p = 0, then [tex]\hbar p \cos{\theta}[/tex]exp[tex][i([/tex]p·x[tex]-Et)][/tex] would also be 0.
Can someone clear this up for me? Thanks.
In section 1.3.1 on Parity, it states the following:
When dealing with p·x in the exponent, which should p and x be treated as - vectors or operators?If the particle in an eigenfunction of linear momentum p, i.e.
[tex]\psi([/tex]x[tex],t)=\psi_p([/tex]x[tex],t)=[/tex]exp[tex][i([/tex]p·x[tex]-Et)][/tex]
Suppose I work it out.
If exp[tex][i([/tex]p·x[tex]-Et)][/tex] is an eigenfunction of momentum, then I assume that means that:
[tex]\hat{p}\psi([/tex]x,t)=[tex]-i\hbar\frac{\partial}{\partial x}[/tex]exp[tex][i([/tex]p·x[tex]-Et)][/tex]
Now if p and x are treated as vectors, then I will have to rewrite p·x as pxcosθ, which would lead to [tex]\hbar p \cos{\theta}[/tex]exp[tex][i([/tex]p·x[tex]-Et)][/tex], 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 [tex]\hbar p \cos{\theta}[/tex]exp[tex][i([/tex]p·x[tex]-Et)][/tex] would also be 0.
Can someone clear this up for me? Thanks.
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