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Linear superposition of single-particle states

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Neitrino
#1
Sep2-05, 09:28 AM
P: 133
Dear all,

I am not sure whether I understand correctly or not.
So from Peskin Schroeder’s book:
[tex]\phi(x)|0>=
\int{\frac{d^3 p}{(2\pi)^3}\frac{1}{2E_p}e^{-ipx}|p>
[/tex]
formula (2.41). Interpreting this formula they say – it’s a linear superposition of single-particle states that have well defined momentum. And also that operator phi(x) acting on the vacuum, creates a particle at position x.
My question – since it is a superposition of single-particle states and creates a particle at position X, So that operator creates many single-particle states with different momentum (since there is integration over p and each single-particle state has different momentum) and all of them (particles with different momentum ) are created at one position X?
Or briefly – many different momentum particles are created at one position X?

Thanks
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vanesch
#2
Sep2-05, 01:57 PM
Emeritus
Sci Advisor
PF Gold
P: 6,236
Quote Quote by Neitrino
My question – since it is a superposition of single-particle states and creates a particle at position X, So that operator creates many single-particle states with different momentum (since there is integration over p and each single-particle state has different momentum) and all of them (particles with different momentum ) are created at one position X?
Or briefly – many different momentum particles are created at one position X?
What's the superposition of a one-particle and a one-particle state ? A two-particle state or another one-particle state ? Answer: another one-particle state. Superpositions of N-particle states are again N-particle states.
So you should view this as ONE particle is created, in a superposition of momentum states, exactly as in NR quantum mechanics, where ONE position state is written as (about the same) superposition of several momentum states.

cheers,
Patrick.


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