What is the missing factor in equations (10.58) and (10.60)?

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Homework Help Overview

The discussion revolves around the equations (10.58) and (10.60) from a physics text, specifically focusing on the structure and potential missing factors in these equations related to field theory. Participants are examining the implications of summing over momentum states and the treatment of positive and negative momentum components.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are analyzing the relationship between equations (10.57) and (10.58), questioning whether a factor of 2 is missing in (10.58) due to the treatment of momentum states. There is also discussion about the implications of summing over both positive and negative momentum components and how this affects the terms derived from (10.57).

Discussion Status

The discussion is ongoing, with participants expressing uncertainty about the correctness of equations (10.63) and (10.64) in relation to (10.60) and (10.61). Some participants are exploring the origins of the commutator in the context of the equations, while others are seeking clarification on the assumptions regarding the momentum components.

Contextual Notes

There is mention of specific constraints regarding the treatment of momentum components in the equations, particularly the distinction between positive and negative values of \vec{p}. Participants are also referencing specific pages and equations from the text to support their arguments.

Jimmy Snyder
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Homework Statement


Equation (10.58) is:
[tex]\phi(t, \vec{x}) = \frac{1}{\sqrt{V}}\Sigma_{\vec{p}}\frac{1}{\sqrt{2E_p}}(a_{p}e^{-iE_pt + i\vec{p}\cdot\vec{x}} + a_p^{\dagger}e^{iE_pt - i\vec{p}\cdot\vec{x}})[/tex]

Homework Equations


Here is equation (10.57)
[tex]\phi_{p}(t, \vec{x}) =\frac{1}{\sqrt{V}}\frac{1}{\sqrt{2E_p}}(a_{p}e^{-iE_pt + i\vec{p}\cdot\vec{x}} + a_p^{\dagger}e^{iE_pt - i\vec{p}\cdot\vec{x}})[/tex]
[tex]+\frac{1}{\sqrt{V}}\frac{1}{\sqrt{2E_p}}(a_{-p}e^{-iE_pt - i\vec{p}\cdot\vec{x}} + a_{-p}^{\dagger}e^{iE_pt + i\vec{p}\cdot\vec{x}})[/tex]

The Attempt at a Solution


The idea is that the second term on the r.h.s. of (10.57) is the same as the first term evaluated for [itex]\vec{p} = -\vec{p}[/itex], which does not effect [itex]E_p[/itex]. Then (10.58) is supposed to be the sum of (10.57) over all values of [itex]\vec{p}[/itex]. My problem is that I think there is a factor of 2 missing on the r.h.s. of (10.58) because each of the terms in (10.57) should appear twice in the sum. What am I missing? The same problem arises on page 176 for equations (10.60) and (10.61) which are sums of (10.55) and (10.56) respectively.
 
Last edited:
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You're argument makes sense, but I don't understand it well enough to say conclusively. 10.57 could be the equation for both p and -p, but again I am not sure.

I'll post back if this becomes clear to me.

I do think 10.63 and 10.64 are wrong if 10.60 and 10.61 are correct, however. Where does the commutator come from if both equations sum over all possible values of a their vector index?
 
Last edited:
ehrenfest said:
I do think 10.63 and 10.64 are wrong if 10.60 and 10.61 are correct, however. Where does the commutator come from if both equations sum over all possible values of a their vector index?
No, I think (10.63) and (10.64) follow from (10.60) and (10.61). The product equals the commutator in this case because the annihilator annihilates the vacuum.
 
jimmysnyder said:

Homework Statement


Equation (10.58) is:
[tex]\phi(t, \vec{x}) = \frac{1}{\sqrt{V}}\Sigma_{\vec{p}}\frac{1}{\sqrt{2E_p}}(a_{p}e^{-iE_pt + i\vec{p}\cdot\vec{x}} + a_p^{\dagger}e^{iE_pt - i\vec{p}\cdot\vec{x}})[/tex]

Homework Equations


Here is equation (10.57)
[tex]\phi_{p}(t, \vec{x}) =\frac{1}{\sqrt{V}}\frac{1}{\sqrt{2E_p}}(a_{p}e^{-iE_pt + i\vec{p}\cdot\vec{x}} + a_p^{\dagger}e^{iE_pt - i\vec{p}\cdot\vec{x}})[/tex]
[tex]+\frac{1}{\sqrt{V}}\frac{1}{\sqrt{2E_p}}(a_{-p}e^{-iE_pt - i\vec{p}\cdot\vec{x}} + a_{-p}^{\dagger}e^{iE_pt + i\vec{p}\cdot\vec{x}})[/tex]

The Attempt at a Solution


The idea is that the second term on the r.h.s. of (10.57) is the same as the first term evaluated for [itex]\vec{p} = -\vec{p}[/itex], which does not effect [itex]E_p[/itex]. Then (10.58) is supposed to be the sum of (10.57) over all values of [itex]\vec{p}[/itex]. My problem is that I think there is a factor of 2 missing on the r.h.s. of (10.58) because each of the terms in (10.57) should appear twice in the sum. What am I missing? The same problem arises on page 176 for equations (10.60) and (10.61) which are sums of (10.55) and (10.56) respectively.

But in each term of equation 10.57, all the components of [itex]\vec{p}[/itex] are supposed to be positive 9again, this is true for each of the two terms of 10.57). But in 10.58 the components of p are allowed to be negative. So in the sum of 10.58 here is what happens: when the p's are positive, one generates the pieces corresponding to the first term of 10.57. When the p components in 10.58 are negative, one generates the pieces coresponding to the second term of 10.57.

Does that make sense?
 
nrqed said:
But in each term of equation 10.57, all the components of [itex]\vec{p}[/itex] are supposed to be positive.
Is that implied by something in the text? Just above equation (10.58) he says:
phi includes contributions from all values of [itex]\vec{p}[/itex]
 

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