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

  • Thread starter Jimmy Snyder
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In summary, the conversation discusses equations (10.57) and (10.58) and their relationship. The first equation includes contributions from all values of \vec{p} while the second equation has restrictions on the values of p. It is noted that there may be a factor of 2 missing in (10.58) due to the possibility of negative p values, and this issue is further discussed in equations (10.60) and (10.61). The conversation ends with a clarification on the relationship between the terms in (10.58) and the terms in (10.57).
  • #1
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.
 
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  • #2
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?
 
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  • #3
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.
 
  • #4
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?
 
  • #5
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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Pages 175 and 176 in Zwiebach's book discuss the concept of D-branes and their role in string theory. This is a crucial topic in understanding the properties and behavior of strings in high-dimensional spacetime.

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D-branes are unique because they have a tension that can be adjusted, unlike other objects such as strings or particles which have a fixed tension. This allows for more flexibility in their interactions and behavior.

3. Can D-branes exist in more than three dimensions?

Yes, D-branes can exist in any number of dimensions. In fact, the number of dimensions in which they exist determines their properties and behavior, making them a fundamental part of string theory in higher dimensions.

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D-branes are closely related to black holes in string theory. In some cases, multiple D-branes can combine to form a black hole, and the properties of the D-branes can affect the behavior of the black hole.

5. How do D-branes contribute to the holographic principle?

The holographic principle states that all the information about a higher-dimensional system can be encoded on a lower-dimensional boundary. D-branes play a crucial role in this principle by providing a way to map the behavior of strings in higher dimensions to a lower-dimensional boundary.

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