Linear momentum of the Klein Gordon field

In summary, the correct answer for ##P_j## should be $$P_j = \int \frac{dp^3}{(2\pi)^3} \frac{1}{2E_{\vec{p}}} \Big(-p_j a(\vec{p})a(-\vec{p})-p_ja^\dagger(\vec{p})a^\dagger(-\vec{p}) + 2p_j a(\vec{p})a^\dagger(\vec{p}) \Big)$$. The factor of 2 in the denominator can be explained by the odd nature of the integrand when changing ##\vec p## to ##-\vec p##. However, this still does not
  • #1
101
15
Homework Statement
I need to find an expression for the linear momentum of the Klein Gordon field in terms of creation and anihilation operators, but I fail to do it.
Relevant Equations
##P_j = \int dx^3 \pi(x) \partial^j\phi(x)##
The correct answer is:
#P = \int \frac{dp^3}{(2\pi)^3}\frac{1}{2E_{\vec{p}} \big(a a^{\dagger} + a^{\dagger}a\big)#

But I get terms which are proportional to ##aa## and ##a^{\dagger}a^{\dagger}##

I hereunder display the procedure I followed:

First:
##\phi = \int \frac{dp^3}{(2\pi)^3}\frac{1}{2E_{\vec{p}}} \big(e^{-ipx}a(\vec{p})+e^{ipx}a^\dagger(\vec{p})\big)##

## \pi = \int \frac{dp^3}{(2\pi)^3}\frac{i}{2} \big(-e^{-ipx}a(\vec{p})+e^{ipx}a^\dagger(\vec{p}) \big)##

So that for the j-th component of the momentum we have

##P_j = \int dx^3 \pi(x) \partial^j\phi(x)=\int \frac{dp^3dp'^3dx^3}{(2\pi)^6}\frac{i}{4E_{\vec{p}}} \Big(-e^{-ipx}a(\vec{p}) + e^{ipx} a^{\dagger}(\vec{p}) \Big) \Big(-ip'_je^{-ip'x}a(\vec{p'}) +ip'_je^{ip'x} a^{\dagger}(\vec{p'}) \Big) = \int \frac{dp^3dp'^3dx^3}{(2\pi)^6}\frac{i}{4E_{\vec{p}}} \Big(- p_je^{-ix(p'+p)}a(\vec{p})a(\vec{p'}) - p_je^{ix(p+p')} a^{\dagger}(\vec{p})a^{\dagger}(\vec{p'})+ p_je^{ix(p-p')}(\vec{p})a^{\dagger}(\vec{p'}) + p_je^{ix(p-p')} a^{\dagger}(\vec{p})a(\vec{p'}) \Big) = \int \frac{dp^3}{(2\pi)^3} \frac{1}{4E_{\vec{p}}} \Big(-p_j a(\vec{p})a(-\vec{p})-p_ja^\dagger(\vec{p})a^\dagger(-\vec{p}) + p_j a(\vec{p})a^\dagger(\vec{p}) + pja^\dagger(\vec{p})a(\vec{p}) \Big)##

Note that there is also a factor 2 that disagrees with the desired solution.

Many thanks in advance.
 
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  • #2
Jufa said:
##P_j = \int dx^3 \pi(x) \partial^j\phi(x)= ... = \int \frac{dp^3}{(2\pi)^3} \frac{1}{4E_{\vec{p}}} \Big(-p_j a(\vec{p})a(-\vec{p})-p_ja^\dagger(\vec{p})a^\dagger(-\vec{p}) + p_j a(\vec{p})a^\dagger(\vec{p}) + pja^\dagger(\vec{p})a(\vec{p}) \Big)##

Note that there is also a factor 2 that disagrees with the desired solution.
Consider ## \int \frac{dp^3}{(2\pi)^3} \frac{1}{4E_{\vec{p}}} \Big(-p_j a(\vec{p})a(-\vec{p})\Big)##

Show that the integrand is odd in ##\vec p##. That is changing ##\vec p## to ##-\vec p## just changes the overall sign of the integrand.

Your factor of 4 occurring in ##4E_{\vec p}## in the denominator looks correct to me. But I might be missing something.

You can write $$ p_j a(\vec{p})a^\dagger(\vec{p}) + p_ja^\dagger(\vec{p})a(\vec{p})$$ as $$2 p_ja^\dagger(\vec{p})a(\vec{p}) + p_j[a(\vec{p}),a^\dagger(\vec{p})] $$ Integration over the 2nd term is trivial if you use the result for the commutator ##[a(\vec{p}),a^\dagger(\vec{p})]##. So, you are left with just an integral over the first term. This has a factor of 2 which can be used to change ##4E_{\vec p}## into ##2E_{\vec p}## in the denominator. You'll end up with a nice compact result for ##P_j##, but it still doesn't agree with the answer that was given to you as shown below:

Jufa said:
The correct answer is:
$$P = \int \frac{dp^3}{(2\pi)^3}\frac{1}{2E_{\vec{p}}} \big(a a^{\dagger} + a^{\dagger}a\big)$$

This doesn't look right. Check to see if you typed this correctly.
 
  • #3
TSny said:
Consider ## \int \frac{dp^3}{(2\pi)^3} \frac{1}{4E_{\vec{p}}} \Big(-p_j a(\vec{p})a(-\vec{p})\Big)##

Show that the integrand is odd in ##\vec p##. That is changing ##\vec p## to ##-\vec p## just changes the overall sign of the integrand.

Your factor of 4 occurring in ##4E_{\vec p}## in the denominator looks correct to me. But I might be missing something.

You can write $$ p_j a(\vec{p})a^\dagger(\vec{p}) + p_ja^\dagger(\vec{p})a(\vec{p})$$ as $$2 p_ja^\dagger(\vec{p})a(\vec{p}) + p_j[a(\vec{p}),a^\dagger(\vec{p})] $$ Integration over the 2nd term is trivial if you use the result for the commutator ##[a(\vec{p}),a^\dagger(\vec{p})]##. So, you are left with just an integral over the first term. This has a factor of 2 which can be used to change ##4E_{\vec p}## into ##2E_{\vec p}## in the denominator. You'll end up with a nice compact result for ##P_j##, but it still doesn't agree with the answer that was given to you as shown below:
This doesn't look right. Check to see if you typed this correctly.
Yes. it is definitely not right. There's a two at the denominator missing and also the momentum p should multiply the whole expression. Thank you very much for your answer, what you told me about the expression being odd rescues everything.
 
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What is linear momentum?

Linear momentum is a physical quantity that describes the motion of an object in a straight line. It is a vector quantity, meaning it has both magnitude and direction. In other words, it is the product of an object's mass and its velocity.

What is the Klein Gordon field?

The Klein Gordon field is a quantum field theory that describes the behavior of spinless particles, such as scalar bosons. It is named after physicists Oskar Klein and Walter Gordon, who first proposed the theory in the 1920s.

How is linear momentum of the Klein Gordon field calculated?

The linear momentum of the Klein Gordon field is calculated by taking the derivative of the field with respect to time and multiplying it by the field's conjugate. This is known as the canonical momentum and is a fundamental quantity in quantum field theory.

What are the units of linear momentum?

The units of linear momentum depend on the system of measurement being used. In the SI system, linear momentum is measured in kilogram meters per second (kg·m/s). In the cgs system, it is measured in gram centimeters per second (g·cm/s).

Why is linear momentum important in physics?

Linear momentum is important in physics because it is a conserved quantity, meaning it remains constant in a closed system. This principle is known as the law of conservation of momentum and is essential for understanding the behavior of objects in motion.

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