# Momentum Operator for the real scalar field

• Markus Kahn
In summary, the conversation discussed a problem and solution involving a straightforward calculation. The problem involved a formula and applying it to every term of a sum. There was some uncertainty about a specific step and how to get rid of a certain term in the final equation. A suggestion was made to examine the calculation and a comment was given on the definition of conjugate momentum of the field. It was then discussed how the second term in the final equation could be shown to be zero through a change of variables and using the commutation relation between operators.
Markus Kahn
Homework Statement
Show that the momentum for the real scalar field reduces to
$$\vec{P} = \int \frac{dp^3}{(2\pi)^3 2e(p)} \vec{p} a_p^\dagger a_p.$$
Relevant Equations
$$\vec{P}=-\int dx^3 \pi \nabla \phi$$
$$\phi(\vec x )= \int \frac{dp^3}{(2\pi)^3 2e(p)} (a_pe^{ipx} + a^\dagger_p e^{-ipx})$$
$$\pi(\vec x )= \int \frac{dp^3}{(2\pi)^3 }\frac{i}{2} (a_p^\dagger e^{-ipx} - a_p e^{ipx})$$
I think the solution to this problem is a straightforward calculation and I think I was able to make reasonable progress, but I'm not sure how to finish this...

\begin{align*} \vec{P}&=-\int dx^3 \pi \nabla \phi\\ &= -\int\int\int dx^3\frac{dp^3}{(2\pi)^3 2e(p)} \frac{du^3}{(2\pi)^3} \frac{i}{2} \left(a_u^\dagger e^{-iux} - a_u e^{iux}\right)\nabla \left(a_pe^{ipx} + a^\dagger_p e^{-ipx}\right)\\ &= -\int\int\int dx^3\frac{dp^3}{(2\pi)^3 2e(p)} \frac{du^3}{(2\pi)^3} \frac{i}{2} (ip)\left(a_u^\dagger e^{-iux} - a_u e^{iux}\right)\left(a_pe^{ipx} - a^\dagger_p e^{-ipx}\right)\\ &= \int\int\int dx^3\frac{dp^3}{(2\pi)^3 2e(p)} \frac{du^3}{(2\pi)^3} \frac{p}{2} \left(a_u^\dagger e^{-iux} - a_u e^{iux}\right)\left(a_pe^{ipx} - a^\dagger_p e^{-ipx}\right)\\ &= \int\int\frac{dp^3}{(2\pi)^3 2e(p)} \frac{du^3}{(2\pi)^3}\times \\& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{p}{2} \int dx^3 \left(a_u^\dagger a_p e^{i(p-u)x} +a_ua_p^\dagger e^{i(u-p)x} -a_u^\dagger a_p^\dagger e^{-i(u+p)x} - a_ua_p e^{i(p+u)x}\right) \end{align*}

I know applied to every term of the sum the following formula:
$$\delta (x-\alpha )={\frac {1}{2\pi }}\int _{-\infty }^{\infty }e^{ip(x-\alpha )}\ dp,$$
which resulted in

\begin{align*}\vec{P}&= \int\int\frac{dp^3}{(2\pi)^3 2e(p)} \frac{du^3}{(2\pi)^3}\times \\& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{p}{2}(2\pi)^3 \left(a_u^\dagger a_p \delta(p-u) +a_ua_p^\dagger \delta(u-p) -a_u^\dagger a_p^\dagger \delta(u+p) - a_ua_p \delta(p+u)\right)\\ &= \int \frac{dp^3}{(2\pi)^3 2e(p)} \frac{p}{2}\times\\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\, \int du^3 \left(a_u^\dagger a_p \delta(u-p) +a_ua_p^\dagger \delta(u-p) -a_u^\dagger a_p^\dagger \delta(u+p) - a_ua_p \delta(u+p)\right) \end{align*}
$$\int f(x)\,\delta(x-a)\,dx =\int f(x)\,\delta(a-x)\,dx.$$
If I continue from here on I arrive at
\begin{align*}\vec P &= \int \frac{dp^3}{(2\pi)^3 2e(p)} \frac{p}{2}\times\\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\, \int du^3 \left((a_u^\dagger a_p+a_ua_p^\dagger )\delta(u-p) -(a_u^\dagger a_p^\dagger + a_ua_p) \delta(u+p)\right)\\ &=\int \frac{dp^3}{(2\pi)^3 2e(p)} \frac{p}{2}\times (a_p^\dagger a_p+a_pa_p^\dagger )-(a_{-p}^\dagger a_p^\dagger + a_{-p}a_p)\\ &=\int \frac{dp^3}{(2\pi)^3 2e(p)} \frac{p}{2}\times (2a_p^\dagger a_p+1 )-(a_{-p}^\dagger a_p^\dagger + a_{-p}a_p)\\ &=\int \frac{dp^3}{(2\pi)^3 2e(p)} p (a_p^\dagger a_p+\frac{1}{2} ) -(a_{-p}^\dagger a_p^\dagger + a_{-p}a_p) \end{align*}
The term corresponding to the ##1/2## in parenthesis can be resolved with renormalization (as far as I understand), so we don't really need to bother with it. But what about the ##(a_{-p}^\dagger a_p^\dagger + a_{-p}a_p)##? How do I get rid of that?

Hi.
Say E(p)=E(-p), the integrand of
$$-\frac{1}{4}\int \frac{dp^3}{(2\pi)^3E(p)}p(a^\dagger_{-p} a^\dagger_p+a_{-p}a_p)$$
is antisymmetric for inverse of p . So the integral vanishes.

Markus Kahn
Hello Markus Kanh!
So, I examined your calculation and I have feeling that something is odd in your definition of conjugate momentum of the field. In the books that I have learning I never see yet that definition. I often see the following definition $$\pi(\vec x)=-i\int \frac {d^{3}p} {(2\pi)^{3} } {\sqrt{E_{p}/ 2}}(a_{p}e^{ipx}-a_{p}e^{-ipx})$$ One example is in the Peskin's book ''An introduction to QFT''. Anyway... this is just one comment.
About your question, we can show that the second term in your last equation is zero##\int \frac p {(2\pi)^{3}2E_p} (a_{-p}^\dagger a_{p}^\dagger+a_{-p}a_{p})\,d^{3}p=^{!}0##.
Consider the following integral $$I=\int \frac p {(2\pi)^{3}2E_p} a_{-p}^\dagger a_{p}^\dagger\,d^{3}p$$ we can change the variable ##p\to-p## That change don't change the value of integral because that integral is evaluated in all p-space. $$I=-\int \frac p {(2\pi)^{3}2E_p} a_{p}^\dagger a_{-p}^\dagger\,d^{3}p$$ summing these two pieces above, we have $$2I=\int \frac p {(2\pi)^{3}2E_p} (a_{-p}^\dagger a_{p}^\dagger-a_{p}^\dagger a_{-p}^\dagger)\,d^{3}p$$ that is $$2I=\int \frac p {(2\pi)^{3}2E_p}[a_{-p}^\dagger,a_{p}^\dagger]\,d^{3}p$$ note that ##[a_{-p}^\dagger,a_{p}^\dagger]=0## in this way we can saying that $$I=0$$ Similarly, we can show that the second term in the your last equation is also zero.

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## 1. What is the momentum operator for the real scalar field?

The momentum operator for the real scalar field is a mathematical operator that acts on a wave function to determine the momentum of a particle in the field. It is represented by the symbol "p" and is defined as the derivative of the wave function with respect to position.

## 2. How is the momentum operator derived for the real scalar field?

The momentum operator for the real scalar field is derived from the classical momentum operator, which is defined as the product of the mass and velocity of a particle. By applying the principles of quantum mechanics, this operator is then modified to account for the wave nature of particles in the scalar field.

## 3. What is the significance of the momentum operator in quantum mechanics?

The momentum operator is a fundamental concept in quantum mechanics, as it allows us to calculate the momentum of a particle in a given field. It is also a key component in the Heisenberg uncertainty principle, which states that the more precisely we know the momentum of a particle, the less precisely we can know its position.

## 4. How is the momentum operator used in practical applications?

The momentum operator is used in a variety of practical applications, such as in calculating the energy spectrum of a quantum system and determining the probability of a particle occupying a certain energy state. It is also used in the field of quantum field theory to study the behavior of particles in fields.

## 5. Can the momentum operator be applied to other fields besides the real scalar field?

Yes, the momentum operator can be applied to other quantum fields, such as the electromagnetic field and the Higgs field. In fact, it is a key concept in understanding the behavior of particles in these fields and is essential in many areas of modern physics, including particle physics and cosmology.

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