3
$\begingroup$

I am asked to show that the normal ordered momentum operator for free theory is

$$\hat{p^\mu} = \int \frac{d^3 p}{(2 \pi)^3} p^\mu \: a_p^\dagger \:a_p.$$

The free theory Lagrangian is given by

$$\mathcal{L}=\frac{1}{2} \partial_\mu \phi \partial^\mu \phi - \frac{1}{2}m^2\phi.$$

I have shown that for this Lagrangian the energy-momentum tensor is

$$T^{\mu \nu} = \partial^\mu \phi \partial^\nu \phi - \eta^{\mu \nu}\mathcal{L}.$$

From here we can show that the energy is

$$p_0 = \int d^3x T^{00} = \int d^3x \left( \frac{1}{2} \phi^2 + \frac{1}{2} (\nabla \phi) ^2 + \frac{1}{2}m^2 \phi^2\right),$$

and the three-momentum is

$$p_i = \int d^3x T^{0i} = \int d^3x \:\dot{\phi} \partial^i \phi.$$

When quantizing these expressions, I am trying to work in the Schrodinger picture. In order to show the first expression, I will work with the spatial and temporal parts of the four-momentum separately due to the difference in their expressions in the classical field theory. At this point I'm only trying to show that the spatial part of the normal ordered momentum is consistent with the first equation of this post. Quantizing gives us

$$\hat{\phi}(x) = \int \frac{d^3 p}{(2 \pi)^3 \sqrt{2E_p}} \left[ a_p e^{i \vec{p} \cdot \vec{x}} + a_p^\dagger e^{-i \vec{p} \cdot \vec{x}} \right].$$

I am told that this expression actually contains the time part in the exponentials, but since this is the Schrodinger picture we choose to define this operator at some fixed time which we conveniently choose to be $t=0.$ Given that, the time derivative of the above expression is

$$\dot{\phi}(x)= -i \int \frac{d^3p}{(2 \pi)^3} \sqrt{\frac{E_p}{2}} \left[ a_p e^{i \vec{p} \cdot \vec{x}} - a_p^\dagger e^{-i \vec{p} \cdot \vec{x}} \right],$$

and

$$\partial^i \phi = i \int \frac{d^3p}{(2 \pi)^3 \sqrt{2 E_p}} p^i \left[ a_p e^{i \vec{p} \cdot \vec{x}} - a_p^\dagger e^{-i \vec{p} \cdot \vec{x}} \right].$$

Per my rough calculation this means that

$$\dot{\phi} \partial^i \phi=\int \frac{d^3p d^3k}{(2 \pi)^6} \sqrt{\frac{E_p}{4 E_k}} k_i \left[ a_p a_k e^{i(p+k)x} - a_p a_k^\dagger e^{i(p-k)x} - a_p^\dagger a_k e^{i(k-p)x} + a_p^\dagger a_k^\dagger e^{-i(p+k)x} \right].$$

Then, using the exponential definition of the delta function and integrating through x and p I found

$$p^i = \int d^3x \dot{\phi} \partial^i \phi= \int \frac{d^3k}{(2 \pi)^3} \frac{1}{2} k_i \left[a_k a_{-k} - a_k a_k^\dagger - a_k^\dagger a_k + a_k^\dagger a_{-k}^\dagger \right].$$

I appear to have made a mistake in sign somewhere but I'm not too worried about that -- I can go back and find it on my own time. That said,, this is something like what I want except for the first and last terms in square brackets. If I do this in the Heisenberg picture I will get an exponentially decaying factor attached to the first and last terms which I think will make those two terms disappear but I feel like I should be able to get the same result in the Schrodinger picture, and I can't quite understand why I'm not.

$\endgroup$
4
$\begingroup$

The first and last terms strictly and completely disappear. They are equal to zero because $k_i a_k a_{-k} $ (just like its hermitian conjugate) is an odd function of $\vec k$ – a function obeying $f(-\vec k) = -f(\vec k)$ – and the integral of an odd function vanishes because the contribution from the region around $\vec k$ is exactly cancelled by the contribution from the vicinity of $-\vec k$.

Without words, you may divide the $k_i a_k a_{-k}$ term as $$\frac 12 (k_i a_k a_{-k} + k_i a_k a_{-k} ) $$ and write the result as a sum of two integrals. In one of them, you may use the substitution $k\to (-k)$ which changes $k_i a_k a_{-k}$ to $-k_i a_{-k} a_k =-k_i a_k a_{-k}$ which is the same thing as the term you didn't change so they exactly cancel.

$\endgroup$
  • $\begingroup$ I see. In that case why doesn't the same argument make the middle terms disappear? I.e. if $f(k)=k_ia_ka_k^\dagger$ then $f(-k)=-k_ia_{-k}a_{-k}^\dagger,$ which if my calculations are correct means $-k_ia_{-k}a_{-k}^\dagger = -k_ia_ka_k^\dagger$. It seems to me that replacing $k$ with $-k$ leaves the creation and annihilation operators unchanged because of their definition: $$\left( \begin{array}{cc}a_k^\dagger \\ a_k \end{array} \right) = \int \! d^3\!x \: e^{-ikx}\left[ \sqrt{\frac{E_k}{2}} \phi(x) \mp i \sqrt{\frac{1}{2E_k}} \pi(x) \right].$$ $\endgroup$ – Dargscisyhp Dec 2 '14 at 18:52
  • $\begingroup$ Since $E_k=E_{-k}$ the terms inside the bracket are left unchanged under a change of sign for $k.$ Since the integral is over the entire space, it shouldn't matter is the exponential has a $-ikx$ exponent or an $ikx$ exponent, so it seems to me that if you change $k \to -k$ the definition of the creation and annihilation operators are left unchanged. $\endgroup$ – Dargscisyhp Dec 2 '14 at 18:55
  • $\begingroup$ The two middle terms are, first of all, the same up to a c-number that cancels. And this term $k a^\dagger_k a_k$ doesn't cancel because this function of $k$ isn't odd. Substituting $-k$ for $k$ gives $-k a^\dagger_{-k} a_{-k}$ which is something else because the creation/annihilation operators are taken at a different value of the momentum. It's essential for the first and last term to cancel that one of the subscripts is $+k$ and the other is $-k$ and they get mapped to one another. If the signs are the same, there is no potential for cancellation. You are doing an elementary mistake. $\endgroup$ – Luboš Motl Dec 2 '14 at 19:04
  • $\begingroup$ Your comment that it doesn't matter whether individual factors in individual terms (in the formula for the creation operator, for example) have $+ikx$ or $-ikx$ in the exponent doesn't make the slightest sense. It's just like saying that 5-3 is the same thing as 5+3. $\endgroup$ – Luboš Motl Dec 2 '14 at 19:06
  • $\begingroup$ I think I see where I was getting confused. If we call the term in the bracket $f_k$ and since $f_{-k}=f_k$, I thought we could write something like $\int_{-\infty}^0 e^{ikx} f_k = \int_0^\infty f^{-ikx}f_k$ and everything would come out equal. But since $f_k$ is actually a function of $x,$ the equality does not hold, is that correct? You're right, that is a silly mistake. $\endgroup$ – Dargscisyhp Dec 3 '14 at 5:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.