Tong QFT sheet 2, question 6: Normal ordering of the angular momentum operator

In summary: This last integral is zero because the integral on the left and right hand side are the same (since momentum is conserved).
  • #1
Gleeson
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Homework Statement
I've been studying Tong's QFT notes and am trying to do problem sheet 2, question 6 here: https://www.damtp.cam.ac.uk/user/tong/qft.html


We are asked to take the classical angular momentum of the field,

##Q_i = \epsilon_{ijk}\int d^3x \left(x^j T^{0k} - x^k T^{0j}\right)##,

and show that after normal ordering, the quantum operator ##Q_i## can be written as


##Q_i = -i \epsilon_{ijk}\int \frac{d^3p}{(2 \pi)^3}a_{\vec{p}}^{\dagger}\left(p^j \frac{\partial}{\partial p_k} - p^k \frac{\partial}{\partial p_j} \right) a_{\vec{p}} ##,


and hence to show that a stationary one-particle state has zero angular momentum.
Relevant Equations
None.
My attempt/questions:

I use ##T^{0i} = \dot{\phi}\partial^i \phi##, ##\dot{\phi} = \pi##, and antisymmetry of ##Q_i## to get:

##Q_i = 2\epsilon_{ijk}\int d^3x [x^j \partial^k \phi(\vec{x})] \pi(\vec{x})##.

I then plug in the expansions for ##\phi(\vec{x})## and ##\pi(\vec{x})## and multiply it out to get that:

##Q_i = -2i \epsilon_{ijk} \int \frac{d^3xd^3p d^3q}{(2\pi)^6}\sqrt{\frac{E_{\vec{q}}}{E_{\vec{p}}}} x^j i p^k\left(a_{\vec{p}}a_{\vec{q}} e^{i(p + q)x} - a_{\vec{p}}a_{\vec{q}}^{\dagger} e^{i(p - q)x} - a_{\vec{p}}^{\dagger}a_{\vec{q}} e^{i(q - p)x} + a_{\vec{p}}^{\dagger}a_{\vec{q}}^{\dagger} e^{-i(p + q)x} \right)##,

where ##x##, ##p##, and ##q## are all 3-vectors, but I'm dropping the arrow.Therefore
##Q_i = -2i \epsilon_{ijk} \int \frac{d^3p d^3q}{(2\pi)^3} \sqrt{\frac{E_{\vec{q}}}{E_{\vec{p}}}} p^k \left( a_pa_q \int \frac{d^3x}{(2\pi)^3} \frac{\partial}{\partial p_j}e^{i(p + q)x} - a_pa_q^{\dagger} \int \frac{d^3x}{(2\pi)^3} \frac{\partial}{\partial p_j}e^{i(p - q)x} + a_p^{\dagger}a_q \int \frac{d^3x}{(2\pi)^3} \frac{\partial}{\partial p_j}e^{i(q - p)x} - a_p^{\dagger}a_q \int \frac{d^3x}{(2\pi)^3} \frac{\partial}{\partial p_j}e^{-i(p + q)x}\right)##,

so

##Q_i = - 2 i \epsilon_{ijk}\int \frac{d^3p d^3q}{(2 \pi)^3} \sqrt{\frac{E_{\vec{q}}}{E_{\vec{p}}}}p^k \left((a_pa_q - a_p^{\dagger}a_q)\frac{\partial}{\partial p_j}\delta(p + q) + (a_p^{\dagger}a_q - a_p a_q^{\dagger})\frac{\partial}{\partial p_j}\delta(p - q) \right)##.

- I can add more intermediate lines of explanation to the above if needed

- Is this correct so far? If not, what is wrong?

- I believe I need to integrate by parts now, to get the derivatives off the delta functions. It looks like the ratio of Energies should eventually cancel due to a delta function, but I can't see how the next couple of lines should go. Won't the derivative end up on the delta Energy, and the Energy term will need to be derived before the delta function can act on it?

- If someone could walk me through the next couple of lines I would very much appreciate it.

Thanks
 
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  • #2
Fix LaTeX
 
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  • #3
I am trying to.
 
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  • #4
What does it mean that ##Q_i## is antisymmetric? it is a rank-1 tensor
 
  • #5
I mean that we can rewrite the first line as:

##Q_i = \epsilon_{ijk}\int d^3x \left(x^j T^{0k} - x^k T^{0j}\right)##

##Q_i = \epsilon_{ijk}\int d^3x \left(x^j T^{0k} + x^j T^{0k}\right)##

##Q_i = 2\epsilon_{ijk}\int d^3x \left(x^j T^{0k}\right)##.

Sorry, I shouldn't have referred to this as antisymmetry of ##Q_i##.
 
  • #6
Gleeson said:
the Energy term will need to be derived
you mean "differentiated"?

I have not checked my own solution to this yet since it was +10 years ago since I did it. Its lying around somewhere...

Can you perform the integration by parts on the first term and see what you get?
 
  • #7
Yes I did mean "differentiated".

Starting from my last line above, after I integrate by parts, I get:

##Q_i = -2i \epsilon_{ijk} \int \frac{d^3q}{(2\pi)^3} \left( \frac{\partial}{\partial p_j} \left[\sqrt{\frac{E_q}{E_p}} p^k (a_pa_q - a_p^{\dagger}a_q^{\dagger}) \delta(p + q)\right] - \frac{\partial}{\partial p_j} \left[\sqrt{\frac{E_q}{E_p}} p^k (a_pa_q - a_p^{\dagger}a_q^{\dagger})\right] \delta(p + q) + \frac{\partial}{\partial p_j} \left[\sqrt{\frac{E_q}{E_p}} p^k (a_p^{\dagger}a_q - a_p a_q^{\dagger}) \delta(p- q)\right] - \frac{\partial}{\partial p_j} \left[\sqrt{\frac{E_q}{E_p}} p^k (a_p^{\dagger}a_q - a_p a_q^{\dagger})\right] \delta(p - q) \right)##.

It's not clear to me how to proceed from here.
 
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  • #8
Anyone?
 
  • #9
That integration by part does not look correct.
Let me do a VERY schematic one for you, omitting ##2\pi##, ## i ## etc. just focusing on what's really going on at the core of things.

##\displaystyle \int \mathrm{d}^3p \mathrm{d}^3q \sqrt{\dfrac{E_{\vec q}}{E_{\vec p}}} p^k \dfrac{\partial}{\partial p^j} \delta(\vec p+\vec q) ##
## =\displaystyle \int \mathrm{d}^3q \sqrt{\dfrac{E_{\vec q}}{E_{\vec p}}} p^k \delta(\vec p+\vec q) - \int \mathrm{d}^3p \mathrm{d}^3q \delta(\vec p+\vec q) \dfrac{\partial}{\partial p^j} \left( \sqrt{\dfrac{E_{\vec q}}{E_{\vec p}}} p^k\right) ##
Perform the integration over momentum q:
The first term equals zero.
The second term, the delta "function" will make the energy ratio becomes 1 and you are left with a kronecker delta.
 
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  • #10
Thank you.

I have a typo in my integration by parts. There is a $$d^3p$$ missing from the integrals. I had actually just rewritten the previous expression using the product rule, and hadn't actually integrated anything yet.

I don't see how you are getting your first term. I understand that this is a "boundary term" and we are going to be arguing that it is zero, but I don't understand how you got that term, and I don't see what is wrong with, say, the first term I have above. And therefore I also don't understand why this term is zero.

Regarding your second term, I think it is basically the same as I had above (if I edited my above post to put in the missing $$d^3p$$). What I don't understand is why you are allowed evaluate the delta function before taking the derivative. In this expression, we have the delta function times the derivative of a function. So why is it valid to first evaluate the delta function and then take the derivative?
 
  • #11
Gleeson said:
I have a typo in my integration by parts.
Then nevermind my previous reply.
I'll think of something else.
Will post here later.
 
  • #12
I have been stuck on this question for a quite a while now. Here is my current effort.

Starting from:
$$\phi(\vec{x}) = \int d^3p \sqrt{\frac{1}{2 E_p}}\left(a_p e^{i px} + a_p^{\dagger}e^{-ipx}\right),$$

$$\pi(\vec{x}) = \int d^3q (-i) \sqrt{\frac{E_q}{2 }}\left(a_q e^{i qx} - a_q^{\dagger}e^{-iqx}\right),$$

$$T^{0k} = \pi(\vec{x})\partial^k\phi(\vec{x}),$$

and $$Q_i = \epsilon_{ijk}\int d^3x \left(x^j T^{0k} - x^k T^{0j} \right) = 2\epsilon_{ijk}\int d^3x \left(x^j T^{0k} \right),$$ I am trying to calculate the normal ordered angular momentum operator.

All the $x$, $p$, and $q$ are 3-vectors. I didn't always put the vector symbol.

My attempt so far is:

$$Q_i = -2i \epsilon_{ijk} \int \frac{d^3xd^3p d^3q}{(2\pi)^6}\sqrt{\frac{E_{\vec{q}}}{E_{\vec{p}}}} x^j i p^k\left(a_{\vec{p}}a_{\vec{q}} e^{i(p + q)x} - a_{\vec{q}}^{\dagger}a_{\vec{p}} e^{i(p - q)x} - a_{\vec{p}}^{\dagger}a_{\vec{q}} e^{i(q - p)x} + a_{\vec{p}}^{\dagger}a_{\vec{q}}^{\dagger} e^{-i(p + q)x} \right), $$

where I have just plugged in the above definitions, multiplied out, and normal ordered. Then,

$$Q_i = -2i \epsilon_{ijk} \int \frac{d^3p d^3q}{(2\pi)^3} \sqrt{\frac{E_{\vec{q}}}{E_{\vec{p}}}} p^k \left( a_pa_q \int \frac{d^3x}{(2\pi)^3} i x^j e^{i(p + q)x} - a_q^{\dagger} a_p \int \frac{d^3x}{(2\pi)^3} i x^j e^{i(p - q)x} - a_p^{\dagger}a_q \int \frac{d^3x}{(2\pi)^3} i x^j e^{i(q - p)x} + a_p^{\dagger}a_q^{\dagger} \int \frac{d^3x}{(2\pi)^3} i x^j e^{-i(p + q)x}\right),$$

$$Q_i = -2i \epsilon_{ijk} \int \frac{d^3p d^3q}{(2\pi)^3} \sqrt{\frac{E_{\vec{q}}}{E_{\vec{p}}}} p^k \left( a_pa_q \int \frac{d^3x}{(2\pi)^3} \frac{\partial}{\partial p_j}e^{i(p + q)x} - a_q^{\dagger} a_p \int \frac{d^3x}{(2\pi)^3} \frac{\partial}{\partial p_j}e^{i(p - q)x} + a_p^{\dagger}a_q \int \frac{d^3x}{(2\pi)^3} \frac{\partial}{\partial p_j}e^{i(q - p)x} - a_p^{\dagger}a_q^{\dagger} \int \frac{d^3x}{(2\pi)^3} \frac{\partial}{\partial p_j}e^{-i(p + q)x}\right),$$


$$Q_i = -2i \epsilon_{ijk} \int \frac{d^3p d^3q}{(2\pi)^3} \sqrt{\frac{E_{\vec{q}}}{E_{\vec{p}}}} p^k \left( a_pa_q \frac{\partial}{\partial p_j} \int \frac{d^3x}{(2\pi)^3} e^{i(p + q)x} - a_q^{\dagger} a_p \frac{\partial}{\partial p_j} \int \frac{d^3x}{(2\pi)^3} e^{i(p - q)x} + a_p^{\dagger}a_q \frac{\partial}{\partial p_j} \int \frac{d^3x}{(2\pi)^3} e^{i(q - p)x} - a_p^{\dagger}a_q^{\dagger} \frac{\partial}{\partial p_j} \int \frac{d^3x}{(2\pi)^3} e^{-i(p + q)x}\right),$$

so

$$Q_i = - 2 i \epsilon_{ijk}\int \frac{d^3p d^3q}{(2 \pi)^3} \sqrt{\frac{E_{\vec{q}}}{E_{\vec{p}}}}p^k \left((a_pa_q - a_p^{\dagger}a_q^{\dagger})\frac{\partial}{\partial p_j}\delta(p + q) + ( a_p^{\dagger} a_q- a_q^{\dagger}a_p )\frac{\partial}{\partial p_j}\delta(p - q) \right).$$

Invoking the product rule, I get:

$$Q_i = -2i \epsilon_{ijk} \int \frac{d^3 p d^3q}{(2\pi)^3} \left( \frac{\partial}{\partial p_j} \left[\sqrt{\frac{E_q}{E_p}} p^k (a_pa_q - a_p^{\dagger}a_q^{\dagger}) \delta(p + q)\right] - \frac{\partial}{\partial p_j} \left[\sqrt{\frac{E_q}{E_p}} p^k (a_pa_q - a_p^{\dagger}a_q^{\dagger})\right] \delta(p + q) + \frac{\partial}{\partial p_j} \left[\sqrt{\frac{E_q}{E_p}} p^k (a_p^{\dagger}a_q - a_q^{\dagger}a_p ) \delta(p- q)\right] - \frac{\partial}{\partial p_j} \left[\sqrt{\frac{E_q}{E_p}} p^k (a_p^{\dagger}a_q - a_q^{\dagger}a_p )\right] \delta(p - q) \right).$$

Is this correct so far? If not, can someone please tell me what is wrong above?
 
  • #13
Gleeson said:
so

$$Q_i = - 2 i \epsilon_{ijk}\int \frac{d^3p d^3q}{(2 \pi)^3} \sqrt{\frac{E_{\vec{q}}}{E_{\vec{p}}}}p^k \left((a_pa_q - a_p^{\dagger}a_q^{\dagger})\frac{\partial}{\partial p_j}\delta(p + q) + ( a_p^{\dagger} a_q- a_q^{\dagger}a_p )\frac{\partial}{\partial p_j}\delta(p - q) \right).$$

I don't think there should be a factor of 2 out front in ##Q_i##. The expansions of ##\phi(x)## and ##\pi(x)## in terms of the creation and destruction operators contain the normalization factors ##\frac{1}{\sqrt{2 E_{p}}}## or ##\frac{1}{\sqrt{2 E_{q}}} ##. So, ##\frac{1}{\sqrt{2 E_p}} \frac{1}{\sqrt{2 E_q}}## gives a factor of 1/2 that cancels the 2 in front, unless I'm overlooking something else.

Let ##Q^{aa}_i## denote the ##a_{\vec{p}} a_{\vec{q}}## contribution. $$Q^{aa}_i =
- i \epsilon_{ijk}\int \frac{d^3p d^3q}{(2 \pi)^3} \sqrt{\frac{E_q}{E_p}}p^k a_pa_q\frac{\partial}{\partial p_j}\delta(p + q).$$

Do the partial integration for ##p_j##. The boundary term will set ##p_j = \pm \infty## which makes ##E_p## infinite, which is not physically allowable. Assuming that this means that we can ignore the boundary term, we have $$Q^{aa}_i = + i \epsilon_{ijk}\int \frac{d^3p}{(2 \pi)^3} p^k \frac{\partial}{\partial p_j} \left(\frac{a_p}{\sqrt{E_p}}\right) \int d^3q \sqrt{E_q} a_q\delta(p + q)$$ Do the ##d^3q## integration $$Q^{aa}_i = + i \epsilon_{ijk}\int \frac{d^3p}{(2 \pi)^3} \sqrt{E_p} \, a_{-p} \, p^k \frac{\partial}{\partial p_j} \left(\frac{a_p}{\sqrt{E_p}}\right) $$

If this looks OK, then continue on and see if you can show ##Q^{aa}_i = 0##.
 
  • #14
TSny said:
The expansions of ##\phi(x)## and ##\pi(x)## in terms of the creation and destruction operators contain the normalization factors ##\frac{1}{\sqrt{2 E_{p}}}## or ##\frac{1}{\sqrt{2 E_{q}}} ##. So, ##\frac{1}{\sqrt{2 E_p}} \frac{1}{\sqrt{2 E_q}}## gives a factor of 1/2 that cancels the 2 in front, unless I'm overlooking something else.
the mode expansion of
##\pi(\vec{x}) = \int d^3q (-i) \sqrt{\frac{E_q}{2 }}\left(a_q e^{i qx} - a_q^{\dagger}e^{-iqx}\right)##
has ##\sqrt{\frac{E_q}{2 }}## as normalization
 
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  • #15
Gleeson said:
I mean that we can rewrite the first line as:

##Q_i = \epsilon_{ijk}\int d^3x \left(x^j T^{0k} - x^k T^{0j}\right)##

##Q_i = \epsilon_{ijk}\int d^3x \left(x^j T^{0k} + x^j T^{0k}\right)##

##Q_i = 2\epsilon_{ijk}\int d^3x \left(x^j T^{0k}\right)##.

Sorry, I shouldn't have referred to this as antisymmetry of ##Q_i##.
The correct angular-momentum operator is, of course,
$$\hat{\vec{L}} = \int_{\mathbb{R}^3} \mathrm{d}^3 x [\vec{x} \times \hat{\vec{\Pi}}(t,\vec{x})],$$
where
$$\hat{\Pi}^i(t,\vec{x})=\hat{T}^{0i}(t,\vec{x})$$
is the momentum-density operator.

The expression in terms of annihilation and creation operators is a bit lengthy to derive, but it's a rather straight-forward calculation using basic Fourier-transformation formulas. The only trick is to write the ##\vec{x}## as a derivative of the mode function wrt. ##\vec{p}##.
 
  • #16
TSny said:
I don't think there should be a factor of 2 out front in ##Q_i##. The expansions of ##\phi(x)## and ##\pi(x)## in terms of the creation and destruction operators contain the normalization factors ##\frac{1}{\sqrt{2 E_{p}}}## or ##\frac{1}{\sqrt{2 E_{q}}} ##. So, ##\frac{1}{\sqrt{2 E_p}} \frac{1}{\sqrt{2 E_q}}## gives a factor of 1/2 that cancels the 2 in front, unless I'm overlooking something else.

Let ##Q^{aa}_i## denote the ##a_{\vec{p}} a_{\vec{q}}## contribution. $$Q^{aa}_i =
- i \epsilon_{ijk}\int \frac{d^3p d^3q}{(2 \pi)^3} \sqrt{\frac{E_q}{E_p}}p^k a_pa_q\frac{\partial}{\partial p_j}\delta(p + q).$$

Do the partial integration for ##p_j##. The boundary term will set ##p_j = \pm \infty## which makes ##E_p## infinite, which is not physically allowable. Assuming that this means that we can ignore the boundary term, we have $$Q^{aa}_i = + i \epsilon_{ijk}\int \frac{d^3p}{(2 \pi)^3} p^k \frac{\partial}{\partial p_j} \left(\frac{a_p}{\sqrt{E_p}}\right) \int d^3q \sqrt{E_q} a_q\delta(p + q)$$ Do the ##d^3q## integration $$Q^{aa}_i = + i \epsilon_{ijk}\int \frac{d^3p}{(2 \pi)^3} \sqrt{E_p} \, a_{-p} \, p^k \frac{\partial}{\partial p_j} \left(\frac{a_p}{\sqrt{E_p}}\right) $$

If this looks OK, then continue on and see if you can show ##Q^{aa}_i = 0##.
Thank you. You are right about my factor of 2. In my initial definition of ##Q_i##, it is correct. But in the subsequent lines the factor of 2 should have been cancelled.

I tried to do what you suggested. I think if the boundary terms are neglected like you said, then the next line is:

$$ Q_i = i \epsilon_{ijk}\int d^3p \frac{\partial}{\partial p^j}\left(\frac{p^k a_p}{\sqrt{E_p}}\right)a_{(-p)}\sqrt{E_p} - i \epsilon_{ijk}\int d^3p \frac{\partial}{\partial p^j}\left(\frac{p^k a_p^{\dagger}}{\sqrt{E_p}}\right)a_{(-p)}^{\dagger}\sqrt{E_p} + i \epsilon_{ijk}\int d^3p \frac{\partial}{\partial p^j}\left(\frac{p^k a_p^{\dagger}}{\sqrt{E_p}}\right)a_{p}\sqrt{E_p} - i \epsilon_{ijk}\int d^3p a_{p}^{\dagger}\sqrt{E_p}\frac{\partial}{\partial p^j}\left(\frac{p^k a_p^{\dagger}}{\sqrt{E_p}}\right)$$

Does this seem right? It is not clear to me how to proceed from here.
 
  • #17
Gleeson said:
$$ Q_i = i \epsilon_{ijk}\int d^3p \frac{\partial}{\partial p^j}\left(\frac{p^k a_p}{\sqrt{E_p}}\right)a_{(-p)}\sqrt{E_p} - i \epsilon_{ijk}\int d^3p \frac{\partial}{\partial p^j}\left(\frac{p^k a_p^{\dagger}}{\sqrt{E_p}}\right)a_{(-p)}^{\dagger}\sqrt{E_p} + i \epsilon_{ijk}\int d^3p \frac{\partial}{\partial p^j}\left(\frac{p^k a_p^{\dagger}}{\sqrt{E_p}}\right)a_{p}\sqrt{E_p} - i \epsilon_{ijk}\int d^3p a_{p}^{\dagger}\sqrt{E_p}\frac{\partial}{\partial p^j}\left(\frac{p^k a_p^{\dagger}}{\sqrt{E_p}}\right)$$
Yes, I think that looks good except that in the derivatives, the ##\large \frac{\partial}{\partial p^j}## should be ##\large \frac{\partial}{\partial p_j}##, with lower index ##j## on ##p##. Otherwise, you would need to change some signs since ##p^j = -p_j##. Also, ##\large \frac{\partial}{\partial p_j}## is more appropriate since it puts the index ##j## in the correct position for contraction with the ##j## in ##\epsilon_{ijk}##. Also, in Tong's final expression for ##Q_i## the derivatives involve lower indices ##p_j## and ##p_k##.

Consider the first integral $$Q^{aa}_i = i \epsilon_{ijk}\int d^3p \frac{\partial}{\partial p_j}\left(\frac{p^k a_p}{\sqrt{E_p}}\right)a_{(-p)}\sqrt{E_p} $$

Note ##k \neq j##, otherwise ## \epsilon_{ijk} = 0##. So you can pull out the ##p^k## from the derivative.

Now consider $$\frac{\partial}{\partial p_j}\left(\frac{a_p}{\sqrt{E_p}}\right) = \frac{\partial}{\partial p_j}\left(\frac{1}{\sqrt{E_p}}\right)a_p+\frac{\partial a_p}{\partial p_j}\frac{1}{\sqrt{E_p}}$$

Show that $$\frac{\partial}{\partial p_j}\frac{1}{\sqrt{E_p}} =-\frac{\partial}{\partial p^j}\frac{1}{\sqrt{E_p}} = +\frac{p^j}{2E_p^{5/2}}$$ Note ##j## is in the upper position in the last expression so that it can be contracted with the ##j## in ##\epsilon_{ijk}##.

Proceed to show ##Q_i^{aa} = 0##.

I am leading you through the way that I did the calculation. I end up with the desired result for ##Q_i## except I get an overall sign difference from Tong's expression. I have not done many of these types of calculations, so I'm certain that my way is not the slickest. But I think it gets us there. Feel free to drop this approach if other helpers provide a better way. I replied to your thread only because you didn't seem to be getting many responses (until now).
 
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  • #18
Thank you very much.

I think where you have ##\frac{p^j}{2 E_p^{\frac{5}{2}}} ##, it should be ##\frac{p^j}{E_p^{\frac{3}{2}}} ##. This doesn't affect the answer.

Also, for the record, the last expression I had for ##Q_i## had typos. It was missing the ##(2 \pi )^3##, and the last of the four terms had a dagger that wasn't supposed to be there.

I have done what you suggested, and also got the final expression, also with a minus sign difference from the expected expression. Perhaps there is a typo in the question.

Terms end up being zero from combinations of contraction of symmetric with antisymmetric indices, or from relabeling dummy indices and substituting ## p \to -p##. There is then one final integration by parts needed to get the derivative onto the ##a_p## term only.

I presume this is the final steps you had?
 
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  • #19
Assuming I have the correct steps.

If anyone has a more efficient method, please let me know.Also, I would appreciate more detail on why the boundary terms are zero. Both for the initial place where I was stuck above when we are taking the derivative off the delta functions, and then also for the very last step where we take the derivative off the ##a_p^{\dagger}## and put it on ##a_p##.

I know that these things are often assumed to be zero. But I would like to at least know explicitly what assumptions are being made so that these terms are zero. And why these are reasonable assumptions. I'd like to do better than assuming these boundary terms are zero because otherwise I get the wrong answer.
 
  • #20
Gleeson said:
why the boundary terms are zero.
You can write the creation and annihilation operators as linear combinations of the fields and the fields are assumed to vanish at infinity.
 
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  • #21
Thank you. That is the kind of thing I was hoping for. I think that justifies to me the final integration by parts with the ##a_p^{\dagger}## and ##a_p##.

How about this step:
TSny said:
Let ##Q^{aa}_i## denote the ##a_{\vec{p}} a_{\vec{q}}## contribution. $$Q^{aa}_i =
- i \epsilon_{ijk}\int \frac{d^3p d^3q}{(2 \pi)^3} \sqrt{\frac{E_q}{E_p}}p^k a_pa_q\frac{\partial}{\partial p_j}\delta(p + q).$$

Do the partial integration for ##p_j##. The boundary term will set ##p_j = \pm \infty## which makes ##E_p## infinite, which is not physically allowable. Assuming that this means that we can ignore the boundary term, we have..

?
 
  • #22
Gleeson said:
I think where you have ##\frac{p^j}{2 E_p^{\frac{5}{2}}} ##, it should be ##\frac{p^j}{E_p^{\frac{3}{2}}} ##. This doesn't affect the answer.

I wrote
TSny said:
$$\frac{\partial}{\partial p_j}\frac{1}{\sqrt{E_p}} =-\frac{\partial}{\partial p^j}\frac{1}{\sqrt{E_p}} = +\frac{p^j}{2E_p^{5/2}}$$

The 5/2 power looks right to me. The left side of the equation has dimensions of momentum to the (-3/2) power. The far right side also has these dimensions.

Gleeson said:
I have done what you suggested, and also got the final expression, also with a minus sign difference from the expected expression. Perhaps there is a typo in the question.
Good.

Gleeson said:
Terms end up being zero from combinations of contraction of symmetric with antisymmetric indices, or from relabeling dummy indices and substituting ## p \to -p##. There is then one final integration by parts needed to get the derivative onto the ##a_p## term only.

I presume this is the final steps you had?

Sounds a lot like my steps. I would need to see the details to be sure.

I had to use the commutation relation ##[a_p, a_p^{\dagger}] = 1## at some point.
 
  • #23
Thank you.

I mistakenly used that ##E_p = p^2 + m^2##, instead of ##E_p = \sqrt{p^2 + m^2}##, so both your power on ##E_p## and your factor of ##2## are correct. I should have noticed dimensional mismatch. I think I've been less diligent since things were set to ##1##. But this is not a good argument.

I didn't use that commutator at any stage. I think the need to use it is avoided by the request to "normal order". And in fact, if we use that commutator we have infinity, because the ##1## is actually a delta function evaluated at ##0## instead of a ##1##.
 
  • #24
Gleeson said:
I didn't use that commutator at any stage. I think the need to use it is avoided by the request to "normal order". And in fact, if we use that commutator we have infinity, because the ##1## is actually a delta function evaluated at ##0## instead of a ##1##.
Yes, you are right. Thanks.
 
  • #25
malawi_glenn said:
You can write the creation and annihilation operators as linear combinations of the fields and the fields are assumed to vanish at infinity.
But our boundary terms deal with infinite momentum rather than spatial infinity.
 
  • #26
TSny said:
But our boundary terms deal with infinite momentum rather than spatial infinity.
fourier transform
 
  • #27
malawi_glenn said:
fourier transform
OK. Paserval's theorem says that if the fields are square integrable in physical space, then the transform is square integrable in momentum space. Is that the sort of thing you are hinting at?
 
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  • #28
TSny said:
OK. Paserval's theorem says that if the fields are square integrable in physical space, then the transform is square integrable in momentum space. Is that the sort of thing you are hinting at?
This is basically what you do with the fields in qft at this stage yes
 

Related to Tong QFT sheet 2, question 6: Normal ordering of the angular momentum operator

1. What is the purpose of normal ordering in the angular momentum operator?

The purpose of normal ordering is to rearrange the terms in the angular momentum operator in a specific way that allows for easier calculations and a more concise representation of the operator.

2. How is normal ordering performed in the angular momentum operator?

Normal ordering is performed by moving all creation operators to the left of all annihilation operators in the operator expression. This results in a simplified form of the operator that can be used in calculations.

3. Why is normal ordering important in quantum field theory?

Normal ordering is important in quantum field theory because it helps to eliminate infinities that arise in calculations. By rearranging the operator in a specific way, the infinities can be isolated and removed, resulting in a more accurate and meaningful calculation.

4. Can normal ordering be applied to other operators besides the angular momentum operator?

Yes, normal ordering can be applied to any operator in quantum field theory. It is a general technique used to simplify and remove infinities in operator expressions.

5. How does normal ordering affect the commutation relations of the angular momentum operator?

Normal ordering does not affect the commutation relations of the angular momentum operator. The commutation relations remain the same, but the operator expression is rearranged in a way that makes it easier to work with and eliminates infinities.

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