Proving [itex][a_k^\dagger, a_q^\dagger]=0[/itex]

  • Thread starter mjordan2nd
  • Start date
In summary: And thank you Dr. Claude for the clarification. I think I understand now. So my final expression is indeed zero and there is no discrepancy. In summary, the Dirac delta function only has a meaning in integrals and when integrated with a function that vanishes at zero, the result is also zero. This explains why my final expression, when integrated, gives zero and there is no discrepancy.
  • #1
mjordan2nd
177
1
I am trying to prove the commutation relations between the creation and annihilation operators in field theory. I was already able to show that [itex][a_k, a_q^\dagger]=i\delta(k-q)[/itex]. I want to show that [itex][a_k^\dagger, a_q^\dagger]=0,[/itex] but I am stuck on the last step. I think my confusion has to do with interpreting the delta function and I was hoping someone can help me out. So far I have shown that

[tex]a_k^\dagger = \int \! d^3\!x \: e^{-ikx}\left[ \sqrt{\frac{E_k}{2}} \phi(x) - i \sqrt{\frac{1}{2E_k}} \pi(x) \right],[/tex]

where [itex]\phi(x)[/itex] and [itex]\pi(x)[/itex] are the field and conjugate momentum operators respectively. Based on this I was able to show that

[tex][a_k^\dagger, a_q^\dagger]=\int \! d^3y \: d^3x \: e^{-i(kx+qy)} \left[ -i \sqrt{\frac{E_k}{E_q}} [\phi(x), \pi(y)] + i \sqrt{\frac{E_q}{E_k}} [\phi(y),\pi(x)] \right].[/tex]

Using the commutation relation [itex][\phi(x), \pi(y)]=i \delta^3(x-y)[/itex] this becomes

[tex][a_k^\dagger, a_q^\dagger]=\int \! d^3y \: e^{-i (k+q) y} \left[ \sqrt{\frac{E_k}{E_q}} -\sqrt{\frac{E_q}{E_k}} \right].[/tex]

And I believe the integral over y gives us a delta function. If I have done everything correctly then this should be

[tex][a_k^\dagger, a_q^\dagger]=\delta^3(k+q) \left[ \sqrt{\frac{E_k}{E_q}} -\sqrt{\frac{E_q}{E_k}} \right].[/tex]

Since I already know the commutation relations I know that the final expression should equal zero for any arbitrary value of [itex]k[/itex] and [itex]q[/itex]. But my final expression doesn't really say that for [itex]k=-q[/itex]. I don't really understand where this discrepancy is coming from, though. Is there some way to interpret this delta function that I'm not understanding or have I made a computational error somewhere? Thanks
 
Physics news on Phys.org
  • #2
mjordan2nd said:
But my final expression doesn't really say that for k=−q.
Actually it does, because the square bracket vanishes in that case and ##\delta(0)\cdot 0 =0##.
 
Last edited:
  • Like
Likes bhobba
  • #3
I don't really understand why. I guess the [itex]\infty \cdot 0 = 0[/itex] is what's throwing me off. Is there a way to put this on firmer mathematical footing?
 
  • #4
Just take it a step back to the integral. The integral of 0 is 0, even if the integral is over all of space. :)
 
  • #5
mjordan2nd said:
[tex][a_k^\dagger, a_q^\dagger]=\delta^3(k+q) \left[ \sqrt{\frac{E_k}{E_q}} -\sqrt{\frac{E_q}{E_k}} \right].[/tex]Since I already know the commutation relations I know that the final expression should equal zero for any arbitrary value of [itex]k[/itex] and [itex]q[/itex]. But my final expression doesn't really say that for [itex]k=-q[/itex].
Hint: what's the formula for ##E_k## ... ? ;)
 
  • #6
mjordan2nd said:
I don't really understand why. I guess the [itex]\infty \cdot 0 = 0[/itex] is what's throwing me off. Is there a way to put this on firmer mathematical footing?

You need to study distribution theory - get the following book - you won't regret it:
https://www.amazon.com/dp/0521558905/?tag=pfamazon01-20

The Dirac Delta function only has a meaning in integrals. When integrated it gives the value of the function before it at zero. In your case that's zero, so, when integrated gives zero. Hence, on a distribution basis is the same as zero.

Its a bit strange on first acquaintance, but once you get used to it its easy - which is why I really do suggest getting the book I mentioned - it will make everything clear - and much much more besides eg Fourier transforms become a doodle.

Thanks
Bill
 
Last edited by a moderator:
  • Like
Likes vanhees71 and Demystifier
  • #7
mjordan2nd said:
I don't really understand why. I guess the [itex]\infty \cdot 0 = 0[/itex] is what's throwing me off. Is there a way to put this on firmer mathematical footing?
The ##\delta##-function has a mathematical meaning only inside an integral:
$$\int dx\, \delta(x) f(x) =f(0)$$
which is ##0## if ##f(0)=0##.

Another, more heuristic way to think of a product ##\delta(k-q) f(k,q)## is to replace the Dirac ##\delta(k-q)## with the Kronecker ##\delta_{kq}##. In this way you can put
$$\delta_{kq}f(k,q)=f(k,k)=0$$
without doing the integral/sum.
 
  • #8
Thank you for the replies. I will check out that book, Bill!
 

1. What is the significance of proving [ak, aq] = 0?

This relation is known as the commutation relation between creation operators in quantum mechanics. It states that the order in which the creation operators are applied does not affect the final result. It is an important property that allows for the simplification of calculations in quantum systems.

2. How is the commutation relation derived?

The commutation relation between two operators can be derived using the canonical commutation relations, which describe the fundamental properties of quantum operators. By applying these relations to the creation and annihilation operators, we can obtain the commutation relation [ak, aq] = 0.

3. What is the physical interpretation of [ak, aq] = 0?

This commutation relation reflects the fact that in quantum mechanics, particles are indistinguishable and cannot be counted as separate entities. Instead, the creation and annihilation operators represent the process of adding or removing a particle from a quantum system.

4. Is the commutation relation always true for any quantum system?

No, the commutation relation [ak, aq] = 0 only holds for systems with a finite number of particles. In systems with an infinite number of particles, such as in quantum field theory, the commutation relation is modified to take into account the continuous creation and annihilation of particles.

5. How is the commutation relation used in practical applications?

The commutation relation between creation operators is a fundamental property that is used in various calculations and equations in quantum mechanics and quantum field theory. It allows for the simplification of mathematical expressions and plays a crucial role in the development of quantum theories and their applications in fields such as particle physics and condensed matter physics.

Similar threads

  • Quantum Physics
Replies
4
Views
1K
  • Quantum Physics
Replies
3
Views
753
  • Quantum Physics
Replies
0
Views
619
Replies
1
Views
1K
  • Quantum Physics
Replies
4
Views
728
Replies
24
Views
1K
Replies
27
Views
2K
  • Quantum Physics
Replies
2
Views
837
  • Quantum Physics
Replies
4
Views
845
  • Quantum Physics
Replies
9
Views
1K
Back
Top