Quantum operators and commutation relations

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CAF123
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Homework Statement


Given the mode expansion of the quantum field ##\phi## and the conjugate field one can derive $$\mathbf P = \int \frac{d^3 \mathbf p}{(2\pi)^3 2 \omega(\mathbf p)} \mathbf p a(\mathbf p)^{\dagger} a(\mathbf p)$$ By writing $$e^X = \text{lim}_{n \rightarrow \infty} \left(1+\frac{X}{n}\right)^n$$ calculate ##[\mathbf P, a(\mathbf q)^{\dagger}]## and hence evaluate $$\exp(-i \mathbf P \cdot \mathbf z) a(\mathbf q)^{\dagger} \exp(i \mathbf P \cdot \mathbf z),$$ with ##\mathbf z## a constant vector.

Homework Equations


##[AB,C] = A[B,C] + [A,C]B##

Commutation relations for the creation and annihilation operators.

The Attempt at a Solution


I think I can show that ##[\mathbf P, a(\mathbf q)^{\dagger}] = \mathbf q a(\mathbf q)^{\dagger}## but why in particular is this definition of the exponential useful? I can use it to rewrite the exponential factors appearing in that expression but that gives me $$\text{lim}_{n \rightarrow \infty} \text{lim}_{r \rightarrow \infty} \left(1- i \frac{\mathbf P \cdot \mathbf z}{n}\right)^n a(\mathbf q)^{\dagger} \left(1+ i \frac{\mathbf P \cdot \mathbf z}{r}\right)^r$$ but I can't see an easy way to proceed.

It seems to me that I could make more progress using the Baker Campbell Hausdorff formula for non commuting exponentials
 

Answers and Replies

  • #2
strangerep
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  • #3
CAF123
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Hi strangerep,
I think you can just use a well-known commutator->derivative formula, which comes up quite regularly in the quantum forum. See, e.g., these threads:

https://www.physicsforums.com/threads/noncommuting-series-expansion.862416

https://www.physicsforums.com/threads/coulomb-potential-as-an-operator.856374/page-2#post-5373021
Is this to help me in computing ##[\mathbf P,a(q)^{\dagger}]?## I think I managed to do this using the integral representation of the spatial momentum in terms of c/a operators. My result was ##qa(q)^{\dagger}##. So this tells me that ##a(q)^{\dagger}## and ##\mathbf P## do not commute and so the expression ##\exp(-iP \cdot z) a(q)^{\dagger} \exp(i P \cdot z)## can't be simplified trivially. i thought about using the BCH formula for ##e^{-X} A e^X## and operators X and A but I am instructed to use the limit definition of the exponential and I can't see how to progress using such a definition.
(I presume your c/a operators are bosonic?)
Yup, I should have said ##\phi## is a real scalar field.
 
  • #4
TSny
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Can you use your commutation relation for ##[\mathbf P,a(q)^{\dagger}]## to move ##a(q)^{\dagger}## all the way to the left in the expression ##\left( 1-i P \cdot z \right)a(q)^{\dagger}##?

Repeat for ##\left( 1-i \Large{\frac{P \cdot z}{2}}\right)^2a(q)^{\dagger}##.
 
  • #5
CAF123
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Hi TSny,

I suppose you are maybe hinting at me trying a few cases and seeing if there is a generalisation? $$(1-iP \cdot z) a(q)^{\dagger} = ((a(q)^{\dagger} - i((a(q)^{\dagger} P \cdot z + q \cdot z a(q)^{\dagger}))$$ Then $$\left(1-i \frac{P \cdot z}{2}\right)^2 a(q)^{\dagger} = (1-iP \cdot z) a(q)^{\dagger} - \frac{1}{4} (P \cdot z)^2 a(q)^{\dagger}$$ It seems like the first term will be common but the second term is not?

Thanks!
 
  • #6
TSny
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Hi TSny,

I suppose you are maybe hinting at me trying a few cases and seeing if there is a generalisation?
Yes.
$$(1-iP \cdot z) a(q)^{\dagger} = ((a(q)^{\dagger} - i((a(q)^{\dagger} P \cdot z + q \cdot z a(q)^{\dagger}))$$
Simplify by factoring ##a(q)^{\dagger}## out to the left on the RHS.
Then $$\left(1-i \frac{P \cdot z}{2}\right)^2 a(q)^{\dagger} = (1-iP \cdot z) a(q)^{\dagger} - \frac{1}{4} (P \cdot z)^2 a(q)^{\dagger}$$ It seems like the first term will be common but the second term is not?
Write $$\left(1-i \frac{P \cdot z}{2}\right)^2 a(q)^{\dagger} = \left(1-i \frac{P \cdot z}{2}\right) \left(1-i \frac{P \cdot z}{2}\right)a(q)^{\dagger}$$ Try moving ##a(q)^{\dagger}## all the way to the left (one set of parentheses at a time).
 
  • #7
CAF123
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Simplify by factoring ##a(q)^{\dagger}## out to the left on the RHS.
$$a(q)^{\dagger}(1-i(P+q)\cdot z)$$
Write $$\left(1-i \frac{P \cdot z}{2}\right)^2 a(q)^{\dagger} = \left(1-i \frac{P \cdot z}{2}\right) \left(1-i \frac{P \cdot z}{2}\right)a(q)^{\dagger}$$ Try moving ##a(q)^{\dagger}## all the way to the left (one set of parentheses at a time).
I get that $$a(q)^{\dagger}\left(1-\frac{i}{2}(P+q)\cdot z \right)^2$$

So, we end up with $$\text{lim}_{n,r \rightarrow \infty} a(q)^{\dagger} \left(1-\frac{i}{n}((P+q)\cdot z)\right)^n \left(1+\frac{i}{r} P \cdot z \right)^r = a(q)^{\dagger} \exp(-i(P+q)\cdot z) \exp(i P \cdot z) = a(q)^{\dagger} \exp(-iq \cdot z)$$

Similarly, I can compute e.g $$I = \exp(-iP \cdot z) \phi(x) \exp(i P \cdot z)$$ where $$\phi(x) = \int \frac{d^3 q}{(2\pi)^3 2 w(q)} (a(q) e^{-iq \cdot x} + a(q)^{\dagger} e^{iq \cdot x})$$ and using the results above gives the result that ##I## is ##\phi(t, \mathbf x - \mathbf z)## so that ##I## is a space translation operator for the field ##\phi## in the ##-z## direction. The last result makes sense to me, so I guess my answer to the first part was also correct?
 
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  • #8
TSny
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That looks good. You've shown that the momentum operator is the generator of translations in space. I think most people would refer to ##\exp(-iP \cdot z)## as the translation operator rather than ##I##.

Yes, I believe you worked the first part correctly.
 
  • #9
strangerep
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Is this to help me in computing ##[\mathbf P,a(q)^{\dagger}]?##
No. That part's easy.

I think I managed to do this using the integral representation of the spatial momentum in terms of c/a operators. My result was ##qa(q)^{\dagger}##. So this tells me that ##a(q)^{\dagger}## and ##\mathbf P## do not commute and so the expression ##\exp(-iP \cdot z) a(q)^{\dagger} \exp(i P \cdot z)## can't be simplified trivially.
I'm not sure whether you need any more hints from me -- maybe TSny's hints are enough?
 

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