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Quantum operators and commutation relations

  1. Apr 4, 2016 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    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.

    2. Relevant equations
    ##[AB,C] = A[B,C] + [A,C]B##

    Commutation relations for the creation and annihilation operators.

    3. 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
     
  2. jcsd
  3. Apr 4, 2016 #2

    strangerep

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    Last edited: Apr 4, 2016
  4. Apr 5, 2016 #3

    CAF123

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    Hi strangerep,
    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.
    Yup, I should have said ##\phi## is a real scalar field.
     
  5. Apr 5, 2016 #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}##.
     
  6. Apr 5, 2016 #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!
     
  7. Apr 5, 2016 #6

    TSny

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    Yes.
    Simplify by factoring ##a(q)^{\dagger}## out to the left on the RHS.
    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).
     
  8. Apr 5, 2016 #7

    CAF123

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    $$a(q)^{\dagger}(1-i(P+q)\cdot z)$$
    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?
     
    Last edited: Apr 5, 2016
  9. Apr 5, 2016 #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.
     
  10. Apr 5, 2016 #9

    strangerep

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    No. That part's easy.

    I'm not sure whether you need any more hints from me -- maybe TSny's hints are enough?
     
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