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I Creation and annihilation operator commutation confusion

  1. Oct 29, 2016 #1
    In Quantum Field Theory by Lancaster, equation 3.14

    $$ [\hat{a_i},\hat{a_j}^\dagger]=\delta{ij}$$
    is introduced as "we define". Yes, example 2.1, where the creation and annihilation operators applied to harmonic operator states, there is a nice simple proof that this is true (although instead of the dirac delta, it's just 1.)

    The trouble I am having is that these are operators and they have to operate on something. The identity above says that the commutator must be true no matter what it operates on. Let's say the commutator acts on a single particle in state |i> and for this example, let i=j, so we should get 1. Then we'd have:

    $$ \hat{a_i}\hat{a_i}^\dagger|i> - \hat{a_i}^\dagger\hat{a_i}|i>$$
    After the rightmost operators act we should get:

    $$ \hat{a_i}|2i> - \hat{a_i}^\dagger|0>$$

    Here my |2i> represents a state with 2 particles in state i (there was one there to start, and the creation operator created another). Letting the remaining operators act we should get:

    $$|i> - |i> = 0$$

    I was expecting 1, not 0.

    Thanks for looking.
     
  2. jcsd
  3. Oct 29, 2016 #2

    PeterDonis

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    2016 Award

    Staff: Mentor

    Not quite: You are assuming that ##{a}^{\dagger}_i | i > = | 2i >##; but actually ##{a}^{\dagger}_i | i > = \sqrt{2} | 2i >##. (Make sure you understand why this is the case.) Similarly, ##a_i | 2i > = \sqrt{2} | i >##. The two factors of ##\sqrt{2}## applied to the first term should make things work out.
     
  4. Oct 29, 2016 #3
    Got it. Thanks Peter!
     
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