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Difference wave mch. and field operators.

  1. Sep 21, 2012 #1
    Hi there

    I've recently started studying quantum field theory and I'm trying to understand the field operators.

    One thing that bugs me is the difference between field operators and wave mechanics operators. For instance, let's take the kinetic energy operator in wave mechanics for a single particle
    [tex] T_1(\mathbf{r}) = -\frac{\hbar^2}{2m} \nabla^2 + V_1(\mathbf{r}) [/tex]
    And then we have the standard field operator, usually written as
    [tex] \psi(\mathbf{r}) = \sum_k c_k u_k(\mathbf{r}) [/tex]
    where [itex]c_k[/itex] are lowering and raising operators from single particle QM and [itex]u_1(\mathbf{r})[/itex] are single particle wavefunctions in state [itex]k[/itex].

    I've been told that they are two totally different things. But I'm not sure in what manner. For example, what would their commutator give?

    Cheers,
    S
     
  2. jcsd
  3. Oct 1, 2012 #2
    The field operators create (or annihilate) particles in the position eigenstates. In second quantization the fields are the operators and the kinetic and potential energies are complex coefficients. The roles are in some sense reversed from first quantization.

    I'm not really sure what you are asking about the commutator. Are you asking what [tex] [\nabla^2,\psi(\mathbf{r})] [/tex] gives? I don't think this commutator makes sense, since the field operators are operators in an abstract occupation-number Hilbert space (they depend on the creation and annihilation operators) whereas the kinetic energy operator in first quantization is an operator in a Hilbert space of complex wave functions. If one wants to calculate the commutator of a field operator and the kinetic energy operator, the kinetic energy operator should be expressed in terms of the field operators first.
     
  4. Oct 1, 2012 #3
    Yes yes, this is exactly what I'm wondering about. It's part of understanding why [itex][N,H] = 0[/itex] where
    [tex] H = T + V = \int d^3 r \psi^\dagger(\mathbf{r}) T_1(\mathbf{r}) \psi(\mathbf{r}) + \frac{1}{2} \iint d^3r \ d^3r' \psi^\dagger(\mathbf{r}) \psi^\dagger(\mathbf{r}') V_2(\mathbf{r}, \mathbf{r}')\psi(\mathbf{r}') \psi(\mathbf{r}) [/tex]
    [tex] N = \int \psi^\dagger(\mathbf{r}) \psi(\mathbf{r}) \ d^3r [/tex]

    Let's just focus on the kinetic term
    [tex] [N,H] = [N,T] + [N,V] = 0 [/tex]
    [tex] [N,T] = \left[ \int \psi^\dagger(\mathbf{r}) \psi(\mathbf{r}) \ d^3r \ , \ \int \psi^\dagger(\mathbf{r}') T_1(\mathbf{r}') \psi(\mathbf{r}') d^3 r' \right] [/tex]
    [tex] = \iint d^3r d^3 r' \left[ \psi^\dagger(\mathbf{r}) \psi(\mathbf{r}) \ , \ \psi^\dagger(\mathbf{r}') T_1(\mathbf{r}') \psi(\mathbf{r}') \right] = \iint d^3r d^3 r' T_1(\mathbf{r}') \left[ \psi^\dagger(\mathbf{r}) \psi(\mathbf{r}) \ , \ \psi^\dagger(\mathbf{r}') \psi(\mathbf{r}') \right] [/tex]

    I've already verified that
    [tex] \int d^3r \ \left[ \psi^\dagger(\mathbf{r}) \psi(\mathbf{r}) \ , \ \psi^\dagger(\mathbf{r}') \psi(\mathbf{r}') \right] = 0 [/tex]

    Which means that everything fits if I can allow myself to extract [itex]T_1(\mathbf{r})[/itex] from the commutator. This means that in Fock space the wave mechanical operators commute for Fock states (or don't act on them, I'm confused what this means) whereas second quantization operators don't necessarily commute.
     
  5. Oct 1, 2012 #4
    You cannot just extract [itex]T_1[/itex] from the commutator. To show that [itex][N,T] = 0[/itex], you can use, for example, the commutator relation [tex][AB,C]=A[B,C]+[A,C]B[/tex] and once you get commutator terms like [tex]\psi^\dagger(\mathbf{r}')[\nabla'^2 \psi(\mathbf{r}'),\psi^\dagger(\mathbf{r}) \psi (\mathbf{r})][/tex] then you can extract the [itex]\nabla'^2[/itex], since the other field operators in the commutator depend on [itex]\mathbf{r}[/itex], not on [itex]\mathbf{r}'[/itex]. Finally after using the commutation relations for the field operators, you get two canceling terms, and thus [itex][N,T] = 0[/itex].
     
    Last edited: Oct 1, 2012
  6. Oct 10, 2012 #5
    Hmmmm.... I think I get what you're saying there. If I get terms like
    [tex] \left[ \frac{\partial^n}{\partial x^n} , f(x') \right] [/tex]
    then this is automatically vanishes because I'm differentiating with respect to another variable?
     
  7. Oct 18, 2012 #6
    Yes, those commute, but that commutator makes sense only if the two operators operate in the same space. What I mean is that you can write, for example,
    [tex]\psi^\dagger(\mathbf{r}')[\nabla'^2 \psi(\mathbf{r}'),\psi^\dagger(\mathbf{r}) \psi (\mathbf{r})]=\psi^\dagger(\mathbf{r}')\nabla'^2[\psi(\mathbf{r}'),\psi^\dagger(\mathbf{r}) \psi (\mathbf{r})][/tex]
    and further use the commutator relation to prove that [itex][N,T] = 0[/itex].
     
    Last edited: Oct 18, 2012
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