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So far in our lecture we defined creation operators $a^{\dagger}_{n}$ in the following way, that we said:

Somebody got you a antisymmetric or symmetric N- particle state and now $a^{\dagger}_{n}$ puts another particle in state n, so that we end up with a symmetric/antisymmetric N+1-particle state. This interpretation is somehow clear to me in the sense that these $a^{\dagger},a$ operators avoid the cumbersome slater determinants and so on. Despite, we are still dealing with well-defined symmetrized/antisymmetrized product states that become extended or reduced by one state, which are hidden behind this notation.

Now, we also defined field-operators in QM by $\psi^{\dagger}(r) = \sum_{i;\text{all states}} \psi_i^*(r) a_i^{\dagger}.$ We said that they create a particle at position $r$. Somehow, it is not clear to me what this means:

To create a particle at an exact position $r_0$ in QM would mean that we now have an additional state $\psi_i(r) = \delta(r-r_0)$ in our slater determinant. I doubt that this is the idea behind this. But, since the $a_i^{\dagger}$ operators act on $N$-particle state and map to $N+1$ particle states, the same must be true for $\psi^{\dagger}(r)$. Nevertheless, I have difficulties interpreting the result.

If anything is unclear, please let me know.

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The $\psi_i$ in your sum do not need to be delta functions. You can think for example as them being energy eigenfunctions $$ \mathcal{H} \psi_i(r) = E_i \psi_i(r) $$ thus creating a particle at $r$ means that you obtain a superposition of all the possible ways a particle can be at $r$ (in this particular choise of basis): $$ \underbrace{\psi^\dagger ( r )}_{\text{operator}} | 0 \rangle = \sum_i \overbrace{\psi_i^*(r)}^{\text{complex numbers}} | i \rangle$$ where $|0\rangle$ is the vacuum state (or ground state if you want) and $|i \rangle$ is the Fock state with one particle in the n-th mode. You can think of this equation as stating the for each $i$, $\psi_i^*(r)$ is the probability amplitude of finding the particle at the position $r$ if you know it is in the state $i$.

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  • $\begingroup$ the interpretation of creating a superposition of all the possible ways a particle can get to the position $r$ looks meaningful to me. I mean what we do is, if I understood you correctly, that we create a particle in any eigenstate and look for the probability amplitude that this particle is at position $r$. What I don't see is how this notion is related to the actual creation of a particle at position $r$. If you think about it, then these are two different things. Could you try to explain, what we want to model with this field operator? $\endgroup$ – Solid State Physicist Nov 14 '14 at 23:21
  • $\begingroup$ It really depends on the context. The "particle" interpretation is not always suitable, more generally you can think of these operators as creating/annihilating quantum states. In the context of QFT these states are indeed (usually) particle states and $|0\rangle$ the state with no particles, and hence the terminology. But for example in NRQM this is often not true, and the "vacuum state" is in this case just the ground state of the system. They "create"/"destroy" states in the sense that they send a given Fock space into another with one additional/less state of that particular kind. $\endgroup$ – glS Nov 14 '14 at 23:34
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Think of it as a change of basis. $a_i^\dagger$ creates a particle in the state $|i\rangle$. Now, this state $|i\rangle$ can be written in terms of the position states $|r\rangle$ as $$ |i\rangle=\int dr\, \psi_i(r)|r\rangle,$$ thus creating a particle in this state is equivalent to create a particle in a superposition of position state with the appropriate weight $\psi_i(r)$. Equivalently, a particle localized in $|r\rangle$ can be described as being in a superposition of state $$|r\rangle=\sum_i \psi_i^*(r)|i\rangle,$$ and thus creating a particle in the state $|r\rangle$, the operator $\psi^\dagger(r)$ is defined by the operator $\sum_i \psi_i^*(r)\,a_i^\dagger$.

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  • $\begingroup$ sorry, but this answer is highly confusing. you seem to sum over positions. Notice, that position are not discrete! Thus, I have severe troubles to understand your $|r \rangle$ 's. $\endgroup$ – Solid State Physicist Nov 14 '14 at 21:26
  • $\begingroup$ @TobiasHurth: that's just notations (think about a discretized version of space). But I just changed to integral, if that makes you feel better. $\endgroup$ – Adam Nov 15 '14 at 0:50

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