- #1
acegikmoqsuwy
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Hi,
I was reading a book about second quantization and there were a few things that I didn't quite understand entirely.
This is what I understood so far:
Given an operator ##\mathcal A## and two orthonormal bases ##|\alpha_i\rangle## and ##|\beta_i\rangle## for the Hilbert space, ##\mathcal H##, of interest, we can write $$\mathcal A = \sum\limits_{ij} \mathcal A_{ij} |\alpha_i\rangle \langle\beta_j|$$ where ##\mathcal A_{ij} = \langle \alpha_i|A|\beta_j\rangle.##
Now if we let ##a_i## and ##a_i^{\dagger}## be the annihilation and creation operators for the state ##|\alpha_i\rangle## and similarly ##b_i## and ##b_i^{\dagger}## be those for ##|\beta_i\rangle## (satisfying the appropriate commutation relations) then we can define an operator, ##A##, on the Fock space corresponding to ##\mathcal H## given by $$A = \displaystyle\sum\limits_{ij} \mathcal A_{ij} a_i^{\dagger} b_j.$$
The first question I have is this: what is the actual interpretation of ##A##? I can see that, whereas ##\mathcal A## is only defined on a one-particle Hilbert space, ##A## is a map from a many-particle space to itself (in particular, preserving the number of particles). Is it supposed to be the case that ##A## reduces to ##\mathcal A## in the single particle case? Is the action of ##A## on a state of particles supposed to somehow "simultaneously" be an application of ##\mathcal A## to all the particles at once?
Second: In my book, it is written that for an operator (on single particle Hilbert space) depending on position, we can write the second quantization in a similar form, but using the creation and annihilation operators ##\hat \psi^{\dagger}(x)## and ##\hat \psi(x)## for position. The equation it gives for second quantization of a potential function operator ##V(x)## is (assuming that the position spectrum is not discrete) $$\hat V = \int \text d^3 x \hat \psi^{\dagger}(x) V(x) \hat \psi(x).$$ This appears to be slightly different in the order that the terms are laid out when compared to the expression for ##A## above. Now, a few other manipulations performed in the book lead me to guess that somehow the operators ##\hat \psi^{\dagger}(x)## and ##\hat \psi(x)## commute with the operator ##V(x)##. Why is this the case? I know that ##\hat \psi^{\dagger}(x)## and ##\hat \psi(x)## are operators on the Fock space, but isn't ##V(x)## also an operator on Fock space in a sense?
I was reading a book about second quantization and there were a few things that I didn't quite understand entirely.
This is what I understood so far:
Given an operator ##\mathcal A## and two orthonormal bases ##|\alpha_i\rangle## and ##|\beta_i\rangle## for the Hilbert space, ##\mathcal H##, of interest, we can write $$\mathcal A = \sum\limits_{ij} \mathcal A_{ij} |\alpha_i\rangle \langle\beta_j|$$ where ##\mathcal A_{ij} = \langle \alpha_i|A|\beta_j\rangle.##
Now if we let ##a_i## and ##a_i^{\dagger}## be the annihilation and creation operators for the state ##|\alpha_i\rangle## and similarly ##b_i## and ##b_i^{\dagger}## be those for ##|\beta_i\rangle## (satisfying the appropriate commutation relations) then we can define an operator, ##A##, on the Fock space corresponding to ##\mathcal H## given by $$A = \displaystyle\sum\limits_{ij} \mathcal A_{ij} a_i^{\dagger} b_j.$$
The first question I have is this: what is the actual interpretation of ##A##? I can see that, whereas ##\mathcal A## is only defined on a one-particle Hilbert space, ##A## is a map from a many-particle space to itself (in particular, preserving the number of particles). Is it supposed to be the case that ##A## reduces to ##\mathcal A## in the single particle case? Is the action of ##A## on a state of particles supposed to somehow "simultaneously" be an application of ##\mathcal A## to all the particles at once?
Second: In my book, it is written that for an operator (on single particle Hilbert space) depending on position, we can write the second quantization in a similar form, but using the creation and annihilation operators ##\hat \psi^{\dagger}(x)## and ##\hat \psi(x)## for position. The equation it gives for second quantization of a potential function operator ##V(x)## is (assuming that the position spectrum is not discrete) $$\hat V = \int \text d^3 x \hat \psi^{\dagger}(x) V(x) \hat \psi(x).$$ This appears to be slightly different in the order that the terms are laid out when compared to the expression for ##A## above. Now, a few other manipulations performed in the book lead me to guess that somehow the operators ##\hat \psi^{\dagger}(x)## and ##\hat \psi(x)## commute with the operator ##V(x)##. Why is this the case? I know that ##\hat \psi^{\dagger}(x)## and ##\hat \psi(x)## are operators on the Fock space, but isn't ##V(x)## also an operator on Fock space in a sense?