Understanding second quantization

  • #1

Main Question or Discussion Point

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?
 

Answers and Replies

  • #2
atyy
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What book are you using?
 
  • #3
vanhees71
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If the Hilbert space you describe is a single-particle Hilbert space (which is a proper subspace of the Fock space the annihilation and creation operators are defined on), then what you have there is a single-particle operator. The potential you describe is then an external potential (e.g., due to an electrostatic potential acting on charged particles). For an interaction potential like the Coulomb potential between two charged particles, the corresponding operator consists of two annihilation and two creation operators. It's thus a two-particle operator, which comes about because the Coulomb potential describes interactions between all possible pairs of particles in a many-body system. This "ladder" of operators goes on in principle, i.e., you can also have three-body, four-body, etc. operators.

An example is nuclear physics, where you describe the strong interaction by an effective low-energy description of QCD, e.g., using chiral perturbation theory. This leads to many-particle forces in the many-nucleon system, i.e., to strong correlations between two and more particles, necessary to describe, e.g., nuclei as many-body bound states of nulceons (protons and neutrons).
 
  • #4
atyy
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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?
Take a look at Eq 1.26 to Eq 1.29 and Eq 1.60 in the book by Bruus and Flensberg: http://www.phys.lsu.edu/~jarrell/COURSES/ADV_SOLID_HTML/Other_online_texts/Many-body quantum theory in condensed matter physics Henrik Bruus and Karsten Flensberg.pdf

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?
Take a look at Eq 1.73 in the book by Bruus and Flensberg: http://www.phys.lsu.edu/~jarrell/COURSES/ADV_SOLID_HTML/Other_online_texts/Many-body quantum theory in condensed matter physics Henrik Bruus and Karsten Flensberg.pdf
 
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