# Understanding second quantization

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• acegikmoqsuwy
In summary, the conversation discusses the concepts of second quantization and creation and annihilation operators in quantum mechanics. It explains how an operator can be defined on a one-particle Hilbert space and how it can be extended to a many-particle space using creation and annihilation operators. It also discusses the interpretation of these operators and how they relate to external potentials in many-body systems. The conversation references the book by Bruus and Flensberg for further explanation and equations.
acegikmoqsuwy
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?

What book are you using?

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).

bhobba
acegikmoqsuwy said:
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

acegikmoqsuwy said:
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

Last edited:
vanhees71

## 1. What is second quantization?

Second quantization is a mathematical framework used in quantum mechanics to describe systems with an arbitrary number of particles. It allows for the representation of particles as creation and annihilation operators, which simplifies the mathematical calculations involved in quantum systems.

## 2. How is second quantization used in physics?

Second quantization is used to describe quantum systems with a varying number of particles, such as in quantum field theory or many-body systems. It allows for the representation of particles as excitations of a quantum field, rather than individual particles, making it a powerful tool in understanding the behavior of complex systems.

## 3. What is the difference between first and second quantization?

First quantization is used to describe systems with a fixed number of particles, such as in non-relativistic quantum mechanics. Second quantization extends this concept to systems with a varying number of particles, allowing for a more comprehensive description of quantum systems.

## 4. What are creation and annihilation operators?

Creation and annihilation operators are mathematical operators used in second quantization to represent the creation and destruction of particles in a quantum system. The creation operator adds a particle to the system, while the annihilation operator removes a particle. These operators follow specific mathematical rules and are used to simplify calculations in complex quantum systems.

## 5. Why is second quantization important in understanding quantum systems?

Second quantization is important because it allows for the description of quantum systems with a varying number of particles, which is often the case in real-world systems. It also simplifies the mathematical calculations involved in quantum mechanics, making it a powerful tool in understanding the behavior of complex systems.

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