# Tagged Questions

Second quantization or canonical quantization in quantum field theory and many-body systems is the collective organizing and accounting of an infinity of quantum excitations and their interactions through quantum field operators.

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### One-electron reduced density matrix: Argument for positive semidefiniteness

I cannot follow an argument for positive-semidefiniteness of the one-electron density matrix given in "Molecular Electronic-Structure Theory" by Helgaker/Jorgensen/Olsen. First some definitions: $F(M)$...
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### How is intensity defined for quantized EM fields?

Classically intensity is defined as $$I \equiv \frac{1}{2} c \epsilon_0 E^2,$$ but when you perform a second quantization this definition becomes a bit ambiguous since the $E^2$ could be ...
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### Is Hamiltonian a differential operator in second quantization?

Normally, a free particle Hamiltonian is written $$\hat{H} = - \frac{\hbar^2}{2m} \Delta$$ which is a differential operator because Laplacian $\Delta$ is. On the other hand, in second ...
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### second quantization lost particle number information?

In first quantization, we can read the particle number from the Hamiltonian. $$H=\sum_{i=1}^N \hat{T}(x_i)$$ Converting this to second quantization form, the particle number of the system is lost: ...
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### What is the form of the kinetic energy operator on a one-dimensional (real space) lattice? (In second quantization)

I'm trying to figure out how one would write down the Hamiltonian of a free fermion system (eventually in second quantization) on a one dimensional lattice and I'm having trouble both coming up with ...
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### Why are photons bosonic?

I am studying the quantization of the electromagnetic field. My text quantizes by changing amplitudes to ladder operators, by putting in an action and by imposing bosonic commutation relations upon ...
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### Spin zero photons

As I understand it, the reason why there is no Spin 0 Photon is because the polarisation of an EM field lives in two dimension. Hence we only have two basis vectors, yielding two pairs of ladder ...
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### Where can I find a detailed derivation of the form of two body operators in the second quantization?

I've been looking around online for a couple hours now and I can't find a very informative derivation of the form for two body operators in the second quantization. Is there a resource online (...
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### What is the physical interpretation of second quantization?

One way that second quantization is motivated in an introductory text (QFT, Schwartz) is: The general solution to a Lorentz-invariant field equation is an integral over plane waves (Fourier ...
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### Fourier transform of a set of L fermions operators

I have a set of L fermion creation and annihilation operators: $\lbrace{\hat{C}^+_1,...,\hat{C}^+_L\rbrace}$ and $\lbrace{\hat{C}^-_1,...,\hat{C}^-_L\rbrace}$. Every $\hat{C}^+_l,\hat{C}^-_l$ ...
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### Creation and annihilation form of hamiltonian to derive a relation between the ac current applied to the crystal and the oscillations of the crystal

in the book "many-particle physics" by G.Mahan in piezoelectric subsection, it uses the second quantization formalism to derive the relation for hamiltonian of the electron-phonon interaction. so ...
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### Multiply creation operator by a phase factor

A basic question, but I'm not completely confident what I'm doing is legit. I can multiply a creation operator by an arbitrary phase factor and it doesn't change any physics. True? I have a ...
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### Fourier transform of random variables

My question is concerning Fourier transforms of random variables. So if the question itself is too heavy a task but you know of any good resources to learn this topic that would also be very much ...
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### Why does trying to remove a non-existing electron from a state give zero?

Setup Creating an electron that is already in a basis set is zero (Pauli's principle): a_i^+ | \chi_i \cdots \chi_k \cdots \chi_l \rangle = | \chi_i \chi_i \cdots \chi_k \cdots \...
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### What is the physical interpretation of a field operator

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 $a^{\dagger}_{n}$ ...