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## Hot answers tagged second-quantization

28

$\renewcommand{ket}{|#1\rangle}$ Item #4 in your list is best thought of as the definition of the word "particle". Consider a classical vibrating string. Suppose it has a set of normal modes denoted $\{A, B, C, \ldots\}$. To specify the state of the string, you write it as a Fourier series $$f(x) = \sum_{\text{mode } n=\in \{A,B,C,\ldots \}} c_n [\text{... 25 Suppose you have a system described by a Hilbert space H, for example a single particle. The Hilbert space of two non-interacting particles of the same type as that described by H is simply the tensor product$$H^2 := H \otimes H$$More generally, for a system of N particles as above, the Hilbert space is$$H^N := \underbrace{H\otimes\cdots\otimes H}_{...

21

You are correct, Bogoliubov transformations are not unitary in general. By definition, Bogoliubov transformations are linear transformations of creation/annihilation operators that preserve the algebraic relations among them. The algebraic relations are mainly the commutation/anticommutation relations which define the bosonic/fermionic operators. Nowhere ...

15

It doesn't, really. You are definitely free to introduce a new vector field $p^\mu$ and add an interaction $\bar \psi \!\!\not\! p\,\psi$ to your Lagrangian, together with a kinetic term for $p^\mu$. Whether you do this or not, you must include the standard kinetic term $\bar \psi \!\!\not\! \partial\,\psi$ anyway. In this case the new term $\bar \psi \!\!\... 14$\newcommand{\bx}{\mathbf{x}} \newcommand{\psih}{\hat{\psi}} $This is worked out on p. 20 of Fetter and Walecka. I'll add just a little extra detail here. Your particle density operator$n(\bx)=\sum_\alpha\delta(\bx-\bx_\alpha)$is in first-quantized form and$\hat{n}(\bx)=\psih^\dagger(\bx)\psih(\bx)$is second-quantized. When a general one-body ... 14 Great answers, but just for completeness maybe it will be illustrative to have an example. Suppose your$H^1$contains some single-particle states$|a\rangle$,$|b\rangle$, etc. The Fock space removes the limitation on being a single particle, and is composed of$H^0$(which is 1-dimensional),$H^1$,$H^2 = H \otimes H$, etc. This allows states like the ... 14 It's just a convention to write the Dirac Hamiltonian in terms of electron and positron operators. The operator$b_p^{s\dagger}$creates a "hole", i.e. a positron, which is the same as annihilating an electron. So we could just as well define$c_p^s = b_p^{s\dagger}$and write the Hamiltonian in terms of$a$and$c$instead of$a$and$b$. Then everything is ... 13 For simplicity, let us talk about a scalar field$\phi : \mathbb{R}^4 \rightarrow \mathbb{R}$. The action for a free scalar field is $$S[\phi] = \frac{1}{2}\int_{\mathbb{R}^4} \partial_\mu\phi\partial^\mu\phi - m^2\phi^2$$ and its classical equations of motion is the Klein-Gordon equation $$(\partial_\mu\partial^\mu + m^2) \phi = 0$$ Now that looks ... 12 It is very complicated to construct a consistent quantum theory from scratch. One of the most general methods to do so is to take a classical theory, and imitate some of its ingredients - the most important being the symplectic structure and the generators of symmetries. This is called quantisation, because the quantum theory is built using a classical one. ... 11 Unitarity of a quantum mechanical transformation is not determined by how it mixes creation and annihilation operators. (It doesn't matter what kind of matrix—orthogonal, symplectic, or unitary—is involved in the mixing!) Rather, one should examine whether the transformation is associated with a unitary operator acting on the Hilbert space. The ... 11 An axiomatic definition of normal order can be found in the book "Solitons: Differential Equations, Symmetries and Infinite Dimensional Algebra" by T. Miwa, M. Jimbo and E. Date at page 44-46. This is the best definition I have found so far and goes as follow for products of bosonic operators. Call$\mathcal{A}$the set of linear combinations of formal ... 10 The presumed equivalence between the canonical quantization and the Fock space representation is only a particular case. The canonical formalism provides only with canonical Poisson brackets. The first step according to Dirac's axioms is to replace the Poisson brackets by commutators and since these commutators satisfy the Jacobi identity, they can be ... 10 As a matter of fact one could also discuss commutation relations at different time: $$[\phi(x),\phi(y)] = i E(x,y) I\quad (1)$$ for free fields$E$is the so-called causal propagator or advanced minus retarded fundamental solution that depends on the free field equation satisfied by$\phi$. The point is that, passing to considering interacting fields, at ... 10 For a free theory, say for one scalar field for simplicity, which gives a a linear differential equation for the field$\phi$, one can cast the Hamiltonian $$H=\frac{1}{2}\int d^3x \dot\phi^2+(\partial_i \phi)^2+ m^2\phi^2$$ in this form (basically by taking a Fourier Transform): $$H=\mathrm{const}+\int \frac{d^3 k}{(2\pi)^3} \omega(k)a^\dagger(k)a(k)\,,... 10 Indeed, the ordinary Schrödinger equation can be second-quantized, yielding non-relativistic QFT. To do so, one first rewrites the Schrödinger equation as a "classical" field theory: the phase space of this field theory is the Hilbert space \mathcal{K} of the first-quantized theory with (complex) coordinates a_k(\psi) = \left\langle e_k \middle| \psi \... 10 Diagonalizing the Hamiltonian means you want to bring it into the form H=\omega b^\dagger b, and it is pretty obvious that b should be a linear combination of a and a^\dagger, and b should satisfy the canonical commutation of annihilation operators, namely [b,b^\dagger]=1, [b,b]=0. Now let's write b=ua+va^\dagger (this is called the ... 10 There are quite a number of notions of quantisation. It is a theorem by van Hove & Groenewald that the naive notion of quantisation that is usually taught where position and momentum are mapped to the usual operators cannot hold in all generality, that is the classical Poisson bracket is not mapped perfectly to the commutation relations we would expect. ... 9 Take the Hamiltonian you write at end of your question as the example(where your statement about the eigenvector is not right), the procedure to obtain the eigenvalue and eigenvectors of other Hamiltonian is nearly the same.$$H=\sum_{ij}A_{ij}a_i^\dagger a_j$$Writing in matrix form you have(suppose it is a n\times n matrix):$$H=\alpha^\dagger A \... 8 In the statistical mechanics of the grand canonical ensemble, one needs to allow for superpositions and mixtures of of states with different particle number. Thus one is naturally led to considering the tensor product of the$N$-particle spaces with arbitrary$N$. It turns out (and is very relevant for nonequilibrium statistical mechanics) that one can ... 8$\def\rr{{\bf r}} \def\ii{{\rm i}}$There is indeed a continuity equation for the particle density$\rho(\rr)=\Psi^\dagger(\rr)\Psi(\rr),$where the field operator$\Psi^\dagger(\rr)$creates a particle at position$\rr. To derive it, you need only the canonical commutation relations for the field \begin{align} [\Psi(\rr),\Psi^\dagger(\rr')]& = \delta(\... 8 the actual non-relativistic part is the Schrödinger equation Indeed. Hence, people tried to come up with a Lorentz-invariant evolution equation for the wave function such as the Klein-Gordan and Dirac equations (as mentioned by anna v). However, interpretation of solutions of these equations as probability densities are problematic (in case of the Klein-... 7 The (first) quantization is a sound mathematical procedure: usually it associates to a function of two variablesa(x,\xi):\mathbb{R}^d \times \mathbb{R}^d \to \mathbb{C}$, an operator$a(x,D_x)$($D_x$is the derivation multiplied by$-i$) on$L^2(\mathbb{R}^d)$. There are various types of quantization (e.g. Weyl, Wick, Anti-Wick, Born-Jordan) that deal ... 7 As stated in some of the comments, the weighting functional inside the functional integral is sometimes called a wave functional. This type of functional representation can be called the functional Schroedinger representation. Please see the following review by Roman Jackiw. For a general interacting field theory, the exact solution for this wave functional ... 7 The point is that the quantization procedure is usually only valid for real-valued physical observables. All versions of treat the classical observables as real functions on phase space (things get more complicated for fermions, which I will ignore for this issue), and associate quantum observables to those. For instance, the harmonic oscillator annihilation ... 7 There is a conceptually simple (but fiddly) way of relating this expansion to the usual Fourier expansion. TL;DR: Requiring$\phi$to satisfy the Klein-Gordon equation divides the nonzero Fourier components into two classes, corresponding to particles and antiparticles. For a general complex scalar field defined on spacetime, $$\phi(x) = \int\frac{d^4 p}{(... 7 That's essentially what we mean by a quasiparticle: a degree of freedom that has been otherwise decoupled from the rest of the system and which behaves exactly like a particle as far as quantum mechanics can tell. Since the (anti)commutation relations are the crucial part of the quantum mechanics of the relevant particles, those need to be carried as well. ... 7 In Schrodinger picture, as long as you don't have any time-varying external fields, the creation and annihilation operators are time-independent like every other operator. In Heisenberg picture, they should be time-dependent, like every other operator. This is physically intuitive. The creation operator at time t is$$a^\dagger(t) = e^{- i H t} a^\dagger(... 6 Starting with some background information from Wikipedia, we have that under time reversal the position is unchanged while the momentum changes sign. In quantum mechanics we can express the action of time reversal on these operators as$\Theta\,\mathbf{x}\,\Theta^\dagger = \mathbf{x}$and$\Theta\,\mathbf{p}\,\Theta^\dagger = -\mathbf{p}$. It is worth ... 6 The expression $$\hat{H}=\sum_j \varepsilon_j\,a_j^\dagger a_j$$ is not the most general expression for free particles hamiltonian because it implies that you already found the eigenvalues$\varepsilon_j$and diagonalized$\hat{H}$, i.e. already solved the Schrödinger equation. Maybe you should look at the problem in a different basis. Let say$\{\vert j\...

6

The state $\Psi^+(\mathbf{r})|0\rangle$ means the particle created in the point $\mathbf{r}$, it is not very convenient to work with because wave function of this particle is the Dirac delta. So if you need a practical recipe for calculation, you can try to smear out this state in space creating a particle with the wave function $f(\mathbf{r})$, i.e. ...

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