# Tag Info

28

$\renewcommand{ket}[1]{|#1\rangle}$ 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


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. ...

Only top voted, non community-wiki answers of a minimum length are eligible