Answer to Question #122520 in Quantum Mechanics for Meli

Question

How many dimensions does hillbert space of 10 identical particles?

Answer 1

A Hilbert space is a vector space with an inner product such that the norm defined by $\begin{vmatrix} f\\ \end{vmatrix}$ = square root of (f, f) turns H into a complete metric space. If the metric defined by the norm is not complete, then H is known as an inner product space.

Examples of finite-dimensional Hilbert spaces include:

Real numbers Rⁿ with (v, u) the vector dot product of $\nu$ and u

Complex numbers Cⁿ with (v, u), the vector dot product of $\nu$ and u and the complex conjugate of u.

As a consequence of Zorn's lemma, every Hilbert space admits an orthonormal basis; any two orthonormal bases of the same space have the same cardinality, called Hilbert dimension of the space. For example, I² (B)

has an orthonormal basis indexed by B, its Hilbert dimension is the cardinality of B (which may be a finite integer or a countable or uncountable cardinal number).

As a consequence of Parseval's identity, if {ek}k ∈ B is an orthonormal basis of H, then the map Φ : H →l2(B) ( defined by Φ(x) = ⟨x, ek⟩k∈B is an isometric isomorphism of Hilbert spaces: it is a bijective linear mapping such that

al basis indexed by B, its Hilbert dimension is the cardinality of B (which may be a finite integer, or a countable or uncountable cardinal number).

As a consequence of Parseval's identity, if {e_k}_{k ∈ B} is an orthonormal basis of H, then the map Φ : H → l²(B) defined by Φ(x) = ⟨x, e_k⟩_k∈B is an isometric isomorphism of Hilbert spaces: it is a bijective linear mapping such that${\displaystyle {\bigl \langle }\Phi (x),\Phi (y){\bigr \rangle }_{l^{2}(B)}=\left\langle x,y\right\rangle _{H}}$

for all x, y ∈ H. The cardinal number of B is the Hilbert dimension of H. So, every Hilbert space is isometrically isomorphic to a sequence space  l2(B) for some set B.

The sequence space l2 consists of all infinite sequences z = (z1, z2, …) of complex numbers such that the series

${\displaystyle \sum _{n=1}^{\infty }|z_{n}|^{2}}$

converges.

The inner product on I^2 is defined by

${\displaystyle \langle \mathbf {z} ,\mathbf {w} \rangle =\sum _{n=1}^{\infty }z_{n}{\overline {w_{n}}}\,,}$

with the latter series converging as a consequence of Cauchy-Schwarz inequality.

Completeness of the space holds provided that whenever a series of elements from I^2 converges absolutely, then it converges to an element of I^2.

Consider two particles occupying N distinct states |1>, 2>,....,| N>. The Hilbert space dimension of this two-particle system if the two particles are identical bosons will be just N and N^2 - N due to Pauli exclusion if they are in different states.

Assume the states to be N*1column matrices. The two particle state is an N*N

matrix. For bosons, we need a symmetrical matrix. It has N(N+1)/2 distinct terms. For fermions, we need an anti-symmetrical matrix that has N(N-1)/2 distinct terms.