Physics, asked by bhavnad8119, 11 months ago

Can interacting Hamiltonians always be written in second quantized form?

Answers

Answered by xyz592
0
As we have seen, we can formulate the quantum mechanics for a single particle in an abstract Hilbert
space, although in practice we (almost) always end up working in some given representation, in terms
of wavefunctions. Can we generalize this approach to many-body systems? Yes, we can quite easily,
if the number N of particles is fixed. In this case, the many-body state in the ~r-representation is (we
use again x = (~r, σ)):
hx1, . . . , xN |Ψ(t)i = Ψ(x1, . . . , xN ,t)
which is the amplitude of probability that at time t, particle 1 is at x1 = (~r1, σ1), etc. Note that
if the number of particles is not fixed (for instance, in a grand-canonical ensemble) we’re in some
trouble deciding what h| to use. But let us ignore this for the moment, and see how we get in trouble
even if the number of particles is fixed.
We almost always decompose wavefunctions in a given basis. For a single particle, we know
that any wavefunction Ψ(x,t) can be decomposed in terms of a complete and orthonormal basis
φα(x) = hx|αi as Ψ(x,t) =
P
α
cα(t)φα(x) – this reduces the problem to that of working with the
time-dependent complex numbers cα(t).
If we have a complete basis for a single-particle Hilbert space, we can immediately generate a
complete basis for the N-particle Hilbert space (since the particles are identical), as the products of
one-particle basis states:
Ψ(x1, . . . , xN ,t) =
X
α1,...,αN
cα1,...,αN
(t)φα1
(x1)φα2
(x2)· · · φαN
(xN )
However, since the particles are identical, we know that the wavefunctions must be symmetric
(for bosons) or antisymmetric (for fermions) to interchange of any particles:
Ψ(x1, . . . , xN ,t) = ξ
PΨ(xP1
, . . . , xPN
,t)
where
Ã
1 2 . . . N
P1 P2 . . . PN
is any permutation, P is its sign (number of transpositions), and from now on we use the notation:
ξ =
(
−1, for fermions
+1, for bosons
(2.1)
It follows immediately that c...,αi,...,αj ,... = ξc...,αj ,...,αi,... and therefore c...,αi,...,αi,... = 0 for fermions: we
cannot have two or more fermions occupying identical states σi (the Pauli principle).
Because of this requirement, the physically meaningful many-body fermionic (bosonic) wavefunc-
tions are from the antisymmetric (symmetric) sector of the N-particle Hilbert space, and we only
need to keep the properly symmetrized basis states, which we denote as:
φα1,...,αN
(x1, . . . , xN ) =
1
q
N!
Q
i ni
!
X
P ∈SN
ξ
P φα1
(xP1
)φα2
(xP2
)· · · φαN
(xPN
) (2.2)
The factor in front is just a normalization constant; ni
is the total number of particles in the same
state αi (only important for bosonic systems), and the summation is over all possible permutations,
of which there are N!/
Q
i ni
! distinct ones.
Then, any many-body wavefunction can be written as:
Ψ(x1, . . . , xN ,t) =
X
α1,...,αN
cα1,...,αN
(t)φα1,...,αN
(x1, . . . , xN )





















Answered by Anonymous
0

Explanation:

Can interacting Hamiltonians always be written in second quantized form?

Yes

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