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erite423@yahoo.se
science forum beginner

Joined: 10 Jul 2006
Posts: 3

Posted: Mon Jul 10, 2006 3:41 pm    Post subject: Density operator in second quantization

I have some questions concerning density operators and the second
quantization formalism:

1. What is the density operator for a many-electron system in second
quantization?

2. How do I trace out degrees of freedom to form reduced density
operators?

3. Can the operator |vac> <vac|, mapping the vacuum state to itself
and all others to zero, be expressed in terms of creation and
annihilation operators?

Some further explanation of what makes me ask these questions: I am
used to working with the bracket notation in which the density
operator takes forms like

rho = |psi> <psi| (pure state case)
rho = 1/Z exp(-H/T) (thermal equilibrium at temperature T)
rho = w_1 |1> <1| + w_2 |2> <2| + ... (in general)

If b_n = (a_n)^+ is a creation operator, a simple example of the pure
state case might be

|psi> = b_1 |vac>
rho = b_1 |vac> <vac| a_1

and a simple example of the thermal mixture might be

H = h_{11} b_1 a_1 + h_{22} b_2 a_2 + ...
rho = C_{11} b_1 a_1 + C_{12} b_1 a_2 + C_{21} b_2 a_1 + ...

In the first case the creation and annihilation operators are always
separated by the operator |vac> <vac|, which seems to have no analogue
in the second case.

The books I've looked in did not say much about these issues. All I've
been able to find out is that the 1-particle density matrix can be
written

D_{pq} = <psi| b_p a_q |psi>

But I'm not sure how to start from a many-particle density operator,
trace out all particles except one, take matrix elements, and finally
arrive at the matrix D_{pq}.

Erik
Igor Khavkine
science forum Guru

Joined: 01 May 2005
Posts: 607

Posted: Tue Jul 11, 2006 11:38 pm    Post subject: Re: Density operator in second quantization

erite423@yahoo.se wrote:
 Quote: I have some questions concerning density operators and the second quantization formalism: 1. What is the density operator for a many-electron system in second quantization?

It's defined in the same way as in ordinary quantum mechanics. Pure
states are obtained by an outer product like |phi><phi|, while mixed
states are obtained by taking convex linear combinations and limits of
pure states. The difference now is that the states |psi> must be taken
from the multi-particle Fock space instead of the single particle
subspace.

However, I must warn you about a possible clash in terminology. When
dealing with many particle systems (which is where second quantization
comes in), there a physical observable that's called the "density
operator". It's defined as rho(x) = :psi(x)* psi(x):, which is the
normal ordered product of field operators. For example, when dealing
with a multi-electron system, it represents the total charge density at
point x. What you are talking about is still conventionally referred to
as a "density matrix", despite the fact that it's not really a matrix
in the textbook sense of the word.

 Quote: 2. How do I trace out degrees of freedom to form reduced density operators?

Exactly the same way as you are used to. The first requirement for
taking a partial trace is the possibility of expressing the state space
as a tensor product. For example, when dealing with the quantization of
a linear field (or the second quantization of some single-particle
theory, which are equivalent), the state space can be expressed as a
tensor product of the states associated to short wavelength field modes
and long wavelength field modes (with an arbitrarily placed cutoff).
Then either the short- or longwavelength modes can be traced out.

 Quote: 3. Can the operator |vac>

Let N = sum_k a*_k a_k be the particle number operator. Its spectrum
consists of the natural numbers 0, 1, 2, ... . Take an analytic
function f(x) which satisfies f(0) = 1 and f(n) = 0, for integer n > 0.
Then P_vac = f(N) will be the projection operator you want. For
example, take f(x) = sin(Pi*x)/x = Pi - (Pi*x)^2/3! + (Pi*x)^4/5! - ...
.. Substitute N for x and you have the expression you wanted in terms of
the a_k and a*_k.

Hope this helps.

Igor
erite423@yahoo.se
science forum beginner

Joined: 10 Jul 2006
Posts: 3

Posted: Mon Jul 17, 2006 3:32 pm    Post subject: Re: Density operator in second quantization

Igor Khavkine wrote:
 Quote: erite423@yahoo.se wrote: 1. What is the density operator for a many-electron system in second quantization? It's defined in the same way as in ordinary quantum mechanics. Pure states are obtained by an outer product like |phi>

OK, thanks!

 Quote: What you are talking about is still conventionally referred to as a "density matrix", despite the fact that it's not really a matrix in the textbook sense of the word.

It has many well-established names ("statistical operator", "state
operator", "density operator", "density matrix"). I suppose "density
matrix" is used because the distinction between the matrix D_ij and
the operator sum_ij |i> D_ij <j| can be left implicit when working in
a single, orthonormal basis.

 Quote: 2. How do I trace out degrees of freedom to form reduced density operators? Exactly the same way as you are used to. The first requirement for taking a partial trace is the possibility of expressing the state space as a tensor product.

This is where I have trouble translating results from
first-quantization to second-quantization. In first-quantization, I
can take an N-particle density operator and "trace out" one of the
particles, leading to a reduced (N-1)-particle density operator. But
what is the corresponding mathematical operation in second
quantization?

For example, let |n,m> denote the tensor product state |n>|m> and
consider the 2-electron state

|psi> = 1/sqrt(2) ( |1,2> - |2,1> )

giving the 2-particle density operator

rho = 1/2 ( |1,2> <1,2| - |1,2> <2,1| - |2,1> <1,2| + |2,1> <2,1| )

Tracing out e.g. the second of the identical particles gives the
1-particle density operator

P = 1/2 ( |1> <1| + |2> <2| )

(By convention one would probably also get rid of the factor 1/2 to
make the trace of P equal to the number of identical particles.)

When I try to work out the same simple example in second-quantization
I get stuck. The state would be

|psi> = a*_1 a*_2 |vac>

and the 2-particle density operator might be

rho = a*_1 a*_2 |vac> <vac| a_2 a_1

But I don't know the general procedure that, e.g., extracts the
second-quantized P from the second quantized rho.

 Quote: 3. Can the operator |vac> 0. Then P_vac = f(N) will be the projection operator you want. For example, take f(x) = sin(Pi*x)/x = Pi - (Pi*x)^2/3! + (Pi*x)^4/5! - ... . Substitute N for x and you have the expression you wanted in terms of the a_k and a*_k.

Thanks!

Regards,
Erik
Igor Khavkine
science forum Guru

Joined: 01 May 2005
Posts: 607

Posted: Mon Jul 17, 2006 11:30 pm    Post subject: Re: Density operator in second quantization

erite423@yahoo.se wrote:
 Quote: Igor Khavkine wrote: erite423@yahoo.se wrote: 2. How do I trace out degrees of freedom to form reduced density operators? Exactly the same way as you are used to. The first requirement for taking a partial trace is the possibility of expressing the state space as a tensor product. This is where I have trouble translating results from first-quantization to second-quantization. In first-quantization, I can take an N-particle density operator and "trace out" one of the particles, leading to a reduced (N-1)-particle density operator. But what is the corresponding mathematical operation in second quantization?

The key to grasping what to do here is abstraction. Forget about the
distinction between first and second quantized systems, just think of
them as generic quantum systems on some Hilbert space H. Forget about
particles, just think about H as a tensor product H'(x)H'' for some H'
and H''. This information is all that you need to take a partial trace.
Let states |i'> and |i''> respectively span the two factor spaces H'
and H''. Then their tensor products |i'>|i''> span H. Take any density
matrix of the form

rho = sum_{i,j} D_{i,j} |i'>|i''><j'|<j''|.

If all the bases used are orthonormal, then the partial trace over H''
is obtained from the formula

rho' = sum_k'' <k''| rho |k''>
= sum_k sum_{i,j} |i'><j'| <k''|i''><j''|k''>.

Now, that you know what to do, the tricky part is to express H as a
tensor product in any given situation. That's where the subtleties come
in.

 Quote: For example, let |n,m> denote the tensor product state |n>|m> and consider the 2-electron state |psi> = 1/sqrt(2) ( |1,2> - |2,1> ) giving the 2-particle density operator rho = 1/2 ( |1,2> <1,2| - |1,2> <2,1| - |2,1> <1,2| + |2,1> <2,1| ) Tracing out e.g. the second of the identical particles gives the 1-particle density operator P = 1/2 ( |1> <1| + |2> <2| )

Right, what you've done here is written |psi> as an element of the
tensor product H(x)H, where H is spanned by the |n> states (I'm
assuming that the |m> states are the same as the |n> ones, just under a
different label). Note, using this Hilbert space is equivalent to the
physical assumption that you are dealing with a joint QM system of two
*distinguishable* subsystems, each described by the Hilbert space H.

 Quote: When I try to work out the same simple example in second-quantization I get stuck. The state would be |psi> = a*_1 a*_2 |vac and the 2-particle density operator might be rho = a*_1 a*_2 |vac>

One thing you have to remember about second quantization is that a
second quantized system describes a system of a *variable* number of,
in some cases, *indistinguishable* particles. You also have to remember
that its Hilbert space is the Fock space which contains states with an
arbitrary number of particles. Thus your desired intent to "trace out
one particle" is somewhat ill posed. Exactly which particle do you
mean? Can you write the Fock space as a tensor product, with a
one-particle subspace that you intend to trace over as a factor? As you
seem to have already found, there is no natural way to do that.

But there are many natural ways to split the Fock space into a tensor
product. Here's an exercise to deepen your understanding of the second
quantization construction. Can you find a few ways to write the Fock
space as a tensor product? I'll give more details after I see what you
come up with.

Hope this helps.

Igor
erite423@yahoo.se
science forum beginner

Joined: 10 Jul 2006
Posts: 3

Posted: Tue Jul 18, 2006 5:48 pm    Post subject: Re: Density operator in second quantization

Igor Khavkine wrote:
 Quote: erite423@yahoo.se wrote: Right, what you've done here is written |psi> as an element of the tensor product H(x)H, where H is spanned by the |n> states (I'm assuming that the |m> states are the same as the |n> ones, just under a different label). Note, using this Hilbert space is equivalent to the physical assumption that you are dealing with a joint QM system of two *distinguishable* subsystems, each described by the Hilbert space H.

Here I disagree somewhat. A properly (anti-)symmetrized state
describes the joint state of indistinguishable subsystems, as long as
we agree to only ever use the properly (anti-)symmetrized subspace of
the tensor product and as long as we make sure our operators respect
this. The particle labels are of no significance, except for
formalizing the constraint that the state is (anti-)symmetric under
the interchange of two particle labels.

 Quote: One thing you have to remember about second quantization is that a second quantized system describes a system of a *variable* number of, in some cases, *indistinguishable* particles. You also have to remember that its Hilbert space is the Fock space which contains states with an arbitrary number of particles. Thus your desired intent to "trace out one particle" is somewhat ill posed.

My desired intent is not to "trace out one particle". My desired
intent is to do "that which is the second quantization analogue of the
first-quantization procedure of tracing out one particle".

 Quote: Exactly which particle do you mean? Can you write the Fock space as a tensor product, with a one-particle subspace that you intend to trace over as a factor? As you seem to have already found, there is no natural way to do that. But there are many natural ways to split the Fock space into a tensor product. Here's an exercise to deepen your understanding of the second quantization construction. Can you find a few ways to write the Fock space as a tensor product? I'll give more details after I see what you come up with.

The best I can come up with is to factorize an occupation number
vector like this:

|n_1 n_2 ...> = |n_1> |n_2 ...>

with operators like a*_2 similarly factorized into a*_2 = b*_2 c*_2,

b*_2 |n_1> = (-1)^n_1 |n_1>
c*_2 |n_2 ...> = (1 - n_2) |1+n_2 ...>

Iterating the procedure gives a tensor product decomposition of the
remaining part, so the Hilbert space is now

H = H_1 x H_2 x ...

and |n_1 n_2 ...> = |n_1> |n_2> ...

But the particles that are indexed by first quantization particle
labels are not exactly the same sort of entities as those that live in
e.g. H_1, spanned by |n_1=0> and |n_1=1>. Rather, if |n_1=1> is
occupied it will be occupied by a small "bit" of every particle (as
labeled by first quantization labels) in the system. Are you
suggesting that it as valid to "trace out" each of these new entities
as it is to trace out a (first quantized) particle? In that case I
guess that

sum_k <n_k=1| rho |n_k=1>

will do what I want. Up to a phase factor (cancelled by the
ket-vector), the action of <n_k=1| should equal action of the
annihilation operator a_k. Indeed, the procedure for reducing the
N-particle rho_N to the (N-1)-particle rho_{N-1} suggested by all
this, i.e.

rho_{N-1} = sum_k a_k rho_N a*_k,

gives a reasonable result in cases similar to my previous toy example
(i.e. rho_2 = a*_1 a*_2 |vac> <vac| a_2 a_1).

Can you confirm that this summation is the second quantization
analogue of "tracing out a particle"?

Erik

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