# Density operator in second quantization

by erite423@yahoo.se
Tags: density, operator, quantization
 P: n/a 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> <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> 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}. Thanks in advance, Erik
 P: n/a erite423@yahoo.se wrote: > 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> 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. > 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. > 3. Can the operator |vac> and all others to zero, be expressed in terms of creation and > annihilation operators? 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
 P: n/a Igor Khavkine wrote: > 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> states are obtained by taking convex linear combinations and limits of > pure states. OK, thanks! > 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 > 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> > 3. Can the operator |vac> > and all others to zero, be expressed in terms of creation and > > annihilation operators? > > 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. Thanks! Regards, Erik
P: n/a

## Density operator in second quantization

erite423@yahoo.se wrote:
> 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.

> 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.

> 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.

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

 P: n/a Igor Khavkine wrote: > 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. > 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". > 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 will do what I want. Up to a phase factor (cancelled by the ket-vector), the action of
 P: n/a erite423@yahoo.se wrote: > 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? rho = 1/Z exp(-H/T), where H is the Hamiltonian written in terms of creation and annihilation operators, is the thermal equilibrium state at temperature T. In the limit T->0 one gets rho = |psi> = trace rho S is the traditional entropy. Degenerate cases (such as general pure states) arise as limits of such states. > 2. How do I trace out degrees of freedom to form reduced density > operators? The 1-particle reduced density matrix is defined as the operator Rho with matrix elements = trace a(x) rho a^*(y) = . Here the trace is the full trace; the c/a operators do the partial tracing out. Similarly for higher reduced density matrices. See for example the statistical mechanics book by Reichl. > 3. Can the operator |vac> and all others to zero, be expressed in terms of creation and > annihilation operators? The natural expression is as limit for T->0 of an equilibrium state for a Hamiltonian which has the vacuum as ground state. > 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} = This holds for a pure state only. For the general case, see above. > 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}. For a pure product state psi(x_1,...,x_n)=psi_1(x_1)...psi_n(x_n), you can identify the formulas easily. The general case then follows by linearity. I think Reichl's book has an appendix deriving the formulas somewhat differently. Arnold Neumaier
 P: n/a Igor Khavkine wrote: > 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 Fock space is naturally embedded in a tensor product, which suffices for justifying the partial trace. Arnold Neumaier
 P: n/a Arnold Neumaier wrote: > erite423@yahoo.se wrote: > The 1-particle reduced density matrix is defined as the operator > Rho with matrix elements > = trace a(x) rho a^*(y) = . > Here the trace is the full trace; the c/a operators do the > partial tracing out. OK, thanks! By now I had come to suspect as much, but it's good to get it confirmed. > Similarly for higher reduced density matrices. See for example > the statistical mechanics book by Reichl. Unfortunately, Reichl ("A Modern Course in Statistical Mechanics", 1998) follows standard practice by introducing density operators in the context of first quantization and relegating second quantization to a very brief and introductory section at the end of the book. Furthermore, the last mention of a density operator/matrix occurs before the first mention of second quantization. Do you know any book which has some explicit discussion of density operators/matrices in second quantization? I know how to mindlessly calculate things using the second quantization formalism, but I'm still a bit puzzled by the appearance of |vac> D_pq > 3. Can the operator |vac> > and all others to zero, be expressed in terms of creation and > > annihilation operators? > > The natural expression is as limit for T->0 of an equilibrium state > for a Hamiltonian which has the vacuum as ground state. Do you know of any textbook/article that takes the T = 0 limit of the equilibrium density operator and shows how the "|vac> > 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} = > > This holds for a pure state only. For the general case, see above. > > > 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}. > > For a pure product state psi(x_1,...,x_n)=psi_1(x_1)...psi_n(x_n), > you can identify the formulas easily. The general case then follows > by linearity. I think Reichl's book has an appendix deriving the > formulas somewhat differently. Appendix B first briefly discusses density matrices and then briefly discusses second quantization without returning to density operators/matrices. A spin density operator, giving the spin density at points in space, is mentioned on p. 789 but it is not the same and does not contain the puzzling "|vac>
 P: n/a erite423@yahoo.se wrote: > For example, a typical 1-particle operator h is related to its matrix > elements h_pq via > > h = sum_pq a*_p h_pq a_q, > > but the 1-particle density operator D is related to the 1-particle > density matrix D_pq via > > D = sum_pq a*_p |vac> D_pq
 P: n/a erite423@yahoo.se wrote: > Arnold Neumaier wrote: > >>The 1-particle reduced density matrix is defined as the operator >>Rho with matrix elements >> = trace a(x) rho a^*(y) = . >>Here the trace is the full trace; the c/a operators do the >>partial tracing out. > > OK, thanks! By now I had come to suspect as much, but it's good to get > it confirmed. > > >>Similarly for higher reduced density matrices. See for example >>the statistical mechanics book by Reichl. > > > Unfortunately, Reichl ("A Modern Course in Statistical Mechanics", > 1998) follows standard practice by introducing density operators in > the context of first quantization and relegating second quantization > to a very brief and introductory section at the end of the > book. I don't have the book at hand, but I learnt it from there - perhaps reading between the lines. Here is the explicit derivation for 1-particle operators: Let a(x_1:N) := a(x_1)...a(x_N) to simplify notation. In general, an N-particle state with wave function psi(x_1:N) is given in 2nd quantization by psihat = integral dx_1:N psi(x_1:N) a^*(x_1:N) |vac>, hence the corresponding density matrix rho = psi psi^* takes the form rhohat = psihat psihat^* = integral dx_1:N dy_1:N rho(x_1:N,y_1:N), where rho(x_1:N,y_1:N) is the rank one operator psi(x_1:N)psi^*(y_1:N)a^*(x_1:N)|vac> = integral dx dy f(x,y) defines the 1-particle density matrix Rho. The form of f in second quantization is fhat = integral dx dy f(x,y) a^*(x) a(y) (exercise: check that it has indeed the desired action on an N-particle state!), hence one has = integral dx dy f(x,y) . and comparison with the definition of Rho gives the formula = = trace a(x) rho a^*(y), as claimed. Authers who fear integrals write instead similar formulas with sums in place of integrals and discrete indices in place of the x,y. Also, one can do the same in momentum space rather than position space, which amounts to a change of basis but generally leads to computationally more tractable formulations. > still a bit puzzled by the appearance of |vac> operator. Among the common operators, the second quantized density > operators are unique in having this vacuum-to-vacuum operator as a > factor. This is because the creation operators make all states from the vacuum. >>>3. Can the operator |vac> >>and all others to zero, be expressed in terms of creation and >>>annihilation operators? >> >>The natural expression is as limit for T->0 of an equilibrium state >>for a Hamiltonian which has the vacuum as ground state. > Do you know of any textbook/article that takes the T = 0 limit of the > equilibrium density operator and shows how the "|vac> the pure state at T = 0 arises? It doesn't have to be done for an > advanced or realistic Hamiltonian, just something simple that > illustrates the general case would be enough. It is a general property: Assume H has discrete spectrum and has |vac> as the unique ground state. In an orthonormal basis of eigenstates psi_k (and psi_1=|vac>) f(H) = sum_k f(E_k) psi_k psi_k^* for every function f defined on the spectrum. In particular (setting the Boltzmann constant to 1), rho(T) = 1/Z(T) exp(-H/T) = sum_k exp(-E_k/T)/Z(T) psi_k psi_k^* Taking the trace (which must be 1) gives Z(T) = sum_k exp(-E_k/T), and in the limit T -> 0, all terms exp(-E_k/T)/Z(T) become 0 or 1, with 1 only for the ground state psi_1 = |vac>. Thus lim_{T->0} rho(T) = psi_1 psi_1^* = |vac>

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