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Density operator in second quantization

by erite423@yahoo.se
Tags: density, operator, quantization
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erite423@yahoo.se
#1
Jul11-06, 04:00 AM
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> <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}.

Thanks in advance,
Erik

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Igor Khavkine
#2
Jul12-06, 04:00 AM
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><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.

> 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> <vac|, mapping the vacuum state to itself
> 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

erite423@yahoo.se
#3
Jul18-06, 04:00 AM
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><phi|, while mixed
> 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 <j| can be left implicit when working in
a single, orthonormal basis.

> > 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.

> > 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?

>
> 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


Igor Khavkine
#4
Jul18-06, 04:00 AM
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

erite423@yahoo.se
#5
Jul19-06, 04:00 AM
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 <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

Arnold Neumaier
#6
Jul25-06, 04:00 AM
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> <psi|
where psi is the ground state of H, assumed nondegenerate.

General density operators at positive temperature have the form
rho = exp(-kbar S),
where kbar is Boltzmann's constant and S is a self-adjoint
entropy operator with discrete spectrum built up from creation
and annihilation operators.
<S> = 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
<x|Rho|y> = trace a(x) rho a^*(y) = <a^*(y)a(x)>.
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> <vac|, mapping the vacuum state to itself
> 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} = <psi| b_p a_q |psi>


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

Arnold Neumaier
#7
Jul25-06, 04:00 AM
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

erite423@yahoo.se
#8
Jul26-06, 04:00 AM
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
> <x|Rho|y> = trace a(x) rho a^*(y) = <a^*(y)a(x)>.
> 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> <vac| in the density
operator. Among the common operators, the second quantized density
operators are unique in having this vacuum-to-vacuum operator as a
factor.

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 <vac| a_q.

As far as I can see, this disanalogy between density operators and
other operators somehow arises when we go from first to second
quantization.

> > 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?

>
> 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> <vac|" factor in
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.

> > 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>

>
> 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> <vac|" factor.

What I seek is a book that doesn't silently drop density operators
from consideration as soon as second quantization is introduced.

Regards,
Erik

Ralph Hartley
#9
Jul27-06, 04:00 AM
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 <vac| a_q.


I think h is a 1-particle operator, and D is a (1-particle density)
operator. Different binding of the adjective.

They differ in their effect when applied to a state with more than one
particle. D always gives 0.

Viewed as a density operator, h says "there is a particle in this
state", while D says "there is exactly one particle and it is in this
state".

If you are confined to the 1-particle subspace, the most general
operator has the form h, so in that sense, h is a "1-particle operator"
(h also may be applied to multi-particle states, but I am not sure if it
should be called a 1-particle operator in that context).

D is an operator on the entire Fock space. It is not a "1-particle
operator" in the above sense, it is an operator that describes a
1-particle state.

Ralph Hartley

Arnold Neumaier
#10
Jul27-06, 04:00 AM
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
>> <x|Rho|y> = trace a(x) rho a^*(y) = <a^*(y)a(x)>.
>>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><vac|a(y_1:N).
You can see how the |vac><vac| terms arise. Using this correspondence
you can do in first quantization whatever you do in second quantization,
and match the results.

If f is a 1-particle operator given by an integral operator with
kernel f(x,y) (the general case follows by taking limits), the formula
<f> = integral dx dy <x|Rho|y> 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
<f> = integral dx dy f(x,y) <a^*(y)a(x)>.
and comparison with the definition of Rho gives the formula
<x|Rho|y> = <a^*(y)a(x)> = 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> <vac| in the density
> 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> <vac|, mapping the vacuum state to itself
>>>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> <vac|" factor in
> 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> <vac|.

This property is also the reason why many systems can be prepared
in a nearly pure state - namely if this state can be realized
as the ground state of a system where the gap between the two smallest
energy levels exceeds a small multiple of the energy E^* := kbar T.

I haven't seen this mentioned in any textbook or article, though it
is a very plausible observation. If anyone knows of a reference for
it, I'd like to know...

If I'd write a textbook on quantum mechanics, it would be very
different from the traditional ones. So far, however, you must be
content with my theoretical physics FAQ at
http://www.mat.univie.ac.at/~neum/physics-faq.txt
For the above topics, see in particular the sections
S1d. Postulates for the formal core of quantum mechanics
S1l. Second quantization

Arnold Neumaier



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