Table Of ContentEntanglement generation via scattering of two particles with
hard–core repulsion
Frank Schmu¨ser∗ and Dominik Janzing†
Institut fu¨r Algorithmen und Kognitive Systeme, Universit¨at Karlsruhe,
Am Fasanengarten 5, 76131 Karlsruhe, Germany.
(Dated: Jan 26, 2006)
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0 Abstract
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n We analyse the entanglement generation in a one dimensional scattering process. The two col-
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J liding particles have a Gaussian wave function and interact by hard–core repulsion. In our analysis
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2 results on the entanglement of two mode Gaussian states are used. The produced entanglement
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depends in a non-obvious way on the parameters ratio of masses and initial widths. The asymp-
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totic wave function of the two particles and its associated ellipse yield additional geometric insight
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into these conditions. The difference to the quantitative analysis of the amount of entanglement
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6 generated by beam splitters with squeezed light is discussed.
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PACS numbers: 03.67.Mn, 03.65.Nk, 03.67.-a
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∗ electronic mail: [email protected]
† electronic mail: [email protected]
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I. INTRODUCTION
In the last decade the study of entanglement has become one of the major topics in
the flourishing field of quantum information theory (for introductions into this topic see
e. g. [1, 2]). Among other things, it has become clear that entanglement is an essential re-
source for many desirable operations in quantum information theory, for example quantum
teleportation [3] or some protocols in quantum cryptography [4]. For a detailed understand-
ing of entanglement it is necessary to quantify it. For pure states ψ this is fairly
1 2
| i ∈ H ⊗H
straightforward. We consider the reduced density operator in (tracing out system 2)
1
H
ρ1 := Tr ( ψ ψ ) (1)
2
| ih |
and define the entropy of entanglement E( ψ ) of the pure state ψ as the von Neumann
| i | i
entropy of this reduced density operator
E( ψ ) := S ρ1 with S(ρ) := Tr(ρ log ρ) . (2)
| i − 2
(cid:0) (cid:1)
The quantity E( ψ ) has the nice property that it (asymptotically) describes the number
| i
of maximally entangled singlet (or Bell) states of 2 qubits that can be obtained from many
copies of the quantum state ψ when it is shared by two parties and each is allowed to
| i
operate on his part only [5]. The quantification of entanglement for mixed states is a much
more difficult business which however does not concern us in this paper, as we are only
dealing with pure states.
In this paper we study entanglement generation in quantum systems with a continuous
degree of freedom, in particular a scattering process which generates correlations in the
two-particle Schr¨odinger wave function. Continuous variable quantum systems are also an
interesting topic in quantum information theory (cf. the recent review [6]). For instance,
continuous versions of quantum teleportation [7] and quantum cryptography [8] can be
found.
In particular we will be dealing with Gaussian states, since properties of these states
can be often obtained in an analytic fashion such that there is also an extensive theoretical
literature about these states [9, 10]. Gaussian wave functions are also a quite natural
assumption. We will investigate how the amount of generated entanglement depends on the
ratio of the widths of the incoming particle wave functions.
2
In quantum optics the entanglement production for photons can be realized experimen-
tally and has been extensively studied, e. g. for squeezed states hitting a beam splitter
[11, 12]. In contrast to that, there are only a few studies of entanglement production for
scattering of two particles [13, 14]. This is surprising as scattering theory itself is a major
topic in quantum mechanics. In [14] entanglement production for scattering of two Gaussian
particles with hard core repulsion (cf. the potential in eq. (3)) was studied, and the present
paper can be understood as a generalization of this paper which yields additional physical
understanding. For, in [14] only the special case was treated, when the two particles have
equal masses and equal initial widths of their wave functions. It was then found that only
transient entanglement, i. e. non–vanishing entanglement during collision, can be produced.
For a more general repulsive potential only a rather small quantity of asymptotic (perma-
nent) entanglement could be obtained with the initial conditions of [14]. In [15] one can
find calculations of entanglement in spin gases, where the position coordinates are treated
classically.
Ourcalculationlooksquitesimilartothecalculationoftheentanglementthatisgenerated
in a beam splitter when at least one input state is squeezed. However, the decisive difference
is that the action of the beam splitter is a rotation that mixes the quadrature amplitudes
of different modes, whereas the scattering process is a reflection with respect to the non-
orthogonal coordinate system given by the relative coordinates.
Wenowoutlinetheorganizationofthispaper. InsectionIIwestatethescatteringprocess
and review the exact solution of the time–dependent Schr¨odinger equation. In section III
we calculate the asymptotic entanglement of the two particles after the scattering. Here
we use extensively the covariance matrix of Gaussian states. A geometric interpretation of
the results about entanglement is given in section IV where we study the wave function of
the two particles after scattering. Therewith we can also understand in detail under which
conditions significant entanglement is produced – and also when no entanglement at all is
generated. In section V we sum up the obtained results and point out some possibilities for
further research.
3
II. THE SCATTERING PROCESS AND ITS EXACT SOLUTION
Here we consider the quantum mechanical scattering process of two particles in one
spatial dimension which interact via the potential V(x x ). As the scattering potential
1 2
−
shall model hard–core repulsion, it is given by
0 , x x > a
1 2
V ( x x ) := | − | (3)
1 2
| − |
, x x a
∞ | 1 − 2| ≤
In contrast to [14] we consider general mass ratios of the two particles. Hence the Hamilto-
nian for the corresponding one dimensional scattering process is
pˆ2 pˆ2 pˆ2 pˆ2
H = 1 + 2 +V ( x x ) = s + r +V ( x ) , (4)
1 2 r
2m 2m | − | 2M 2M | |
1 2 s r
where we have defined the momenta
pˆ := pˆ +pˆ ,
s 1 2
pˆ := µ pˆ µ pˆ (5)
r 2 1 1 2
−
and center of mass and relative coordinate x and x , respectively,
s r
x = µ x +µ x ,
s 1 1 2 2
x = x x , (6)
r 1 2
−
using the mass fractions
µ := m /(m +m ) , µ := m /(m +m ) . (7)
1 1 1 2 2 2 1 2
Besides, in eq. (4) we use the masses M := (m +m )/2 and M := (m m )/(m +m ).
s 1 2 r 1 2 1 2
We thus see that the Hamiltonian (4) decouples in the coordinates (x , x ).
s r
Our starting configuration at t = 0 are two Gaussian wave packets with widths σ2 and
1
σ2. Let the mean value of particle 1 be located at Q a, that of particle 2 at Q a.
2 1 ≫ − 2 ≪ −
The momentum of the first particle 1 shall be K and that of the second particle K. We
−
first define a state f with wave function
0
| i
f (x , x ) = φ x ;Q , K, σ2 φ x ; Q , K, σ2 , (8)
0 1 2 G 1 1 − 1 G 2 − 2 2
(cid:0) (cid:1) (cid:0) (cid:1)
where the Gaussian (located at Q and with momentum K) is defined as
(x Q)2 1
φ ; Q, K, σ2 := α(σ2) exp(iKx) exp − , α(σ2) := . (9)
G − 2σ2 √σ π1/4
(cid:16) (cid:17)
(cid:0) (cid:1)
4
Note that f itself does not define a physical initial condition for any finite Q, since the
0
| i
Gaussians always overlap. In a more formal approach the two wave packets have to be
starting from an “infinite distance”[19].
Nowwewouldliketofindasolution ψ ofthetime–dependent Schr¨odinger equation
t
| i ∈ H
with the Hamiltonian H in eq. (4) that coincides for t approximatively with f :=
t
→ −∞ | i
exp( iH t) f , where
0 0
− | i
p2 p2
H = s + r . (10)
0
2M 2M
s r
is the free Hamiltonian. In the relative coordinate x we have to satisfy the following
r
(boundary) conditions:
(A) For all x [ a, a] we have ψ (x , x ) = 0 , t R.
r t s r
∈ − ∀ ∈
(B) The solution ψ (x , x ) is continuous in x at the boundary of the potential x = a.
t s r r r
(C) Forx [ a, a]thesolutionψ (x , x )obeysthefreeSchr¨odingerequationwithHamil-
r t r s
6∈ −
tonian H .
0
We now construct the solution as follows. For an arbitrary wave function φ(x ) we
r
introduce the unitary operator P that reflects the relative coordinate at x = a, i. e.
a r
(P φ )(x ) := φ(2a x ) . (11)
a r r
| i −
It can be easily seen that the operator 11 P commutes with the free Hamiltonian (10)
a
⊗
[11 P , H ] = 0 . (12)
a 0
⊗
Here it is important to keep in mind that the tensor product structure refers to relative
coordinates and not to the particle Hilbert spaces. Let us define the state
g := (11 P ) f . (13)
t a t
| i ⊗ | i
Because of the commutation relation (12) the evolution t g is also a solution of the
t
7→ | i
free Schr¨odinger equation. Now the difference f g satisfies condition (B). In order to
t t
| i−| i
enforce condition (A) we use a Heaviside function, and thus the general solution that we
were looking for can be written as
ψ (x , x ) = ( f (x , x ) g (x , x ) ) θ(x a) . (14)
t s r t s r t s r r
− −
5
Since we have
lim ψ f = 0 (15)
t t
t→−∞k| i−| ik
and
lim ψ g = 0 (16)
t t
t→∞k| i−| ik
we consider the mapping f g as the scattering process.
t t
| i 7→ | i
It is important to note that due to the hard–core repulsion particle 1 always stays to
the right of particles 2. In other words, there is only reflection and no transmission for this
particular potential.
III. ENTANGLEMENT BETWEEN THE TWO PARTICLES AFTER SCATTER-
ING
In order to calculate the entanglement between the particles after the scattering, we have
to consider solution (14) for times t . Since eq. (16) holds and the free evolution
→ ∞
exp( iH t) does not change the entanglement, we have
0
−
lim E( ψ ) = E( g ) , (17)
t 0
t→∞ | i | i
where g = (11 P ) f (cf. eq. (13)). It is very important to note that the state g is a
0 a 0 0
| i ⊗ | i | i
Gaussian state [9][20] For, the initial state f is obviously Gaussian and the reflection in
0
| i
relative coordinates does not change the Gaussian nature of the state. In order to obtain
the entanglement of g , we associate the following reduced density operator to the state
0
| i
g
0
| i
ρred := Tr ( g g ) (18)
2 0 0
| ih |
The reduced density operator of a Gaussian state is also Gaussian.
The entanglement of a Gaussian state can be determined from its covariance matrix that
we now introduce. The four standard canonical operators of our two particle system are
R := (xˆ , pˆ , xˆ , pˆ )T . (19)
1 1 2 2
The 4 by 4 covariance matrix of a density operator ρ is then given as
1
σ = Tr(ρ R , R ) Tr(ρ R ) Tr(ρ R ) , i, j 1, ..., 4 , (20)
ij i j i j
2 { } − ∈ { }
6
where , denotes the anticommutator. We recall the following fact (cf. e. g. [9]): a linear
{ }
change of canonical operators with a 4 by 4 symplectic matrix F and a displacement D ,
i. e.
R˜ = F R+D , (21)
implies for the covariance matrix σ˜ of the new canonical operators R˜
σ˜ = F σ FT . (22)
In order to obtain the covariance matrix of the state
ρ′ := g g = U ρ U† with ρ := f f , U := 11 P , (23)
0 0 0 0 a
| ih | | ih | ⊗
we start with the covariance matrix for ρ = f f and the standard operator coordinates
0 0
| ih |
R of eq. (19)
σ2/2 0 0 0
1
0 1/(2 σ2) 0 0
σ := Tr RRT ρ Tr(Rρ) Tr RT ρ = 1 , (24)
− 0 0 σ2/2 0
2
(cid:0) (cid:1) (cid:0) (cid:1)
0 0 0 1/(2 σ2)
2
ascanbeeasilycalculated fromformulas(8) and(9). The covariancematrixforρ′ ineq. (23)
in standard canonical coordinates R can be transformed as (using eq. (23))
σ′ = Tr(ρ′ R R ) Tr(ρ′ R ) Tr(ρ′ R )
ij i j − i j
= Tr ρ (U†R U) (U†R U) Tr ρ (U†R U) Tr ρ (U†R U) . (25)
i j i j
−
(cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
We now seek a symplectic transformation which relates the new operators U†RU to the
standard operators R. Then we can use formula (22) and obtain σ′ from σ in (24). Since
the operator U = 11 P acts on center of mass and relative coordinate, we define the vector
a
⊗
S of canonical operators in this frame as
S = (xˆ , pˆ , xˆ , pˆ )T . (26)
s s r r
As can be easily seen from eqs. (6) and (5), we have
µ 0 µ 0
1 2
0 1 0 1
S = F R with F = . (27)
1 0 1 0
−
0 µ 0 µ
2 − 1
7
The identity
U†RU = F−1U†SU (28)
follows immediately. As U is a reflection in the relative coordinate, we can easily calculate
1 0 0 0 0
0 1 0 0 0
U†SU = GS+D with G = , D = . (29)
0 0 1 0 2a
−
0 0 0 1 0
−
Using eq. (28) and (29), we arrive at
U†RU = F−1GF R+F−1D . (30)
Thus the covariance matrix σ′ is related to the covariance matrix σ of eq. (24) by
σ′ = F−1GF σ FT GT F−1 T . (31)
(cid:0) (cid:1)
We write σ′ in the form
A C
σ′ = , (32)
CT B
where A, B and C are 2 by 2 matrices. Then a straightforward computation of the product
of matrices in eq. (31) yields
2µ2σ2 + (∆µ)2σ12 0 2µ2σ2 + (∆µ)2σ22 0
A = 2 2 2 , B = 1 1 2 ,
0 2µ21 + (∆µ)2 0 2µ22 + (∆µ)2
σ22 2σ12 σ12 2σ22
∆µ (µ σ2 µ σ2) 0
C = 1 1 − 2 2 , (33)
0 ∆µ µ22 µ1
σ12 − σ22
(cid:16) (cid:17)
where we used the abbreviation
∆µ = µ µ . (34)
1 2
−
Looking at the definitions of covariance matrix and reduced density operator, it is clear
that the covariance matrix which corresponds the reduced state ρred of eq. (18) is given by
the submatrix A of eq. (33). It is known [16, 17, 18] that the von Neumann entropy – as
defined in eq. (2) – of a one mode Gaussian state ρred is
1 1 1 1
S ρred = d+ log d+ d log d , (35)
2 2 2 − − 2 2 − 2
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
(cid:0) (cid:1)
8
where the quantity d 1/2 is given as
≥
(∆µ)2 µ2σ2 µ2σ2
d2 = det(A) = det(B) = 4µ2µ2 +(∆µ)2 + 1 1 + 2 2 . (36)
1 2 4 σ2 σ2
" 2 1 #
If we insert the (positive) value of d(µ , µ , σ2, σ2) into eq. (35), we get the desired asymp-
1 2 1 2
totic entanglement in terms of the four parameters µ , µ , σ2, σ2.
1 2 1 2
Instead of calculating the von Neumann entropy of the reduced density operator one can
also consider – as done in [14] – the purity of the state ρred
ρred := Tr ρred 2 . (37)
P
(cid:16) (cid:17)
(cid:0) (cid:1) (cid:2) (cid:3)
The smaller ρred , the larger is the generated entanglement. It is known [9] that the
P
purity of a Gau(cid:0)ssian(cid:1)state like ρred is given by
1 1
ρred = = . (38)
P 2 det(A) 2d
(cid:0) (cid:1)
Thus the purity is – as the von Neumann epntropy - characterized by the determinant of the
covariance matrix A.
Wenowdiscuss ourresult fortheentanglement asexpressed ineqs.(35)and(36). Wefirst
note that d2 depends only on the two width ratios σ2/σ2, σ2/σ2 and on the mass fractions
1 2 2 1
µ , µ of the two particles. We see that for a large d - and therfore a large entanglement –
1 2
it is necessary that one of the two width ratios is large. In contrast to that the quantities
0 < µ , µ < 1 enter in formula (36) not as a quotient, so that choosing large mass ratios
1 2
does not generate a large d. We will discuss the conditions for the production of a lot of
entanglement in detail in section IV.
One can easily check that always d 1/2 and thus S ρred 0. If d = 1/2, then
≥ ≥
the entanglement vanishes according to eq. (35). No entangl(cid:0)emen(cid:1)t is generated for the two
conditions (cf. eq. (36))
1. µ = µ = 1/2, i. e. the two scattered particles have equal mass. This result was
1 2
already obtained in [14].
2. µ σ2 = µ σ2, i. e. the heavy particle has a small width, the light particle has a large
1 1 2 2
width.
9
IV. THE WAVE FUNCTION AFTER SCATTERING
In this section we study the wave function g (x , x ) that corresponds to the state g
0 1 2 0
| i
[21]. This will give us more insight into the results in eqs. (35) and (36) for the generated
entanglement. In complete analogy to the calculation of the covariance matrix σ′ in the last
section, we can obtain the wave function of the state g = (11 P ) f in three steps from
0 a 0
| i ⊗ | i
the initial wave function f (x , x ) of eq. (8):
0 1 2
1. Replace in f (x , x ) the coordinates (x , x ) with the coordinates (x , x ) using the
0 1 2 1 2 s r
inverse to relation (6).
2. Apply the operator (11 P ) by changing the variable x to (2a x ).
a r r
⊗ −
3. Go back to the coordinates (x , x ) with relation (6).
1 2
The result of this procedure is
g (x , x ) = φ 2µ x +(µ µ )x ; Q 2µ a, K, σ2
0 1 2 G 2 2 1 − 2 1 1 − 2 − 1
φ(cid:0) 2µ x (µ µ )x ; Q +2µ a, K, (cid:1)σ2 , (39)
× G 1 1 − 1 − 2 2 − 2 1 2
(cid:0) (cid:1)
where the Gaussians φ ( ) have been defined in eq. (9). Since the entanglement of a
G
Gaussian state depends only onits covariance matrix, any displacement of the wave function
inposition or momentum space does not change its entanglement. Thus instead ofg (x , x )
0 1 2
in eq. (39) we introduce the simpler wave function g˜(x , x ) that keeps only the relevant
1 2
quadratic terms in x , x in the exponent
1 2
[(µ µ )x +2µ x ]2
g˜ (x , x ) = α σ2 exp 1 − 2 1 2 2
0 1 2 1 − 2σ2
1
(cid:16) (cid:17)
(cid:0) (cid:1) [2µ x (µ µ )x ]2
α σ2 exp 1 1 − 1 − 2 2 . (40)
× 2 − 2σ2
2
(cid:16) (cid:17)
(cid:0) (cid:1)
This wave function can be rewritten with a quadratic form in the exponent
1
g˜ (x , x ) = α σ2 α σ2 exp xT LT ΣLx , (41)
0 1 2 1 2 − 2
(cid:16) (cid:17)
(cid:0) (cid:1) (cid:0) (cid:1)
where we have introduced xT := (x , x ) and the two matrices
1 2
µ µ 2µ 1/σ2 0
L := 1 − 2 2 , Σ := 1 . (42)
2µ µ µ 0 1/σ2
1 2 − 1 2
10
