Supersymmetric Hamilton Operator and Entanglement
Willi-Hans Steeb and Yorick Hardy
International School for Scientific Computing, University of Johannesburg, Auckland Park 2006, South Africa
Reprint requests to Prof. W.-H. S; E-mail: WHS@NA.RAU.AC.ZA Z. Naturforsch. 61a, 139 – 140 (2006); received February 22, 2006
We study the entanglement of Fermi particles of a supersymmetric Hamilton operator given by a simple Fermi-Bose system.
Key words: Supersymmetric Hamiltonian; Entanglement; Fermi Operators; Bose Operators.
Entanglement has been studied in detail for finite- dimensional quantum systems and to a lesser extent for infinite-dimensional quantum systems (see [1, 2] and references therein). Here we study the entanglement for states of a supersymmetric Hamilton operator [3]
given by Bose operators b†, b and Fermi operators with spin up and spin down, i.e. c†↑, c†↓, c↑, c↓. Let
Q :=b⊗c†↑c†↓ (1) be a linear operator, where b is a Bose annihilation op- erator, c†↑is a Fermi creation operator with spin up, c†↓ is a Fermi operator with spin down and⊗the tensor product [4]. Since c†σc†σ =0,σ∈ {↑,↓}we find that Q2=0. We define the supersymmetric Hamilton oper- ator ˆH as
H :ˆ = [Q,Q†]+≡QQ†+Q†Q.
From (1) we obtain Q†=b†⊗c↓c↑. Let ˆnB:=b†b, ˆ
n↑:=c†↑c↑, ˆn↓:=c†↓c↓ be the number operators. Ap- plying[b,b†] =IBand[cσ,c†σ]+=IFδσ,σwe arrive at
Hˆ = (2 ˆnB+IB)⊗nˆ↑nˆ↓+nˆB⊗(IF−nˆ↑−nˆ↓), where IB is the identity operator in the Hilbert space HB of the Bose operators and IF is the identity op- erator in the Hilbert space HF of the Fermi opera- tors. Straightforward calculation yields[Hˆ,Q] =0 and [Hˆ,Q†Q] =0. Thus the three operators ˆH, Q, Q†Q may be diagonalized simultaneously. Let|nbe the number states (Fock states), where n=0,1,2,...andn|n=1.
0932–0784 / 06 / 0300–0139 $ 06.00 c2006 Verlag der Zeitschrift f ¨ur Naturforschung, T ¨ubingen·http://znaturforsch.com
For the Fermi operators we use the matrix representa- tion [4]
c†↑=1
2σ+⊗I2, c†↓=1
2σz⊗σ+. Thus
c↑=1
2σ−⊗I2, c↓=1 2σz⊗σ−
and
c†↑c†↓=−1
4σ+⊗σ+.
Thus the Fermi operators act in the Hilbert space C4. It follows that
ˆ
n↑=c†↑c↑=10 0 0
⊗I2, nˆ↓=c†↓c↓=I2⊗10 0 0
,
ˆ
n↑nˆ↓=1000⊗1000.
Then a basis in the product Hilbert space is given by
|n ⊗10
⊗10
, |n ⊗10
⊗01
,
|n ⊗01⊗10, |n ⊗01⊗01, where n=0,1,2,.... Now we obtain
Hˆ|n ⊗10⊗10= (n+1)|n ⊗10⊗10. This is an eigenvalue equation with eigenvalue n+1, where n=0,1,2,.... Furthermore
Hˆ|n ⊗10⊗01=0,
140 W.-H. Steeb and Y. Hardy·Supersymmetric Hamilton Operator and Entanglement Hˆ|n ⊗01⊗10=0.
Both states have eigenvalue 0. Finally Hˆ|n ⊗01⊗01=n|n ⊗01⊗01
with eigenvalue n, where n=0,1,2,.... Thus the low- est eigenvalue is 0. The Bell states are given by
|Φ+= 1
√2 1
0
⊗10+01⊗01 ,
|Φ−= 1
√2 1
0
⊗10−01⊗01 ,
|Ψ+= 1
√2 1
0
⊗01+01⊗10 ,
|Ψ−= 1
√2 1
0
⊗01−01⊗10 .
Consider the product states of the number states and the Bell states. Applying the Hamilton operator we find
Hˆ|n⊗ |Φ+=n|n⊗ |Φ++ 1
√2|n⊗10⊗10,
Hˆ|n⊗ |Φ−=n|n⊗ |Φ−+ 1
√2|n⊗10⊗1
0
,
Hˆ|n ⊗ |Ψ+=0, Hˆ|n ⊗ |Ψ−=0.
Consider now the unitary operator U(t) =exp(−i ˆHt). Then we obtain
U(t)|n ⊗10⊗10=e−it(n+1)|n ⊗10⊗10, U(t)|n ⊗01⊗01=e−itn|n ⊗01⊗01 and
U(t)|n ⊗10⊗01=|n ⊗10⊗01, U(t)|n ⊗01⊗10=|n ⊗01⊗10. It follows that
U(t)|n ⊗ |Φ+= 1
√2e−itn|n
⊗
e−it10⊗10+01⊗01 , U(t)|n ⊗ |Φ−= 1
√2e−itn|n
⊗
e−it10⊗10−01⊗01 , and U(t)|n ⊗ |Ψ+=|n ⊗ |Ψ+, U(t)|n ⊗ |Ψ−=
|n ⊗ |Ψ−. Thus the states|n ⊗ |Ψ+and|n ⊗ |Ψ− do not change under the unitary transformation. The Fermi part of the state U(t)|n ⊗ |Φ±is also a Bell state. There is a continuous oscillation between|n ⊗
|Φ+and|n ⊗ |Φ−with periodicity 2π.
[1] W.-H. Steeb and Y. Hardy, Problems and Solutions in Quantum Computing and Quantum Information, World Scientific, Singapore 2006.
[2] Y. Hardy and W.-H. Steeb, Int. J. Theor. Phys. 43, 2207 (2004).
[3] W.-H. Steeb, Problems and Solutions in Theoretical and Mathematical Physics, Vol. II: Advanced Level, 2nd ed., World Scientific, Singapore 2003.
[4] W.-H. Steeb, Matrix Calculus and Kronecker Product with Applications and C++ Programs, World Scien- tific, Singapore 1997.
