Notes on relaxation of the resonant model using a scattering approach
11 Notes on relaxation of the resonant model using a scattering approach Alexander
Semenov S St.,
Department of Chemistry,
University of Pennsylvania,
Philadelphia,
Pennsylvania,
Email: [email protected]
Abstract:
Here we consider a non ‐ perturbative description of the non ‐ interacting resonant model using a scattering approach employed in JCP ) (2020). We showed that such description coincides with the standard NEGF and
Landauer ‐ Buttiker approaches Introduction.
Suppose that a system of infinite size (say, baths connected through a central region) is characterized by a steady ‐ state density matrix ˆ and the total Hamiltonian of the system is ˆ H . Suppose that the
Hamiltonian has undergone a sudden perturbation ˆ V at some time T . What will happen with the density matrix?
The boundary condition: ˆ ˆ( ) t T . (1.1) Thus ˆ ˆˆ ˆ( ) exp( ( ) / ) exp( ( ) / ) t T iH T T iH T T (1.2) where ˆ ˆ ˆ H H V (1.3) In Ref. we introduced the following operator ˆ ˆ( , ) exp iT T H T T (1.4) We proved that the following expression †0 ˆ ˆˆ ˆ(0, ) (0, ) (1.5) corresponds to a steady state density matrix for the Hamiltonian ˆ ˆ ˆ H H V i.e. ˆˆ[ , ] 0 H . Since ˆ ˆ ˆ(0, ) ( , 0) ( , ) T at, t the matrix ˆ will relax to a new steady state ˆ ( ) t . This conclusion seems a bit off: how could a unitary transformation give a relaxation? The answer is that a unitary transformation for a finite size system does not produce any relaxation. That is not true for infinite size system. To prove this more “explicitly”, consider a sudden “kick” of the dot level in non ‐ interacting resonant level model with one lead and compute population of the dot at finite times. This is done in the next section. Relaxation of an observable at finite times Consider non ‐ interacting resonant level model † † * † † ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ k k k d k k k kk k H c c d d V d c V d c (2.1)
The density matrix: †0
1ˆ ˆ ˆexp ( ) k k kk Z (2.2) After the sudden kick (t=0) d d : † † * † †2 2 ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ k k k d k k k kk k H c c d d V d c V d c (2.3)
Time evolution of † ˆ ˆ d d over time: † † 2 0 2 ˆ ˆ ˆ ˆ ˆ{ exp( ) exp( )} d d Tr d d iH t iH t (2.4) First, we need to express † ˆ ˆ d d through the new steady state scattering operators, index corresponds to the green’s functions and scattering states of (2.3): † * †2 2 2 2 ˆ ˆ ˆ ˆ( ) ( ) a rdd k k dd n n k nkn d d G V G V (2.5) Taking into account that † † † †2 2 ˆ ˆˆ ˆ( ) / ( ) ( ) / ( ) r rk n dd k k n dd k c G c G (2.6) we have *2 2 *2 2 ( )ˆ ˆ ˆ{1 ( ) / ( )} ( )ˆ ˆ( )( ) ra a m dd mn n dd n dd n n mm m nra m dd mn dd n d d n mm m n V GG G V iV GG V i (2.7) † † * †2 2† * †2 2 ( )ˆ ˆ ˆ{1 ( ) / ( )} ( )ˆ ˆ( )( ) ar r m dd mk k dd k dd k k mm m kar m dd mk dd k d d k mm m k V GG G V iV GG V i (2.8) Taking into account that † †2 2 2 2 2 2 ˆ ˆ ˆ ˆexp( ) exp( ) exp( ( ) ) k n k n k n iH t iH t i t (2.9) we have for (2.4) † † 2 0 2 * †2 2 2 2 0 ˆ ˆ ˆ ˆ ˆ{ exp( ) exp( )} ˆ ˆ ˆexp( ( ) ) ( ) ( ) { } a rk n dd k k dd n n k nkn d d Tr d d iH t iH ti t G V G V Tr (2.10) And with (2.7) and (2.8): †2 2 0 **2 2 2 2*2 *2 2 2 ˆ ˆ ˆ{ }( ) ( ) ( )( )( ) ( ) ( )( ) ( )( ) ( )( ) ( )( ) ( ) k nkn n a rr an dd n k dd kdd k d d k n dd n d d n kn k k nr ar a m dd m m dd mdd k dd n d d n k mm m n m k
Tr f V G V GG V f G V fi iV G V GG G V V fi i (2.11) Inserting (2.11) into (2.10): † *2 2* *2 2 2 2*2 2 2 2 ˆ ˆ ( ) exp( ( ) ) ( ) ( ) ( )exp( ( ) ) ( ) ( ) ( )( ) ( )exp( ( ) ) ( ) ( ) ( )( ) a rkn n k n dd k k n dd nkn aa r r n dd nk n dd k k n dd n dd k d d k nkn n ka r a kk n dd k k n dd n dd n d d nkn d d f i t G V V G V Gi t G V V G G V fiVi t G V V G G V * ** 2 *2 2 2 2 2 ( ) ( ) ( ) ( )exp( ( ) ) ( ) ( ) ( ) ( )( ) ( ) rdd k kk n r aa r r a m dd m m dd mk n dd k k dd n n dd k dd n d d n k mkn m m n m k
G fi V G V Gi t G V G V G G V V fi i (2.12)
This expressing has large terms. Let’s evaluate them: *2 2*2 2 2 22 ( ) exp( ( ) ) ( ) ( )1( ) ( ) ( ) ( ) ( ) ( ) ( )212 a rkn n k n dd k k n dd nkn a r a rk dd k k k dd k dd ddk dd f i t G V V Gf G V V G G G f dA fd (2.13) (reducing the summation to the integral is a procedure used throughout the manuscript and not showed here for the sake of brevity) The second term: * *2 2 2 22 2 2 2 2 ( )exp( ( ) ) ( ) ( ) ( )( ) ( )1 1( ) ( ) ( ) ( ) ( ) exp( ( ) ) ( ) ( ) ( )2 aa r r n dd nk n dd k k n dd n dd k d d k nkn n ka r a rd d dd dd dd dd
V Gi t G V V G G V fif d G G i t G G di (2.14)
Before proceeding further let’s evaluate the following integral: dd A di (2.15) To do that, we convert the integration to the summation: * a rdd dd k k k dd kk kk k a addk k A d G V V Gi id d d G d Gi (2.16)
Using (2.16) one can perform “sanity check” for t=0: we have for (2.14) at t=0: a r ad d dd dd ddad d dd dd f G G G df G A d (2.17) By analogy the third term: rd d dd dd f G A d (2.18) And the last term: ** 2 *2 2 2 2 22 *2 2 222 ( ) ( )( ) ( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( ) ( )1 1( ) ( )2 ( ( )) r aa r r a m dd m m dd mdd k k dd n n dd k dd n d d n k mkn m m n m kr a a rd d m dd m m dd m m dd m dd mmd d d
V G V GG V G V G G V V fi iV G V G f G Gf
22 2 ( )/ 4 dd A d (2.19)
Summation of (2.19),(2.18),(2.17) and (2.13) leads to:
22 22 2 2 2 22 2 22 22 2 2 ( ) 2( )( ( ))1 ( ) ( ){ 1}2 ( ( )) / 4 ( ( )) / 4( ) 2( )( ( )) ( ) / 41 ( ) ( ){2 ( ( )) d d d d ddd d dd d d d d ddd d f A df A }/ 4( ( )) / 41 ( ) ( ){ }2 ( ( )) / 41 ( ) ( )2 ddd ddd df A df A d (2.20) which is the equilibrium population of the dot as expected at t = ( immediately after the kick the relaxation has not started yet) At times t>0 we assume the wide band approximation in odder tot analytically evaluate time ‐ dependent exponents. Then let’s evaluate the following integral: dd I t i t A di (2.21) We differentiate (2.21) with respect to time:
22 2 22 t dddd d iI t i t A dii i t A di i t d (2.22)
The residue points: / 2 d i (2.23) which means that:
1( ) exp( ( ) )2 ( ) / 41 2 exp( ( ) / 2 ) exp( ( ) / 2 )2 t dd d
I t i i t di i t t i i t t (2.24) Thus exp( ( ) / 2 ) exp( ( ) / 2 )( ) ( ) / 2 / 2 d dd d i t t i t tI t C Ci i (2.25)
With the boundary condition (2.16) we have: ( ) exp( ( ) / 2 ) add d I t G i t t (2.26)
With (2.26) the second term (2.14) becomes: a r ad d dd dd dd dad d dd dd d f G G G i t t dt f G A i t d (2.27)
The third term becomes: a r rd d dd dd dd drd d dd dd d f G G G i t t dt f G A i t d (2.28)
The fourth term ** 2 *2 2 2 2 2* 22 2 2 2 2* * ( ) ( )exp( ( ) ) ( ) ( ) ( ) ( )( ) ( )( ) ( ) ( ) ( )( )( ) r aa r r a m dd m m dd mk n dd k k dd n n dd k dd n d d n k mkn m m n m ka r r add k k dd n n dd k dd n d dkn rm dd m m dn km m n
V G V Gi t G V G V G G V V fi iG V G V G GV G V GV Vi ( ) ( ) exp( ( ) ) exp( ( ) ) ad m m k m m nm k f i t i ti (2.29) In Eq. (2.29) we perform summation over k and n by turning them into integrals: for k dependent part: *2 2 22 2 exp( ( ) )( ) ( )1 1exp( ( ) )2 exp( ( ) / 2 ) a r k mdd k k dd k kk m km dd madd m d m i tG V G V ii t A diG i t t (2.30) for n dependent part: *2 22 2 exp( ( ) )( ) ( )exp( ( ) / 2 ) a r m ndd n n dd n nn m nrdd m d m i tG V G V iG i t t (2.31) And combining (2.31), (2.30) with (2.29) we have for the fourth term:
22 22 2 d d ddd t f A d (2.32) Gathering (2.32),(2.28), (2.27) and (2.13) we finally have : † 2 2 2 2 222 22 22 2 ˆ ˆ12 1 ( ) exp( / 2 ) { exp( ( ) ) exp( ( ) )}21 1( ) exp( ) ( )2 ( ) / 412 1 ( ) exp( / 2 )2 dd r ad d dd dd d dd dd d dddddd d d d A fd t fA G i t G i t dt f A dA fd t
2( ) cos(( ) ) sin(( ) )( ) / 41 1( ) exp( )2 ( ) / 4 d d ddd dd d ddd t tfA dt f A d (2.33) As expected that t † ˆ ˆ d d becomes dd A fd which is an equilibrium dot population at new position d of the dot level. It is also easy to check that at t=0: (2.34) † 2 2 2 2 222 22 2 22 22 2 2 2 22 ˆ ˆ12 2( )1 ( )2 ( ) / 41 1( )2 ( ) / 42( )( ) ( )1 (1 )2 ( ) / 4 ( ) / 42( )1 (2 dd dd d dd dd d dddd d d d ddd d dddd d d A fd fA df A dA f dA f ( ) ( ) ( ) / 4 )( ) / 4( ) / 4 / 41 1( )2 ( ) / 4 2 d d d d ddddd ddd dA f d A fd the dot population is the initial dot population as expected. By analogy, one can compute relaxation of any observables. We can also check that (2.33) is the same expression obtained in Ref Indeed, in this paper: ˆ ˆ 2 d d A fd (2.35) where ( )(1 exp{ ( ) / 2) })1 {1 }/ 2 / 2 d d dd d i tA i i (2.36) We re ‐ write (2.36) as follows: ( )(1 exp{ ( ) / 2) })1 {1 }/ 2 / 2( ) ( ) exp{ ( ) / 2) }1 {1 }/ 2 / 2 / 2( ) exp{ ( ) / 2) }1 (1 )/ 2 / 2 d d dd dd d d d dd d dd d dd d i tA i i i ti i ii ti i (2.37) Thus exp{ ( ) / 2) } exp{ ( ) / 2) }| | (1 ( ) )(1 ( ) )/ 2 / 22( ) cos(( ) ) sin(( ) )(1 ( ) exp( / 2 )( ) / 41( ) exp( ))( ) / 4 d ddd d d d dd dd d ddd d d dd d d i t i tA A i it tA tt (2.38) which, after substituting in (2.35), coincides with (2.33). Currents
Eqs. (2.7) ‐ (2.8) are written for the incoming solution ( obtained using the Lipmann ‐ Schwinger
Equation for the retarded
Green’s function, see the manuscript).
For the outgoing solution we use the advanced GF, which gives *2, , 2 2 , ( )ˆ ˆ ˆ( )( ) ar m dd mn n dd n d d n mm m n V GG V i (3.1) *2, , 2 2 , ( )ˆ ˆ ˆ( )( ) ra m dd mk k dd k d d k mm m k V GG V i (3.2)
Eqs. (3.1) and (3.2) refer to the creation/annihilation of a scattering wave function ( plane, since the problem is To connect our formalism to the Landauer ‐ Buttiker description, recall that the total field operator: † †, ,
1ˆ ˆ( ) 2 k k kk t Dv (3.3) where k p and k v are momentum and velocity, †, ˆ k is just † ˆ k (retarded solution) used above in section and k k nn D p p is the density of states in the momentum space. One can see that †0 , , k k n k kn k nn k kp k k kn k k k kn k n k kn k nk k Tr f p pv v pf f fv mv (3.4)
The total incoming flux of particles: †0 , , inc k n k n I d d Tr (3.5) and the outgoing flux †0 , , out k n k n I d d Tr (3.6) In the equilibrium †0 , , ˆ ˆ ˆ{ } ( ) k n k kn Tr f (see eq. (J9) in the manuscript), thus †0 , , ˆ ˆ ˆ{ } ( ) ( ) k n k kn k n Tr f (3.7) and both fluxes (3.6) and (3.5) are equal, thus † ˆ ˆ 0 N inc out d d d I I Idt In case of the sudden kick outgoing flux: †0 2, 2, out k n k n I t d d Tr (3.8) Consider †0 2, 2, ˆ ˆ ˆ{ } k n Tr : †0 2, 2, *2 2*2 22 2 2 ˆ ˆ ˆ{ } ( )( )2 ( ) ( )( ) ( ) ( ) exp( ( ) )( )( )( ) ( ) ( ) exp( ( ) )( )( ) ( )( k n ar m dd mk k n dd n d d n k k m n km m nra m dd mdd k d d k n n m n km m ka rdd k d d dd n Tr V Gf G V f i tiV GG V f i tiG G **2 ( ) ( )) ( ) ( ) exp( ( ) ) r al dd l m dd md d k n l l m n kl ml k m n
V G V GV V f i ti i (3.9)
Since the first term in this expression will be canceled out by the incoming flux , consider the integration of the second term: *2 2 *2 2 ( )( )( ) ( ) ( ) exp( ( ) )( )( )( ) ( ) exp( ( ) ) ar m dd mk n dd n d d n k k m n km m nar m dd mn dd n d d n m n mm m n V Gd d G V f i tiV Gd G V f i ti (3.10) To evaluate this integral, recall the procedure we used in the section first, we put t to define the boundary conditions and we also assume the wide band approximation in order to evaluate integrals analytically and compare to (2.33): ( )( )( ) ( )( )1 ( )( ) ( )21 1( ) ( ) ( ) ( )2 ar dd mn dd n d d mm m n ar dd mn m dd n d d mm na rd d m dd m m n dd n m n V Gd G V fiGd d G fid G f d G i (3.11) To evaluate the expression above recall that
1( ) 1 1/ 21 1/ 21 1 1/ 2 / 2 / 212 / 2 rn dd n m nn n d m nn n d n mnm d n d m n d mm d d G id i id i ii di i ii i (3.12) (we used here PP f x dx f x f x dxx x ) Thus for (3.11)
1( ) ( ) ( ) ( ) ( )/ 2 a a rm dd m m m dd m dd m mm d i d G f i d G G fi (3.13)
For t we differentiate (3.10) with respect to time: *2 2 22 ( )( )( ) ( ) ( ) ( ) exp( ( ) )( ) exp( ( ) )1 ( ) ( ) ( )21 ( ) ( ) ( ) exp(2 ar m dd mn dd n d d n m k m n m n mm m na r n m n mm dd m m n dd n m na rm dd m m n dd n V Gd G V f i i ti i i td G f d G id G f d G i i ( ) )( ) ( ) exp( ( ) / 2 ) n mam dd m m d m td G f i t t (3.14) Comparing (3.13) and (3.14)finally gives us the expression for the second term in (3.9): ( ) ( ) ( ) ( ) exp( ( ) / 2 ) a rd d m dd m dd m m d m i d G G f i t t (3.15) (note: we could directly evaluate exp( ( ) )( ) r n mn dd n m n i td G i applying several times the residue theorem, however this way is more tedious). By analogy, the third term: ( ) ( ) ( ) ( ) exp( ( ) / 2 ) r ad d m dd m dd m m d m i d G G f i t t (3.16) and for the fourth term we twice apply the procedure outlined in Eqs. (3.11) ‐ (3.16) (it is possible because this term can be factorized into two integrals over k and n , for more details Ref
22 22 2
1( ) exp( ) ( ) / 4 d d dd mm d t f A d (3.17) Thus, combining(3.15), (3.16) and (3.17) for the particle current we have (we have replaced with m ):
22 22 22 2 22 2 2
N inc out d d dddr ad d dd dd da rd d dd dd d
I t I I t f A di d G G f i t ti d G G f i t t (3.18) On the other hand if we differentiate the expression (2.33) with respect to the time variable, we have: † 2 2 2 22 2 2 222 22 2 ˆ ˆ( ) exp( / 2 ) ( ( ) / 2) exp( ( ) )1 ( ) exp( / 2 ) ( ( ) / 2) exp( ( ) )2 1 1 ( ) exp( ) ( )2 ( ) / 41 rd d dd d dd dad d dd d dd dd d ddd d d ddt t fA i G i t dt fA i G i t d t f A d { exp( / 2 ) exp( ( ) )2exp( / 2 ) exp( ( ) )1( ) exp( ) ( ) }( ) / 4 a rdd dd da rdd dd dd d ddd i t fG G i t di t fG G t d t f A d (3.19) which coincides with (3.18) as we expected. We see that at t (3.18) ( ( ) N I t ) goes to zero as also expected. We can also re ‐ write (3.18) in a real form:
22 22 22 2 2 2 2 2 N d d dddd d r add dd d d d d dd dd
I t t f A dtd A A f t t G G (3.20) We can also define expressions for the energy currents: †0 , , ( )1 ˆ ˆ ˆ{ ( ) ( )}2 2 k nout k n k n IE d d Tr t t (3.21) †0 , , ( )1 ˆ ˆ ˆ{ ( ) ( )}2 2 k nin k n k n IE d d Tr t t (3.22) We can also introduce so called incoming and outgoing distribution (see Ref †0 /2, /2, ˆ ˆ ˆ( , ) { ( ) ( )} inc t d Tr t t (3.23) †0 /2, /2, ˆ ˆ ˆ( , ) { ( ) ( )} out t d Tr t t (3.24) And re ‐ write (3.21) and (3.22) as follows ( The
Jacobian from k and n to ( )2 k n and n k is in in IE d t (3.25) out out IE d t (3.26) and for particle currents: †0 , , out k n k n out
I d d Tr t t d t (3.27) in in I d t (3.28) One can also define entropy currents: / / / / / in out in out in out in out in out IS d t t t t (3.29) and the heat current †/ / 0 , , ( )1 1 ˆ ˆ ˆ( ) ( , ) { ( ) ( )}{ }2 2 2 k nout in out in k n k n
IQ t d d d Tr t t (3.30)
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