Superconductors: Quantum circuits

Transcription

1 Superconductors: Quantum circuits J. J. García-Ripoll IFF, CSIC Madrid ( )

2 Mesoscopic QIPC

3 Small systems So far we have only seen small systems to store and process QI Individual atoms As trapped ions As pseudo-crystals Other successful systems which are very similar Molecules For NMR Or for AMO type exp.

4 But not so small Even though each individual register qubit is small, we have a mesoscopic number of them atoms So many atoms that the law of averages works! We can define collective observables L z = 1 2 i i z where quantum fluctuations can be measured.

5 Atomic ensembles Trap a gass of atoms in a glass cell. Polarize them L z = 1 2 z i i ~N /2 Study the fluctuations of transverse components [ L x, L y ]=i! L z i! N /2 Q L x 2/ N P L y / 2/ N Q =0 P =0 Use them for continuous variables QIPC.

6 Atomic ensembles All these systems (ions, atoms, molecules) are ideal negligible or controlled interactions isolated, little or no dissipation. Can we use real-life materials and systems for QIPC in a coherent way? What about interactions What about T? What about decoherence

7 Mesoscopic quantum systems

8 Mesoscopic quantum systems Any system is candidate for QIPC if suitable degrees of freedom exist. What matters Energy scales of the degree of freedom. Temperature. In addition to this Decoherence time scales Dimensionality Controllability

9 Mesoscopic quantum systems A lot of systems in the last years have reached the quantum degenerate limit.

10 Superconductors

11 Superconductivity Electrons are the charge carriers in a solid. They may interact through the phonons. This attraction couples pairs of fermions.

12 Superconductivity Electrons are the charge carriers in a solid. They may interact through the phonons. This attraction couples pairs of fermions. Cooper pair (according to Google)

13 Superconductivity u k k v k c k c k vac Cooper pair (sort of) Electrons are the charge carriers in a solid. They may interact through the phonons. This attraction couples pairs of fermions. Bosonic particles Charge carriers Superfluid Result: a conductor without resistance.

14 Quantum degrees of freedom

15 Electric variables We can use Electric potential, V Current intensity, I I V

16 Electric variables We can use Electric potential, V Current intensity, I Or their integrals I Q V Flux, ϕ t = V t dt Charge, Q t Q= I t dt All of them, macroscopic.

17 Linear response The different circuit elements are described by one relation between variables. I Q V

18 Linear response The different circuit elements are described by one relation between variables. Capacitive elements Q V V =g Q

19 Linear response Q V The different circuit elements are described by one relation between variables. Capacitive elements V =g Q Inductive elements I I = f Linear cases Capacitance(C) Inductance(L) V =Q/C I = / L

20 Energies Where does that come from?

21 Power flow The work or power that flows into a circuital element is given by úh =V I =V úq= ú I I V Using the previous relations Q Q H = end cap 0 g Q dq H H = end ind 0 f d for capacitive and inductive elements.

22 Quadratic approximations If we use only the linear part of f and g we obtain I Q V H cap = Q 2 2C H ind = 2 2 L H These can be understood as the quadratic approximations of the electric and magnetic energies.

23 Circuit quantization How to obtain the quantum degrees of freedom

24 Outline: Lagrangian Quantization prescription: Choose set of quantum variables Find evolution equations Find the originating Lagrangian Compute the Hamiltonian Find canonical variables Impose commutation relations (Solve quadratic H)

25 Step 1: identify variables Identify or define a ground point. Trace unique paths to all other vertices, following the inductors. Each line is associated a flux, charge, potential or intensity variable. Each vertex is given a topology-dependent value, the difference of which forms the branch values.

26 Step 1: identify variables a g b Identify or define a ground point. Trace unique paths to all other vertices, following the inductors. Each line is associated a flux, charge, potential or intensity variable. Each vertex is given a topology-dependent value, the difference of which forms the branch values.

27 Step 2: Kirchoff laws Charge conservation: The sum of intensities arriving to a node is zero b arriving at v I b =0 Voltage law The sum of potential differences along a loop is zero b#loop V b =0

28 Step 2: Kirchoff laws Charge conservation: The sum of charges arriving to a node is constant b arriving at v Q b = Q v Voltage law The sum of fluxes along a loop is constant b loop b = l

29 Step 3: Evolution equations For each active node L 3 Sum of currents arriving from inductors equal to a C 3 b sum of currents going into capacitors. L 1 C 1 C 2 L 2 g

30 Step 3: Evolution equations For each active node a Sum of currents arriving from inductors equal to sum of currents going into capacitors. Along capacitor 1 C 1 Q1 =C 1 V 1 g d I 1 =C 1 dt V 1=C 1 a & g =C 1 a

31 Step 3: Evolution equations For the inductive element 1 L 3 I 1 = a g / L 1 a For the inductive element 3 L 1 g I 3 = a b / L 3 where we have used the Kirchoff law to obtain the upper flux.

32 Step 3: Evolution equations Finally L 3 C 1 a C 3 a b = a a b L 1 L 3 a C 3 b C 2 b C 3 b a = b b a L 2 L 3 L 1 C 1 C 2 L 2 g

33 Step 4: Lagrangian Finally L 3 C 1 a C 3 a b = a a b L 1 L 3 a C 3 b C 2 b C 3 b a = b b a L 2 L 3 L 1 C 1 C 2 L 2 g This are Lagrange equations L=C 1 2 a 2 C 2 2 b a b 2 C 2C a 2 b a b 2 L 1 L 2 2 L 3

34 Step 4: Lagrangian From this Lagrangian a L 3 2 a L=C 1 C 3 b 2 b a b 2 2 C 2 2 C a 2 b a b 2 ] L 1 L 2 2 L 3 L 1 C 1 C 2 L 2 g the equations are given as d 2 x dt x= " L 2 "

35 Step 4: Canonical variables L 3 a C 3 b L 1 C 1 C 2 L 2 g We obtain the conjugate variables, charges, as in our case q x = L x q a =C 1 q b =C 2 a C 3 a b b C 3 a b Note that they mix both fluxes This will give rise to long range capacitive interactions

36 Step 5: Hamiltonian a C 3 b We obtain the conjugate variables, charges, as L 3 H = in our case x q x x L L 1 C 1 C 2 L 2 g H = q 2 a q 2 b q aq b 2 2C a 2C b 2C ab 2 a 2 b a b 2 ] L 1 L 2 2 L 3 C a = C 1C 2 C 1 C 3 C 2 C 3 C 2 C 3

37 Step 5: Quantization L 3 The charge and fux variables are imposed to have a C 3 b [ x, q x ]=i ħ L 1 C 1 C 2 L 2 g They are like position momentum variables. If the Hamiltonian is quadratic, we can diagonalize to normal modes, like for ions.