Simple Cond Matter Problem Set

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Rabi oscilations in cond matter

Transcript of Simple Cond Matter Problem Set

  • Ultra Cold Quantum GasesProblem Set 3

    Elvis BejkoGroup ALudwigMaximiliansUniversitat, Fakultat fur Physik

    WS 2013-14

    1 Rabi oscillations in the dressed state picture

    We numerically solved the Schrodinger equation for the Hamiltonian

    H =~2

    0 00 2

    , (1)where L 21 and 0 e(x )E0~ . Our solution is presented in Fig.1. We notice that as we increase thedetuning , our occupation probability oscillation increases its frequency.1 At resonance we see that our atomundergoes, periodically, population inversion with probability 1. As we detour from resonance, the populationinversion occurs with ever decreasing probability.







    |c 1,2|2










    |c 1,2|2




    Figure 1. The numerical solution to the Schrodinger equation with Hamiltonian given by Eq.(3). We solved it using Python andthe RungeKutta 4 integration scheme. The solid line corresponds to |c2(t)|2, while the dashed one to |c1(t)|2.

    At resonance, the Hamiltonian assumes a simple form

    H =~2

    0 00 0

    . (2)

    Elvis.Bejko@campus.lmu.de1 It is obvious if one looks at the various graphs of Fig.1 vertically.

  • 2t0.0





    |c 2|2

    Excited state occupation probability for various

    Figure 2. The excited state occupation probability as a function of time. The effect of detuning is more obvious in this figure;i.e. with increasing , the amplitude decreases while the frequency increases.

    Finding the eigenvalues and the eigenvectors of the above Hamiltonian, is quite straightforward; we get E1,2 = 0,with corresponding eigenvectors

    |S1,2 = 12


    = 1

    2|g+ 1

    2|e .


    Since the dressed state vectors |S1,2 form a complete set, we can write the initial state, |g, as follows|g = |S1 S1| g+ |S2 S2| g

    =12|S1 1

    2|S2 .


    The Schrodinger equation


    dt| = H | , (5)

    has the formal solution

    |(t) = e it~ H |(0) . (6)It is obvious, from the formal solution, that if the initial state |(0) is an eigenstate of the Hamiltonian,2 then theoperation of the exponential operator on that state is trivial, in the sense that

    |(t) = e itEn~ |(0) . (7)In other words, the time evolution appends a phase in front of the eigenstate. If the initial state is a superpositionof eigenstates of the Hamiltonian, each eigenstate will get a different phase e itEn~ , depending on its energy.

    2 |(0) = |n.

  • 3Having established the above, in the sudden approximation, the evolution of the ground state after we turn onour light field, is given by

    |0(t) |g(t) = 12


    i0t2 |S1 e+

    i0t2 |S2

    ). (8)

    To calculate the population of the excited state at time t; |e| (t)|2, we need to express the excited state as afunction of |S1,2, as we did for the ground state in Eq.(4)

    |e = |S1 S1| e+ |S2 S2| e=

    12|S1+ 1

    2|S2 .


    Now we just need to take the inner product of the above two Eqs (8-9). Taking into account that the crosstermsvanish, due to the fact that the |S1,2 are orthogonal, we get

    e| (t) = 12


    i0t2 e+ i0t2

    ). (10)

    Multiplying with the complex conjugate, we arrive at the desired result

    |c2(t)|2 = 12

    (1 cos0t) , (11)

    which is the solution that matches exactly Figs (1-2)!Now that we have the analytic solution, it is obvious that a pipulse, of time duration = pi/0, will completely

    invert the population of the atom from the ground state to the excited state, because

    |c2(pi/0)|2 = 1. (12)By the same token, a pi/2pulse will send the atom to a superposition state where, the atom is equally probable to bein the excited as it is to be in the ground state!

    2 Magnetic trap: 2D quadrupole

    The magnetic field of an infinitely long currentcarrying wire can be found from Amperes lawB dl = 0I, (13)

    whereby, symmetry arguments lead to

    B =02pir

    I, (14)

    where we denote by the unit vector, in polar coordinates, that circulates around the origin.Let us calculate, explicitly, the components of the magnetic field (cartesian coordinates) induced by the wire in the

    origin of our coordinate system. From elementary geometrical considerations, it is clear that

    = sin x+ cos y= y





    Inserting into Eq.(14), we get

    B1 =0I



    x2 + y2(yx+ xy) , (16)

    where the subscript denotes that we are refering to the magnetic field of the currentcarrying wire at the origin of ourcoordinate system. To create a magnetic quadrupole we place a currentcarrying wire, at distance d along the xaxis,

  • 4x


    Quadrupole magnetic field lines

    Figure 3. The quadrupole magnetic field lines. Plotted with the help of Python and Eqs (16 and 17).

    with the current with the opposite direction. Then we place a currentcarrying wire, with the opposite direction, adistance d along the yaxis. Finally, with the current reversed again, we place a wire at a distance d along the yaxis,with respect to the first wire. The other magnetic fields are

    B2 =0I



    (x d)2 + y2 (+yx (x d)y)

    B3 =0I



    (x d)2 + (y d)2 ((y d)x+ (x d)y)

    B4 =0I



    x2 + (y d)2 (+(y d)x xy)


    The resulting magnetic field lines are shown in Fig.3.If we calculate the total magnetic field at the center of the configuration (x = d/2, y = d/2), it will come out zero,

    as can be seen from Fig.3 and Eqs (16, 17). We canhoweverTaylor expand our magnetic field at the center, tocalculate the magnetic field gradient. We do so for the B1

    B1(r) =:0

    B1 (r0) + JB1(r0) (r r0)T +O((r r0)2), (18)

    where r0 = (x = d/2, y = d/2) and

    JB1 =





    , (19)the socalled Jacobian matrix.

    Some simple differentiations reveal that the offdiagonal terms of the Jacobian are zero, while the diagonal termsare





    d2, and B1y

    y= 0I



    d2. (20)

    Since the same is true for all four magnetic fields, we conclude that, for small deviations from the center of theconfiguration

    Bx =40I



    d2x, while By = 40I



    d2y, (21)

  • 5where x and y measure the distance from the center of the configuration. Indeed, if one looks at the Fig.3, above, itis obvious that for constanst x, if we move toward positive y, the By component becomes negative, hence justifyingthe minus sign of Eq.(21).

    If we add a homogeneous magnetic field B = B0ez to our configuration, then the magnitude of the magnetic field3is

    B =B2x +B

    2y +B

    2z =

    B20 +


    pi2d4(x2 + y2). (22)

    Around the center of our configuration, the second term within the square root of the above equation, is small incomparison with the first. That allows us to Taylor expand the above equation

    B = B0

    1 +


    pi2d2B20(x2 + y2)

    = B0

    (1 +




    pi2d2B20(x2 + y2)



    Now, since the potential created by the magnetic field is gFFB |B|, comparing with 12m2(x2 + y2), we find that

    the trap frequency is given by



    2gFFB |B|mpi2d2B20

    . (24)

    The Zeeman energy for the hyperfine interaction is given by

    EZ = gFFB |B|, (25)where

    gF ' gJ f(f + 1) + j(j + 1) i(i+ 1)2f(f + 1)

    , (26)

    and B =e

    2m. For 87Rb, gF = 1/2. Also, the minimum magnetic field intensity is at the center of our configuration,

    where |B| = B0. With everything known, we calculate the Zeeman energy (mF = 2)

    EZ =1

    2 2 (5.788 105 eV T1) (2 104T)

    ' 1.1 108eV.(27)

    Of course, in general, the Zeeman energy is spatially dependent

    EZ(r) = gFFB

    B20 +


    pi2d4(x2 + y2). (28)

    The reason why we need an extra homogeneous field is so that we can prevent our magnetic field from having a zerominimum. Why is that bad? When the atom traverses the zerofield region, sufficiently fast, its magnetic momentcannot adiabatically follow the rapidly changing magnetic field direction. The change in mutual orientation of theatoms magnetic moment and the magnetic field, as a result, causes the atoms to escape the trap, phenomenonknown also as Majorana losses.4

    3 For not too large deviations from the center of our configuration.4 The rapid change of the magnetic field, may cause the sign of the B to change, thus allowing the atom to escape the trap, as it is no more

    in low energy state.

  • 63 Phase space density

    We know, from classical statistical mechanics, that the probability to find a molecule5 at position r with momentump equals

    P (r,p)d3rd3p d3rd3p



    2m+U(r)). (29)

    The velocity distribution is given by integrating the above equation with respect to d3r. Now the point is that theintegral over all positions is extremely hard to carry out, since the interaction can be extremely complicated. However,in an ideal gas, we assume that the gas is dilute enough that we can actually neglect the interactions between themolecules. In that approximation the velocity distribution is independent of the potential shape

    P (p)d3p = Ce



    )d3p, (30)

    where the proportionality constant, C, comes from the spatial integration and is equal to V N . It is obvious that theabove approximation collapses in the case of a BEC, since there we have the extreme case of a nondilute gas,where most of the atoms condensate in the ground state.