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Ultra Cold Quantum Gases—Problem Set 3 Elvis Bejko—Group A * Ludwig–Maximilians–Universit¨ at, Fakult¨ at f¨ ur Physik WS 2013-14 1 Rabi oscillations in the dressed state picture We numerically solved the Schr¨ odinger equation for the Hamiltonian ˆ H = ~ 2 0 Ω 0 Ω 0 -2δ , (1) where δ ω L - ω 21 and Ω 0 e(x · ˆ ε)E 0 ~ . Our solution is presented in Fig.1. We notice that as we increase the detuning δ, our ‘‘occupation probability’’ oscillation increases its frequency. 1 At resonance we see that our atom undergoes, periodically, population inversion with probability 1. As we ‘‘detour’’ from resonance, the population inversion occurs with ever decreasing probability. Figure 1. The numerical solution to the Schr¨ odinger equation with Hamiltonian given by Eq.(3). We solved it using Python and the Runge–Kutta 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 ˆ H = ~ 2 0 Ω 0 Ω 0 0 . (2) * [email protected] 1 It is obvious if one looks at the various graphs of Fig.1 vertically.

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Page 1: Simple Cond Matter Problem Set

Ultra Cold Quantum Gases—Problem Set 3

Elvis Bejko—Group A∗

Ludwig–Maximilians–Universitat, Fakultat fur PhysikWS 2013-14

1 Rabi oscillations in the dressed state picture

We numerically solved the Schrodinger equation for the Hamiltonian

H =~2

0 Ω0

Ω0 −2δ

, (1)

where δ ≡ ωL − ω21 and Ω0 ≡e(x · ε)E0

~. Our solution is presented in Fig.1. We notice that as we increase the

detuning δ, 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.

t0.0

0.2

0.4

0.6

0.8

1.0

|c1,

2|2

δ=0

t

δ=0.5

t0.0

0.2

0.4

0.6

0.8

1.0

|c1,

2|2

δ=0.7

t

δ=1.5

Figure 1. The numerical solution to the Schrodinger equation with Hamiltonian given by Eq.(3). We solved it using Python andthe Runge–Kutta 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 Ω0

Ω0 0

. (2)

[email protected] It is obvious if one looks at the various graphs of Fig.1 vertically.

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2

t0.0

0.2

0.4

0.6

0.8

|c2|2

Excited state occupation probability for various δ

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

Finding the eigenvalues and the eigenvectors of the above Hamiltonian, is quite straightforward; we get E1,2 = ±Ω0,with corresponding eigenvectors

|S1,2〉 =1√2

±1

1

= ± 1√

2|g〉+

1√2|e〉 .

(3)

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〉

=1√2|S1〉 −

1√2|S2〉 .

(4)

The Schrodinger equation

i~d

dt|Ψ〉 = H |Ψ〉 , (5)

has the formal solution

|Ψ(t)〉 = e−it~ H |Ψ(0)〉 . (6)

It is obvious, from the formal solution, that if the initial state |Ψ(0)〉 is an eigenstate of the Hamiltonian,2 then theoperation of the exponential operator on that state is trivial, in the sense that

|Ψ(t)〉 = e−itEn

~ |Ψ(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 |Ψ(0)〉 = |n〉.

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3

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

|Ψ0(t)〉 ≡ |g(t)〉 =1√2

(e−

iΩ0t2 |S1〉 − e+

iΩ0t2 |S2〉

). (8)

To calculate the population of the excited state at time t; |〈e| Ψ(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〉

=1√2|S1〉+

1√2|S2〉 .

(9)

Now we just need to take the inner product of the above two Eq’s (8-9). Taking into account that the cross–termsvanish, due to the fact that the |S1,2〉 are orthogonal, we get

〈e| Ψ(t)〉 =1

2

(e−

iΩ0t2 − e+

iΩ0t2

). (10)

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

|c2(t)|2 =1

2(1− cosΩ0t) , (11)

which is the solution that matches exactly Fig’s (1-2)!Now that we have the analytic solution, it is obvious that a π–pulse, of time duration τ = π/Ω0, will completely

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

|c2(π/Ω0)|2 = 1. (12)

By the same token, a π/2–pulse 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: 2–D quadrupole

The magnetic field of an infinitely long current–carrying wire can be found from Ampere’s law∮B · dl = µ0I, (13)

whereby, symmetry arguments lead to

B =µ0

2πrIθ, (14)

where we denote by θ 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

θ = − sin θx+ cos θy

= −yrx+

x

ry.

(15)

Inserting into Eq.(14), we get

B1 =µ0I

1

x2 + y2(−yx+ xy) , (16)

where the subscript denotes that we are refering to the magnetic field of the current–carrying wire at the origin of ourcoordinate system. To create a magnetic quadrupole we place a current–carrying wire, at distance d along the x–axis,

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x

y

Quadrupole magnetic field lines

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

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

B2 =µ0I

1

(x− d)2 + y2(+yx− (x− d)y)

B3 =µ0I

1

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

B4 =µ0I

1

x2 + (y − d)2(+(y − d)x− xy)

(17)

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 Eq’s (16, 17). We can—however—Taylor 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 =

∂B1x

∂x

B1x

∂y

∂B1y

∂x

B1y

∂y

, (19)

the so–called Jacobian matrix.Some simple differentiations reveal that the off–diagonal terms of the Jacobian are zero, while the diagonal terms

are

∂B1x

∂x=µ0I

π

1

d2, and

B1y

∂y= −µ0I

π

1

d2. (20)

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

Bx =4µ0I

π

1

d2x, while By = −4µ0I

π

1

d2y, (21)

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where 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 field3

is

B =√B2x +B2

y +B2z =

√B2

0 +16µ2

0I2

π2d4(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 +

16µ20I

2

π2d2B20

(x2 + y2)

= B0

(1 +

1

2

16µ20I

2

π2d2B20

(x2 + y2)

).

(23)

Now, since the potential created by the magnetic field is ∝ gFµFµB |B|, comparing with1

2mω2(x2 + y2), we find that

the trap frequency is given by

ω =

√16µ2

0I2gFµFµB |B|mπ2d2B2

0

. (24)

The Zeeman energy for the hyperfine interaction is given by

∆EZ = gFµFµB |B|, (25)

where

gF ' gJf(f + 1) + j(j + 1)− i(i+ 1)

2f(f + 1), (26)

and µB =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× 10−5 · eV · T−1)× (2× 10−4T)

' 1.1× 10−8eV.(27)

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

∆EZ(r) = gFµFµB

√B2

0 +16µ2

0I2

π2d4(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 zero–field region, sufficiently fast, its magnetic momentcannot adiabatically ‘‘follow’’ the rapidly changing magnetic field direction. The change in mutual orientation of theatom’s 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 −µ ·B to change, thus allowing the atom to escape the trap, as it is no more

in low energy state.

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3 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

h30e−β

(p2

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−β

(p2

2m

)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 ‘‘non–dilute’’ gas,where most of the atoms ‘‘condensate’’ in the ground state. There the velocity distribution is strongly correlated withthe potential shape.

To calculate the density of states for various physical systems, it is useful to define N(ε), as the number of states

with energy less than or equal to ε. For a 1–D harmonic oscillator, ε = ~ω(n+

1

2

). So the number of states between

with energy less than or equal to ε is

N(ε) =ε

~ω− 1

2. (31)

The density of energy states, by definition equals ρ(ε) ≡ dN(ε)

dε, so from the above equation, we get, for the 1–D

harmonic oscillator, ρ(ε) =1

~ω; i.e. independent of energy.

For the 3–D harmonic oscillator, the situation is slightly more complicated, nevertheless the physics is the same.We have

ε = ~ωxnx + ~ωyny + ~ωznz +3

2ω, with ω =

ωx + ωy + ωz3

(32)

Now the number of states with energy less than or equal to ε = εx + εy + εz will be given by an integral of the form

N(ε) =1

~ωx~ωy~ωz

∫ ε

0

dεx

∫ ε−εx

0

dεy

∫ ε−εx−εy

0

dεz

=ε3

6~ωxωyωz.

(33)

The density of states, of the 3–D harmonic oscillator potential, is simply given by differentiating the above expressionwith respect to energy

ρ(ε) =ε2

2~ωxωyωz. (34)

The calculation for the 3–D ‘‘box’’ potential, runs on the same path. If we suppose that the dimensions of the boxpotential are all equal to L, then we have the constraint

k =nπ

L, with n = 1, 2, . . . (35)

5 In the canonical ensemble.

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7

where k is the wavenumber of the wavefunction. Let us calculate the number of states that have wavenumber6 lessthan or equal (in magnitude) to k

N(k) =1

8

43πk

3

( πL )3= g

V

6π2k3, (36)

where the factor 1/8 comes because we take into account only possitive values of k, hence we integrate the firstoctant of a sphere in k–space, and the factor g is the degeneracy, which we put in by ‘‘hand’’. As usual, the densityof states is equal to

ρ(k) =N(k)

dk= g

V

2π2k2. (37)

We want the density of energy states and not the density of k–states. Since, for a particle in a box, it holds that

ε =~2k2

2m, (38)

and since ρ(k)dk = ρ(ε)dε, we get

ρ(ε) = gV

2π2

(2m

~2

)3/2√ε. (39)

4 Bose–Einstein Condensation

As we explained in the Problem 2; the main issue that needed to be solved, at the time the paper was written, wasthe Majorana losses due to the nonadiabatic spin flips of the atoms, as they pass near the center, where the fieldchanges rapidly direction. Avoiding lower confinement (which was the issue with ‘‘TOP’’ trap processes), the groupeffectively ‘‘plugged’’ the magnetic field hole by tightly focusing an intense blue–detuned laser, which generated arepulsive potential for the atoms. Far detuning minimized the heating due to photon scattering.

The main observable of the paper, as always, is demonstrated in the abstract. The group observed a Bose–Einsteincondensation of sodium atoms. In order to do so, they employed magnetic, optical forces,7 and evaporative cooling.How did they understand that they created a BEC? The main experimental signature of the BEC was the bimodal8velocity distribution below the critical temperature; indicating that the atoms are condensing to the ground state.There exists; however, a second experimental evidence of the formation of BEC, that is the sudden change of the areaof the cloud, indicating that a phase transition has occured.

6 Which is directly related with the energy; however, the calculation is easier in the k–phase space.7 Optical force is the ‘‘plugging’’ of the region where the magnetic field was zero.8 Meaning a superposition of two distributions; a broad one due to the normal gas and a narrow one due to the condensate.