E356: Particle bound by harmonic oscillator potential

E356: Particle bound by harmonic oscillator potential + Homogeneous magnetic field Submitted by: August Yitzhak The problem: A particle is bound by the two dimensional harmonic oscillator potential. 1 V (r) = µΩ20 r2 2 Ω0 is the natural frequency of the Oscillator,µ is the mass of the Particle. (1) Write the Energy levels.What is the degenerate of each level ? The way to solve this problem is by separation of variables in cartesian coordinates. Next step is to translate this solution to a solution in polar coordinates. (2) Write the Energy levels after adding the magnetic fields B. (3) What is the generic degeneracy of each level ? (4) What is the magnetic susceptibility if the system is in the ground state ? (5) In our assumption was that the particle moves in two dimensions. What would be the answer in three dimension (the particle is free to move in the z directions),neglecting the diamagnetic term. The solution: (1) In cartesian coordinates the Hemiltonian of two dimensional oscillators is H=


p2y 1 p2x + + µΩ20 x2 + y 2 2µ 2µ 2

and the energy levels are En1 ,n2 = ~Ω0 (n1 + n2 + 1) n1 , n2 = 0, 1, 2, 3.. The degeneracy of the n level is g (En ) = n + 1 we assume that physical system is independent of choosing a coordinate system then the energies and the degenerate in cartesian coordinates are the same as in polar coordinates.we are going to describe the following system in polar coordinates. In polar coordinates the Hemiltonian of two dimensional oscillators is H0 =

L2 1 p2r + z 2 + µΩ20 r2 2µ 2µr 2

Lz = −i~

∂ ∂φ

The eigenvalue of Lz are m = 0 ± 1, ±2, ±3..   1 ∂ ∂ 2 pr = r r ∂r ∂r 1

H (m) =

p2r m2 1 + + µΩ20 r2 2 2µ 2µr 2

For each state with positive m there is an independent state with negative m with the same radial dependence and hence the same energy. So now letting m be any integer.The radial quantum number is ν it is non-negative integer. we translate this to a solution in polar coordinate system Em,ν = ~Ω0 (2ν + |m| + 1) The ground state (m = 0, ν = 0) corresponding to (n1 = 0, n2 = 0) is non-degenerate with energy ~ω. Its ~Ω2 0 for every independent oscillator degree of freedom.The first excited state with energy 2~Ω0 is doubly degenerate,(m = ±1, ν = 0) as was the case previously, (2) After adding the magnetic field B the Hmiltonian is H=

p2 1 e2 B 2 2 eB + µΩ20 (x2 + y 2 ) + (x + y 2 ) − Lz 2µ 2 8µ 2µ

can be written as H = H0 + H1 , H0 is Hmiltonianthe of Oscillator as in previous.H1 belong to the external magnetic field . we defin cyclotron frequency to be ΩB ΩB =

eB µ

And we get H=


1
1 p2 1 + µΩ20 x2 + y 2 + µΩ2B x2 + y 2 − ΩB Lz 2µ 2 8 2

we defin effective frequency to be Ω 1 Ω2 = Ω2B + Ω20 4 H=

p2 1 1 + µΩ2 (x2 + y 2 ) − ΩB Lz 2µ 2 2

So the energies are of Particle in new two dimensional harmonic oscillator potential plus Zemman term − 21 ΩB Lz In cartesian coordinates 1 En1 ,n2 = ~Ω (n1 + n2 + 1) − ΩB Lz 2 and in polar coordinates 1 Em,ν = ~Ω (2ν + |m| + 1) − ΩB m 2 (3) The energies levels are 1 Em,ν = ~Ω (2ν + |m| + 1) − ΩB m 2 if β that is β = ΩΩB cannot be considered as a fraction than the generic degeneration of each level is zero. In the limit Ω0 → 0 the energies levels are 1 Em,ν = ~ΩB (2ν + |m| − m + 1) 2 2

(4) The magnetic susiptability of the system in the ground stat m = 0,ν = 0 is ∂E0,0 ~Be2 =− M =− ∂B 8µ2



B 2 e2 + Ω20 4µ2

−1/2

∂M ~e2 χ= = − ∂B B=0 8µ2 Ω20 the susiptability less then zero. (5) For a motion in three dimension in cartesian coordinates the Hemiltonian is H=


p2y p2x p2 1 + + z + µΩ20 x2 + y 2 + z 2 2µ 2µ 2µ 2

and the energy levels are   3 En1 ,n2 ,n3 = ~Ω0 n1 + n2 + n3 + 2

n1 , n2 , n3 = 0, 1, 2, 3.. For the degree of degeneracy n = n1 + n2 + n3 with ni = 0, 1, 2, ... For a given n we can choose a particular n1 . Then n2 + n3 = n − n1 . There are n − n1 + 1 possible pairs n2 , n3 . n2 can take on the values 0 to n − 1, and for each n2 the value of n3 is fixed. The degree of degeneracy therefore is g (En ) =

n X

(n − n1 + 1) =

n1 =0

n X

(n + 1)−

n1 =0

n X

(n1 ) = (n + 1) (n + 1)−

n1 =0

In cylindrical coordinates the Hemiltonian is H0 =

p2r L2 1 p2 + z 2 + µΩ20 r2 + z 2µ 2µr 2 2µ

H (m) =

m2 1 2 2 p2z p2r + + µΩ r + 2µ 2µr2 2 0 2µ

The energy levels are   3 Em,ν,n = ~Ω0 2ν + |m| + n + 2 After adding the magnetic field B the Hmiltonian is H = H0 + H1 1 1 H1 = mΩ2B r2 − ΩB Lz 8 2 And the energies levels are Em,ν,n

  1 1 = ~Ω (2ν + |m| + 1) − ΩB m + ~Ω0 n + 2 2 3

(n + 1) (n + 2) n (n + 1) = 2 2

If we neglect the diamagnetic term the energy levels are   3 1 Em,ν,n = ~Ω0 2ν + |m| + n + − ΩB m 2 2 if β that is β = 2ν + n .

Ω ΩB

cannot be considered as a fraction the generic degenerate of the energies is

g (E0 ) = g (E1 ) = 1 g (E2 ) = g (E3 ) = 2 and in general g (En ) = int

n 2

+1

the magnetic susiptability of the system in the ground stat is as in two dimensions .

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