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• 1Summary of Quantum and Statistical Mechanics

1.1 One Dimensional Schrodinger Equation

In Quantum Mechanics, the state of a particle in one dimension and in presence of apotential U(x, t), is entirely described by a complex wave function (x, t) obeyingthe time dependent Schrodinger equation

ih(x, t)

t= h

2

2m 2(x, t)

x2+U(x, t)(x, t)

where m is the mass of the particle and h is the Planck constant, h, divided by 2 .If we multiply the Schrodinger equation by the complex conjugate wave function(x, t), take the complex conjugate of the Schrodinger equation and multiply by(x, t), and nally subtract both expressions, we nd the so-called continuity equa-tion

|(x, t)|2 t

+x

[ih2m

((x, t)

(x, t)x

(x, t)(x, t)x

)]= 0.

This equation represents the conservation law for the quantity |(x, t)|2 dx and

allows us to interpret |(x, t)|2 as the probability density function to nd the particlein the point x at a time t. The quantity

J(x, t) =ih2m

((x, t)

(x, t)x

(x, t)(x, t)x

)

is the density ux for such probability. The physical interpretation of |(x, t)|2 setssome conditions on (x, t) that has to be chosen as a continuous not multivaluedfunction without singularities. Also, the derivatives of (x, t) have to be continu-ous, with the exception of a moving particle in a potential eld possessing somediscontinuities, as we will see explicitly in the exercises. If the potential does notdepend explicitly on time,U(x, t) =U(x), the time dependence can be separated out

Cini M., Fucito F., Sbragaglia M.: Solved Problems in Quantum and Statistical Mechanics.DOI 10.1007/978-88-470-2315-4 1, c Springer-Verlag Italia 2012

• 4 1 Summary of Quantum and Statistical Mechanics

from the Schrodinger equation and the solutions, named stationary, satisfy

(x, t) = eih Et(x).

In such a case, the functions = ||2 and J are independent of time. Using theform of (x, t) in the original equation, we end up with the stationary Schrodingerequation [

h2

2md2

dx2+U(x)

](x) = H(x) = E(x).

The operator H is known as the Hamiltonian of the system. Continuous, non mul-tivalued and nite functions which are solutions of this equation exist only for par-ticular values of the parameter E, which has to be identied with the energy of theparticle. The energy values may be continuous (the case of a continuous spectrumfor the Hamiltonian H), discrete (discrete spectrum), or even present a discrete andcontinuous part together. For a discrete spectrum, the associated may be normal-ized to unity

|(x)|2 dx = 1.All the functions corresponding to precise values of the energy are called eigen-functions and are orthogonal. With a continuous spectrum, the condition of or-thonormality may be written using the Dirac delta function

E(x)E (x)dx = (EE ).

The condition of continuity for the wave function and its derivatives is valid even inthe case when the potential energy U(x) is discontinuous. Nevertheless, such con-ditions are not valid when the potential energy becomes innite outside the domainwhere we solve our differential equations. The particle cannot penetrate a region ofthe space where U = + (you can imagine electrons inside a box), and in such aregion we must have = 0. The condition of continuity imposes a vanishing wavefunction where the potential energy barrier is innite and, consequently, the deriva-tives may present discontinuities.

Let U be the minimum of the potential. Since the average value of the energy isE = T +U , and since U > U , we conclude that

E > U

due to the positive value of T , that is the average kinetic energy of the particle. Thisrelation is true for a generic state and, in particular, is still valid for an eigenfunctionof the discrete spectrum. It follows that En > U , with En any of the eigenvalues ofthe discrete spectrum. If we now dene the potential energy in such a way that itvanishes at innity (U() = 0), the discrete spectrum is characterized by all thoseenergy levels E < 0 which represent bound states. In fact, if the particle is in a boundstate, its motion takes place between two points (say x1,x2) so that () = 0.

• 1.2 One Dimensional Harmonic Oscillator 5

This constraints the normalization condition for the states. In Classical Mechanics,the inaccessible regions where E

• 6 1 Summary of Quantum and Statistical Mechanics

represents the n-th order Hermite polynomial. The rst Hermite polynomials are

H0( ) = 1 H1( ) = 2

H2( ) = 4 22 H3( ) = 8 312 .Equivalently, we can describe the properties of the harmonic oscillator with thecreation and annihilation operators, a and a, such that [a, a] = 11. In this case, theHamiltonian becomes

H = h(aa+

1211)= h

(n+

1211)

where n = aa is the number operator with the property n |n= n |n . The relationsconnecting the creation and annihilation operators with the position and momentumoperators are

a =

m2h x+

i2mh

p

a =

m2h x i2mh p

x =

h2m (a

+ a)

p = i

mh2 (a

a).The creation and annihilation operators act on a generic eigenstate as step up andstep down operators

a |n=n+1 |n+1 a |n=n |n1

so that

|n= (a)nn!

|0 n|m= nm.

1.3 Variational Method

The variational method is an approximation method used to nd approximateground and excited states. The basis for this method is the variational principlewhich we briey describe now. Let | be a state of an arbitrary quantum systemwith one or many particles normalized such that

N = |= 1.

The energy of the quantum system is a quadratic functional of |

E = |H|

• 1.3 Variational Method 7

and cannot be lower than the ground state 0; in fact, let us expand the state | inthe eigenfunction basis |n (each one corresponding to the eigenvalue n) of H

|=nn||n

n|n||2 = 1.

Since n 0, we nd that

E = |H|=n|n||2n

n|n||20 = 0.

From this we infer that the energy of the ground state can be found by minimization.Let us start by variating the state |

| |+ |.

The variation of the energy which follows is

E = |H|+ |H|+O( 2).

At the same time, the normalization changes as

N = |+ |+O( 2).

The extremum we are looking for must be represented by functions whose norm is1. This condition can be efciently imposed by introducing a Lagrangian multiplierand minimizing the quantity

Q () = |H11|= (EN).

The variation of Q () must be zero for variations in both and

Q () = |H11|= 0 Q () = ( |1) = 0.

The second equation is the constraint, the rst must be valid for arbitrary variations,leading to

H|= |which is the stationary Schrodinger equation. Multiplying by |, we get the valueof the multiplier, i.e. the energy of the ground state. We remark that the condition ofconstrained minimum, follows from that of unconstrained minimum ( |H|) =0 substituting H with (H 11). This can also be done by making | depend onthe multiplier , whose value is xed imposing

N = ( )|( )= 1.

• 8 1 Summary of Quantum and Statistical Mechanics

1.4 Angular Momentum

From the denition of the angular momentum in Classical Mechanics

LLL = rrr ppp

with rrr (x,y,z), ppp (px, py, pz), we get its quantum mechanical expression, oncethe vectors rrr, ppp are substituted by their correspondent operators. Once the commu-tation rule between rrr, ppp is known, it is immediate to deduce the commutation rulesof the different components of the angular momentum

[Lx, Ly] = ihLz [Ly, Lz] = ihLx [Lx, Lz] =ihLy.

From the theory of Lie Algebras, we know that a complete set of states is deter-mined from a set of quantum numbers whose number is that of the maximum num-ber of commuting operator we can build starting from the generators (in our caseLx, Ly, Lz). One of these operators is the Casimir operator

L2 = L2x + L2y + L

2z

which is commuting with all the generators of the group, i.e. [L2, Li] = 0, i = x,y,z.The other element of this sub algebra is one of the three generators Lx, Ly, Lz: theconvention is to choose Lz. The quantum states are thus labelled by the quantumnumbers l, m such that

L2 |l,m= h2l(l+1) |l,m Lz |l,m= hm |l,m m l m

L+ |l,m= h

(lm)(l+m+1) |l,m+1L |l,m= h

(l+m)(lm+1) |l,m1

where L = (Lx iLy) are known as raising and lowering operators for the z com-ponent of the angular momentum. Other useful relations are

[Lz, L] =hL [L2, L] = 0 [L+, L] = 2hLzL+L = L2 L2z + hLzLL+ = L2 L2z hLz.

When acting on functions of the spherical polar coordinates, the generators and theCasimir operator take the form

Lx = ih(sin

+ cot cos

)

Ly =ih(cos

cot sin

)

• 1.4 Angular Momentum 9

Lz =ih

L2 =h2[

1sin

(sin

)+

1sin2

2

2

].

To write the state |l,m in spherical coordinates it is useful to introduce the sphericalharmonics

Yl,m( ,) = , |l,mwhich enjoy the property

Yl,m( ,) = (1)m

(2l+1)4

(lm)!(l+m)!

eimPml (cos) m 0

Yl,m( ,) = Yl,|m|( ,) = (1)|m|Y l,|m|( ,) m < 0where Pml (cos) is the associated Legendre polynomial dened by

Pml (u) = (1u2)m/2dm

dumPl(u) 0 m l

Pl(u) =1

2l l!dl

dul

[(u21)l

]where Pl(u) is the Legendre polynomial of order l. Some explicit expressions forl = 0,1,2 are

Y0,0( ,) =14

Y1,0( ,) =

34

cos Y1,1( ,) =

38

sinei

Y2,0( ,) =

516

(3cos2 1) Y2,1( ,) =

158

sin cosei

Y2,2( ,) =

1532

sin2 e2i .

Given the two angular momentum operators L1, L2, we now want to deduce thestates of the angular momentum operator sum of the two, L= L1+ L2. This is possi-ble using the states |l1, l2,m1,m2 or |l1, l2, l,m, where h2l(l+1), hm are the eigen-values of the operators L2 = L21 + L

22 and Lz = L1z + L2z respectively. The relation

between these two sets of states is

|l1, l2, l,m= m1,m2

l1, l2,m1,m2|l1, l2, l,m |l1, l2,m1,m2 .

• 10 1 Summary of Quantum and Statistical Mechanics

The total angular momentum gets the values l = l2+ l1, l2+ l11, ..., |l2 l1|, m =m1 +m2 and the coefcients l1, l2,m1,m2|l1, l2, l,m are known as Clebsh-Gordanor Wigner coefcients.

1.5 Spin

From the experiment of Stern and Gerlach or from the splitting of the electron en-ergy levels in an atom it follows that, besides an angular momentum, a movingelectron has a spin. Such quantity has no classical correspondence and can get thetwo values h/2. The generators of this physical quantity are

Sx =h2

x Sy =h2

y Sz =h2

z

where x, y, z are the Pauli matrices . A possible representation of these matrices(the one where z is diagonal) is

x =

(0 11 0

)y =

(0 ii 0

)z =

(1 00 1

).

The states which describe the spin are two dimensional vectors

=

(12

)

and |1|2, |2|2 is the probability to get h/2 out of a spin measurement.

1.6 Hydrogen Atom

The problem of the motion of two interacting particles with coordinates rrr1, rrr2 canbe reduced, in analogy with Classical Mechanics, to the motion of a single particleat distance r from a xed centre. The Hamiltonian of the two particles of masses m1and m2 and interacting with a centrally symmetric potential U(r) is given by

H = h2

2m121

h2

2m222+U(r)

where 21, 22 are the Laplacian operators in the coordinates rrr1, rrr2 and rrr = rrr1 rrr2.

Using the coordinate of the center of mass

RRR =m1rrr1+m2rrr2

m1+m2

the Hamiltonian becomes

H = h2

2(m1+m2)2R

h2

2mr2r +U(r)

• 1.6 Hydrogen Atom 11

where m1+m2 is the total mass and mr =m1m2/(m1+m2) the reduced mass. Then,we seek the solution in the form (rrr1,rrr2) = (RRR)(rrr), where (RRR) describes themotion of a free particle and (rrr) describes the motion of a particle subject to thecentrally symmetric potential U(r)

2 +2mrh2

[EU(r)] = 0.

With the use of spherical polar coordinates, the equation becomes

1r2

r

(r2

r

)+

1r2

[1

sin

(sin

)+

1sin2

2 2

]+

2mrh2

[EU(r)] = 0.

The differential operator dependent on the angular variables coincides with theCasimir operator L2, so that

h2

2mr

[ 1

r2 r

(r2

r

)+

L2

h2r2]+U(r) = E.

In the motion in a central eld, the angular momentum is conserved: let us taketwo arbitrary values l, m for the angular momentum and its projection on the z axis.Given that the differential equation is separable, we look for a solution using theansatz

(rrr) = R(r)Yl,m( ,)

where Yl,m are the spherical harmonics, which are eigenfunctions of both Lz and L2.The above equation becomes

1r2

ddr

(r2

dRdr

) l(l+1)

r2R+

2mrh2

[EU(r)]R = 0.

Since the quantum number m is not appearing in this equation, the solutions will be2l+ 1 degenerate with respect to the angular momentum. The dependence on Yl,mhas been removed by multiplying by Y l,m and integrating over the angular part of thevolume. Let us focus now on the radial part of the wave function and let us performa further change of variables, by setting R(r) = (r)r , to get

d2dr2

+

[2mrh2

(EU(r)) l(l+1)r2

] = 0.

The domain of variation of r is now [0,+], and at the boundary of this region thewave function must vanish to guarantee that it can be normalized, thus leading to adiscrete spectrum. The equation we got after these manipulations looks like a onedimensional Schrodinger equation with potential

Ue f f (r) =U(r)+h2

2mr

l(l+1)r2

.

• 12 1 Summary of Quantum and Statistical Mechanics

For a xed l, the radial part is determined by the quantum number labelling theenergy, since in a one dimensional motion the eigenvalues are not degenerate. Theangular part with quantum numbers l, m, and the energy spectrum, En, determine theparticle motion without ambiguities. To label states with different angular momenta,we use the notation

l = 0(s) 1(p) 2(d) 3( f ) 4(g) . . .

Using the theorem of oscillations we see that the ground state is always an s wavesince the wave function cannot have zeros for the lowest level, while the Yl,m forl = 0 are always oscillating functions with positive and negative values.

To close this section on the theory of the angular momentum, we report its appli-cation to the case of hydrogen-like atoms , i.e. those atoms with one electron withcharge e (and mass me) and nuclear charge Ze. After the center of mass is sepa-rated out, the stationary Schrodinger equation describing the wave function of theelectron becomes

h2

2me

[ 1

r2 r

(r2

n,l,m r

)+

l(l+1)r2

n,l,m] Ze

2

rn,l,m = Enn,l,m

where we have used the wave function for the orbital motion

n,l,m(r, ,) = Rn,l(r)Yl,m( ,)

with n the principal quantum number giving the energy

En =Z2e2

2n2aa =

h2

and Rn,l(r) the radial part of the wave function. Some of the expressions of Rn,l(r)are here reported

R1,0(r) =(

Za

)3/22eZr/a

R2,0(r) =(

Z2a

)3/2(2 Zr

a

)eZr/2a

R2,1(r) =(

Z2a

)3/2 Zr3a

eZr/2a.

1.7 Solutions of the Three Dimensional Schrodinger Equation

Let us now discuss the three dimensional solutions of the Schrodinger equation infull generality. We start from(

h2

2m2U +E

) = 0

• 1.7 Solutions of the Three Dimensional Schrodinger Equation 13

and we will choose an appropriate coordinate system according to the symmetryof the problem. This choice is crucial to get a separable differential equation and asimple form for the potential. The standard example is the hydrogen atom where thechoice of spherical coordinates and a potential which is dependent only on the radiusnaturally lead to variable separation. Once this is done, the left over problem is tosolve a one dimensional differential equation. The coefcients of these equationshave pole singularities. There is only a nite number of coordinate systems leadingto separable equations and their singularities can be classied according to theirnumber and ty...