Quantum Mechanics 1 Lecture 7: Central potentials

Quantum Mechanics 1

Lecture 7: Central potentials

Massimiliano Grazzini

University of Zurich

Outline

- Bound states in 3 dimensions

Coulomb potential and the hydrogen atom Motion in a central potential

Spherical potential well

Motion in a central potential

In Quantum Mechanics the problem of two interacting particles (two-body

problem) can be reduced to the one-particle problem in a completely analogous way as we do it in classical mechanics

Let us consider the Hamiltonian

H = − ℏ 2m

∇

− ℏ 2m

∇

+ V(r)

describing two particles of masses and interacting through a potential

which depends on their distance ^m

^m

^V(r)

The operators and are the Laplace operators of the two particles _∇

_∇

We do the change of variables

r = x

− x

R = m

x

+ m

x

m

+ m

Which implies the following relations for the gradients

m₁

m₂

r

∇

= − ∇ + m

m

+ m

∇

∇

= ∇ + m

m

+ m

∇

Where and are the gradient vectors with respect to and , respectively _{∇ ∇}

_{r R} Plugging the above expressions in the Hamiltonian we obtain

H = − ℏ 2m

∇

+ V(r) − ℏ

2(m

+ m

) ∇

where is the reduced mass _m

m = ( 1

m₁ + 1m₂)

−1 = m₁m₂ m₁ + m₂

The Hamiltonian is thus decomposed into the sum of two independent terms and one can look for a solution of the problem in the factorised form

ψ (x

, x

) = Φ(R)ψ (r)

where describes the motion of the centre-of mass of the system as a free particle with mass while describes the relative motion of the

particles as motion of a particle with mass in a field with spherical symmetry

Φ(R) m

+ m

ψ (r) V(r) m

H

ψ = E

ψ H

Φ = E

Φ

H

H

The Schrödinger equation for _{ψ (r)} reads

− ℏ

2m ∇

ψ = ( E − V(r) ) ψ

The laplacian in spherical coordinates reads (recall MMP !)

∇

= 1 r

∂

∂r ( r

∂

∂r ) + 1 r

[ 1

sin θ ∂

∂θ ( sin θ ∂

∂θ ) + 1 sin

θ ∂

∂ϕ

]

But the square of the angular momentum operator reads (recall Lecture 6) _L

L

= − ℏ

[

1

sin θ ∂

∂θ ( sin θ ∂

∂θ ) + 1 sin

θ ∂

∂ϕ

]

Thus the Schrödinger equation becomes

[ − ℏ 2m

1 r

∂

∂r ( r

∂

∂r ) + L

2mr

+ V(r)

] ψ (r) = E ψ (r)

and we see that the angular dependence is fully controlled by _L

Since we have already studied the eigenfunctions of the angular momentum we can introduce the ansatz

ψ (r) = R(r)Y

(θ, ϕ)

which leads to

[− ℏ² 2m 1

r² ∂

∂r (r² ∂

∂r) + ℏ²l(l + 1)

2mr² + V(r)

]R(r) = ER(r)

What we are doing here is a general consequence of the fact that the Hamiltonian is invariant under rotations, and thus

and

[H, L] = 0 [H, L

] = 0

which imply that the operators _{H, L}

and can be simultaneously diagonalised _L

(1)

Now Eq. contains only one variable but it is not yet in the form of a one- dimensional Schrödinger equation ^{(1) r}

It is convenient to rewrite it using _{R(r) = χ (r)/ r}

The derivatives with respect to take the form _r

1 r

∂

∂r ( r

∂R

∂r ) = 1 r

∂

∂r ( r

1 r ∂χ

∂r − r

χ

r

) = 1 r

∂

∂r ( r ∂χ

∂r − χ ) = 1 r ∂

χ

∂r

and therefore the equation becomes

[ − ℏ 2m

d

dr

+ ℏ

l(l + 1)

2mr

+ V(r)

] χ(r) = E χ(r)

This is a one-dimensional Schrödinger equation with the effective potential

V

(r) = V(r) + ℏ

l(l + 1) 2mr

In complete analogy with classical mechanics where the central potential is also modified by a repulsive centrifugal term

We now study in more detail what can be said about _{χ (r)}

(2)

From the condition of normalizability we have

∫ d

x | ψ (x) |

= ∫

∞

0

drr

1

r

| χ (r) |

< ∞

Which means that for _{r → ∞ χ (r)} must vanish faster than _{1/ r} Moreover, since _{R(r) = χ (r)/ r} and (recall MMP)

∇

1

r = − 4πδ

(x)

unless _{V(r) ∼ δ}

_(x) we must require _{χ(0) = 0} to fulfil the Schrödinger equation We now discuss what can be said about the existence of bound states: we will exploit what we have learned in the study of one-dimensional problems and apply it to the function _χ(r)

The condition means that the equivalent one-dimensional problem has a barrier at _{x = 0} ^{χ (0) = 0}

We start from the case _{l = 0} , which means that _V

_{(r) = V(r)}

V

(x) = V(x) x > 0 V

(x) = ∞ x ≤ 0

We also define the symmetric extension _V ^̂

_{(x) = V}

_{( | x | )}

From the general discussion of Lecture 4 we know that a symmetric potential like

leads to at least one bound state with even

wave function, which cannot be acceptable in our case since

̂ V

(x)

χ (0) = 0

The ground state in our case coincides with the first excited state, that is the lowest odd bound state of _V ^̂

_(x)

Thus, _V(r) must exceed a minimum strength so as we can have bound states The centrifugal term is increasingly repulsive as increases, so if there are no bound states for _{l = 0} , there are certainly no bound states for ^l _{l > 0}

Moreover, since the bound states in the one-dimensional motion are non

degenerate, the value of the energy completely determines the solution of Eq. ₍₂₎

We now discuss the qualitative behaviour of _{χ (r)}

limit

r → 0

If in this limit the centrifugal term in the Schrödinger equation dominates with respect to (this is the case for the Coulomb potential

or for the spherical potential well ^{V(r) − E} ) ^{V(r) = − e}

02

Z /(4πϵ

)r V(r) = − V

θ(r

− r)

In this case the radial equation becomes

[ − ℏ 2m

d

dr

+ ℏ

l(l + 1)

2mr

] χ(r) = 0

whose general solution is

χ (r) = Ar

+ Br

Since we must have _{χ(0) = 0} the second term is not acceptable

Recall MMP: in the case of the Coulomb potential or the square-well the

point is a regular singular point for the differential equation (indeed the

coefficient of ^{r = 0} _{χ (r)} has a double pole at _{r = 0} )

limit

r → ∞

If in this limit the potential term in the radial equation can be neglected and the equation becomes

− ℏ 2m

d

dr

χ (r) = Eχ (r)

For bound states ( _{E < 0} ) we obtain the solutions _{exp{± κr}} with

κ = 1 ℏ 2m(− E )

And to obtain meaningful solutions we have to choose the exponentially decreasing function

χ(r) = C exp{− κr}

It is useful to define a dimensionless variable _{ρ = κr} and rewrite Eq. as ₍₂₎

[ d

dρ

− l(l + 1)

ρ

− V( ρ / κ)

| E | − 1

] χ (ρ) = 0 (3)

Spherical potential well

V(r)

r r

E

We focus on the case and we want to find out when we have bound states ^{l = 0}

− V

The solutions for the radial equation are

χ(r) = Ae

+ Be

χ(r) = Ce

r ≤ r

r > r

Since _{χ (0) = 0} we must have _{A = − B}

The continuity conditions at _{r = r}

give

χ (r

) = χ (r

) ⟹ 2iA sin(qr

) = Ce

− cot(qr

) = κ / q

This condition is exactly the same we found in Lecture 4 (page 21) for the odd

κ = 1

ℏ 2m(− E ) q = 1

ℏ 2m(V

+ E )

Spherical potential well

Following the same step we define _{ζ = 2mV}

_r

_{/ ℏ} and rewrite the equation as

− cot(qr

) = κ

q = ζ

− (qr

)

qr

0 1 2 3 4 5 6

2 4 6 8 10

π /2 π 3π /2 2π

−cot(z)

Example: _{ζ = 1} vs _{ζ = 5}

ζ² − z² z

As for odd solutions in the one-dimensional well we have bound states only if

ζ = 2mV

r

ℏ > π 2

or

V

r

> ℏ

π

8m

Coulomb potential

We now consider the case of an electron in the field of an atomic nucleus In this case the potential reads

V(r) = − e

Z (4πϵ

)r

[ d

dρ

− l(l + 1)

ρ

− V( ρ / κ)

| E | − 1

] χ (ρ) = 0

Our starting point is the radial equation in the variable _{ρ = κr}

(3)

Defining

ρ

= e

Z κ

(4πϵ

) | E | = 2m

| E | Ze

(4πϵ

)ℏ

The equation becomes

[ d

dρ

− l(l + 1)

ρ

+ ρ

ρ − 1

] χ( ρ) = 0 (4)

We have seen that the expected behavior of the solution is

χ( ρ) ∼ ρ

ρ → 0 χ( ρ) ∼ e

ρ → ∞

therefore we look for a solution in the form _{χ( ρ) = ρ}

_e

_{w( ρ)}

If we replace in Eq. we obtain the following equation for _{(4) w( ρ)}

ρw′ ′ ( ρ) + 2(l + 1 − ρ)w′ (ρ) + (ρ

− 2l − 2)w( ρ) = 0

We look for a solution as a series expansion _{w( ρ) = ∑}

k=0

a

ρ

and from the differential equation we obtain

∞

∑

a

[ k(k − 1)ρ

+ 2(l + 1)kρ

− 2k ρ

+ (ρ

− 2l − 2)ρ

] = 0

[ (k + 1)k + 2(l + 1)(k + 1) ] a + [ − 2k + (ρ − 2(l + 1)) ] a = 0

which gives

This is a recursive equation which allows us to find the coefficient _a

from _a

In the large limit we have _{k a}

a

∼ 2 k

Which is the same that can be obtained for the exponential function

e

= ∑

k=0

1

k ! (2ρ)

a

a

= 2

/(k + 1)!

2

/ k ! = 2 k + 1 ∼ 2 k

Such behaviour for _{w( ρ)} is not acceptable, so the series must terminate for some

k = N

[ (k + 1)k + 2(l + 1)(k + 1) ] a

+ [ − 2k + (ρ

− 2(l + 1)) ] a

= 0

Looking at the recursion relation

We conclude that

ρ

= 2(N + l + 1), N = 0,1,2...

Where is called radial quantum number _N

From the definition of _ρ

ρ

= 2m

| E | Ze

(4πϵ

)ℏ = 2(N + l + 1)

we finally obtain

E = − Z

e

m

(4πϵ

)

ℏ

1 2n

in agreement with what we saw in Lecture 1

n ≡ N + l + 1 = 0,1,2....

We see that the energy depends only on the combination , which is called principal quantum number ^{n = N + l + 1}

For a given fixed the angular momentum quantum number can take the

values ^{n l}

l = 0,1...., n − 1

Remembering that for each there are possible value of we conclude that the degeneracy of each level is ^{l 2l + 1 m}

n−1

∑ (2l + 1) = 2 n(n − 1)

2 + n = n

E = − Z

e

m

(4πϵ

)

ℏ

1

2n

= − mc

(Zα)

2n

By introducing the fine structure constant

α = e

4πϵ

ℏc ∼ 1 137

the expression for the energy levels can be rewritten as

The frequencies of the emitted photons when the atom passes from a state of energy to a state of energy is _E

_E

ν

= E

− E

h = mc

(Z α)

2h (

1

n

− 1 n

)

The frequencies depend on the reduced mass of the nucleus-electron system _m Such property permitted in 1931 the

discovery of deuterium (Harold Urey), which is an isotope of hydrogen with a

nucleus composed by a proton and a neutron

m

m

= 1 + m

/ M

1 + m

/ M

ratio of reduced masses

masses of

hydrogen and deuterium nuclei

We recall that the radial function was defined to be

We now want to find the explicit solutions of the radial equation

χ ( ρ) = ρ

e

w( ρ)

R(r) = χ(r)/ r ρ = κr

ρ w′ ′ ( ρ) + 2(l + 1 − ρ)w′ ( ρ) + (ρ

− 2l − 2)w(ρ) = 0

and that _{w( ρ)} fulfils the differential equation

which, multiplied by _1/2 becomes

(2ρ) d

w

d(2ρ)

+ ( (2l + 1) + 1 − 2ρ ) dw

d(2ρ) + ( (n + l ) − (2l + 1) ) w( ρ) = 0

The solutions of this differential equation are the associated Laguerre polynomials

L

(2ρ)

Let us recall how the Laguerre polynomials arise

(5)

Associated Laguerre polynomials

Recall MMP: The Laguerre polynomials are defined as the mutually orthogonal polynomials with respect to the scalar product

⟨ f | g⟩ = ∫

∞

0

f (x)g(x)e

dx

L₀(x) = 1 L₁(x) = 1 − x

L₂(x) = x² − 4x + 2 . . . .

They can be constructed as a basis of through the Gram Schmidt

procedure* ^L

2

[0, + ∞]

They can be obtained through the formula _L

_{(x) = e}

_∂

_(x

_e

₎ They fulfil the differential equation

xL′

′ (x) + (1 − x)L′

(x) + pL

(x) = 0

This equation is similar to Eq. but not exactly the same ₍₅₎

*Note that with this definition the Laguerre polynomials are not normalised

The associated Laguerre polynomials are defined from the _L

_(x) as

L

(x) = d

dx

L

(x)

Differentiating the Laguerre differential equation times one can show that the associated Laguerre polynomials fulfill the equation ^k

[ x d

dx

+ (k + 1 − x) d

dx + ( p − k)

] L

(x) = 0

Which coincides with our Eq. with _{(5) k = 2l + 1} and _{p = n + l}

The explicit expression of the associated Laguerre polynomials is

L

(x) = ∑

j=0

(− 1)

( p !)

j !( j + k)!( p − j − k)! x

and they are normalised as

∫

∞

dxx

e

[ L

(x) ]

= (2 p − k + 1)( p !)

( p − k)!

Coulomb potential: wrap up

We can now collect the results we have obtained: the stationary states of the Coulomb potential are given by

ψ

(r, θ, ϕ, t) = e

R

(r)Y

(θ, ϕ)

where _Y

_{(θ, ϕ)} are the spherical harmonics studied in MMP and in Lecture 6 and

R

(r) = χ(r)

r = − [

(n − l − 1)!(2κ)

2n((n + l )!)

]

1/2

(2κr)

e

L

(2κr)

with

κ = 1

ℏ 2m(− E ) E = − Z

e

m

(4πϵ

)

ℏ

1 2n

By using the definition (recall Lecture 1) of the Bohr radius we can express as

a

= 4πϵ

ℏ

e

m

κ κ = 1 2m(− E ) = Z

The parameter controls the exponential suppression of the wave function at large

κ = Z na

r

We can thus say that the typical radial extension of the wave function increases with (as we move to more excited states) and decreases with (as the the

strength of the Coulomb attraction increases) ^{n Z} We report the lowest radial wave functions

R

(r) = 2 ( Z a

)

3/2

e

n = 1 l = 0

n = 2 l = 0

B)

3/2

(1 − Zr

2a_B) e^−Zr/2aB

l = 1 R

(r) = 1

3 ( Z 2a

)

3/2

Zr

a

e

s-orbital

p-orbital s-orbital

The states with _{l = 0} are finite in the origin, while those with _{l > 0} vanish as _r

We have seen that the energy levels for the Coulomb potential exhibit a multiple degeneracy in the quantum numbers and _{l m}

We now would like to further investigate this property

The degeneracy in is a general feature of central potentials: indeed since the Hamiltonian commutes with and it commutes also with ^m _L

_L

_L

Therefore if is an eigenfunction of with eigenvalue , that is _ψ

_{H E}

Hψ

= E ψ

we have

HL₊ψ_lm = L₊Hψ_lm = EL₊ψ_lm

Which means that _L

_ψ

is eigenvector of with the same eigenvalue _{H E}

The degeneracy of the energy eigenvalues with respect to is a specific property

of the Coulomb potential ^l

Indeed the Lenz vector _{A = 1}

2m (p × L − L × p) − e

(4πϵ

)r x

Is an additional conserved quantity: classically this corresponds to the

fact that the major axis of the elliptic orbit is fixed