Ben Gurion University of the Negev
www.bgu.ac.il/atomchip, www.bgu.ac.il/nanocenter
Physics 3 for Electrical Engineering
Lecturers: Daniel Rohrlich, Ron Folman
Teaching Assistants: Daniel Ariad, Barukh Dolgin
Week 10. Quantum mechanics – Schrödinger’s equation for the
hydrogen atom • eigenvalues and eigenstates • atomic quantum
numbers • Stern-Gerlach and spin • Pauli matrices • spin-orbit
coupling
Sources: Feynman Lectures III, Chap. 19 Sects. 1-5;
Merzbacher (2nd edition) Chap. 9;
Merzbacher (3rd edition) Chap. 12;
1-2 ‫ פרקים‬,8 ‫ יחידה‬,‫פרקים בפיסיקה מודרנית‬
Tipler and Llewellyn, Chap. 7 Sects. 2-5.
Schrödinger’s equation for the hydrogen atom
The most common isotope of hydrogen contains just one
proton and one electron. Their potential energy is
V (r p  re )  
e2
4 0 r p  re
,
so Schrödinger’s equation depends on rp and re , i.e. on six
coordinates:
   2 2  2 2
e2
i
 
p 
e 
t  2m p
2me
4

r

r
0
p
e


 (rp , re , t ) .

But a clever change of variables makes the equation simpler:
Schrödinger’s equation for the hydrogen atom
Let r = re – rp and R 
me re  m p r p
me  m p
; then Schrödinger’s
equation becomes
2
2

 
2

e
2
2
i
 
R 
r 
 ,
t  2(me  m p )
2
40 r 
where  
me m p
me  m p
is called the reduced mass. For a proton
and an electron, the reduced mass is essentially the electron
mass, since mp ≈ 2000 me. But for a positronium atom (which
has a positron in the place of the proton), μ = me /2.
Schrödinger’s equation for the hydrogen atom
Let r = re – rp and R 
me re  m p r p
me  m p
; then Schrödinger’s
equation becomes
2
2

 
2

e
2
2
i
 
R 
r 
 ,
t  2(me  m p )
2
40 r 
where  
me m p
me  m p
is called the reduced mass. There are
solutions of the form Ψ = T(t)χ(R)ψ(r), where T(t) is the usual
time dependence and χ(R) is a solution to the free Schrödinger
equation in R (the center-of-mass coordinate), which has no
potential term.
Schrödinger’s equation for the hydrogen atom
The equation for ψ(r) is then
 2 2
e2 
Eψ( r)  
r 
ψ ,
4 0 r 
 2
with the Coulomb energy as a central potential V(r).
We have seen that, for a central potential, Schrödinger’s
equation reduces to
ˆ2 
  2   L
Eψ(r , ,  )  
  2  ψ  V (r )ψ .
 r
2 r
r   
2m r  
2
We replace m by μ and V(r) by 
e2
4 0 r
:
ˆ2 
2    2   L
e2
Eψ(r , ,  )  
r
  ψ
ψ ,



r   2 
4 0 r
2r 2  r 
and obtain the eigenvalues of L̂2; they are l (l  1) 2 . We
solved this equation by expressing ψ(r,θ,φ) as a product of two
functions:
ψ(r,θ,φ) = R(r)Ylm(θ,φ) ,
where
ˆ 2Y m  l (l  1) 2Y m ,
L
l
l
Lˆ zYlm  m Ylm .
Eigenvalues and eigenstates
Hence the equation for R(r) is

2    2  
e2
ER(r )  
r
 l (l  1) R(r ) 
R(r ) .



r 
4 0 r
2r 2  r 

The solutions Rnl(r) of this equation have the form
Rnl (r )  e r / na0 Fnl (r / a0 ) ,
with Fnl (r/a0) a polynomial and n ≥ l + 1. The constant a0 is
called the Bohr radius and equals
a0 
4 0  2
e 2
 0.529 Å  5.29  10 11 m .
Eigenvalues and eigenstates
The energies depend only on n:
En  
e 4
4 0 
2
2 2
2 n

13 .6 eV
n
2
,
where 1 eV = 1.602176 × 10-19 J.
Question: What is the normalization condition for ψnlm(r,θ,φ)?
Eigenvalues and eigenstates
The energies depend only on n:
En  
e 4
4 0 
2
2 2
2 n

13 .6 eV
n
2
,
where 1 eV = 1.602176 × 10-19 J.
Question: What is the normalization condition for ψnlm(r,θ,φ)?
Answer: 1 

2
ψ nlm (r, ,  ) r 2 sin  d d dr .
Eigenvalues and eigenstates
Question: What is the normalization condition for Rn0(r)?
Eigenvalues and eigenstates
Question: What is the normalization condition for Rn0(r)?
Answer: 1 

2
ψ n00 (r ,  ,  ) r 2 sin  d d dr
2


Rn 0 (r )


Rn0 (r ) r 2 dr .
r 2 (4 ) dr
4
2
Eigenvalues and eigenstates
Here are the lowest normalized solutions ψnlm = Rnl(r)Ylm(θ,φ)
of the hydrogen atom Schrödinger equation (see also here):
n
l m
ψnlm
Radial eigenfunctions
Rnl(r) and probability
distributions Pnl(r) for
the lowest eigenstates
of the hydrogen atom.
From here.
Pnl(r)
Rnl(r)
Question: How can
Pn0(r) vanish at r = 0
if Rn0(r) does not?
Radius (a0)
Radial eigenfunctions
Rnl(r) and probability
distributions Pnl(r) for
the lowest eigenstates
of the hydrogen atom.
From here.
Pnl(r)
Rnl(r)
Question: How can
Pn0(r) vanish at r = 0
if Rn0(r) does not?
Answer: On a sphere
of radius r, we have
Pn0(r) = r2 |Rn0(r)|2.
Radius (a0)
For a single electron bound by a nucleus containing Z protons,
the solutions of Schrödinger’s equation are almost unchanged;
the reduced mass μ is even closer to me, while the potential
term is
V (r p  re )  
Ze 2
4 0 r p  re
.
Thus by replacing e2 with Ze2 in the eigenstates and
eigenvalues we obtained for Z = 1, we obtain the Z > 1
eigenstates and eigenvalues.
For example, we obtain the energy eigenvalues
 ( Ze 2 ) 2
2 13.6 eV
E 
 Z
.
2
2
2
2
n
n
40  2 n
Exercise: Show that the energy of the ground state is half the
expectation value of the potential energy in the ground state.
Exercise: Show that the energy of the ground state is half the
expectation value of the potential energy in the ground state.
Solution: The ground state energy is E1  
 e4
4 0 
2
2
2
.
The expectation value of the potential energy is

e2
40 r
   ψ100 (r )
2
e2
40 r
4r dr
2
3
 1  e  2 r/a0 e 2
   
4r 2 dr
 40 r
 a0 
4
e2

e
r

e rdr  
2

40 a0
40 
.
Atomic quantum numbers
According to what we have seen so far, every eigenstate of the
hydrogen atom can be associated with three quantum numbers
n, l and m, where
n = 1, 2, 3, … “principle” quantum number
l = 0, …, n – 1;
m = –l, –l +1, …, l – 1, l .
The degeneracy of the energy eigenvalue En is therefore
1 + 3 + 5 + … + [2(n – 1) + 1] = n2 .
In atomic physics, the l quantum numbers have special names:
s for l = 0, p for l = 1, d for l = 2, f for l = 3, etc. Then 2s
means n = 2 and l = 0; 3p means n = 3 and l = 1; and so on.
Since the energy of an electron in a stationary hydrogen atom
can only be one of the En , where
En  
e 4
4 0 
2
2 2

13 .6 eV
2
,
n
2 n
a stationary atom can absorb and emit photons only if the
photon energy equals Ephoton = En – En’ . And since the energy
of a photon is related to its frequency ν by Ephoton = hν, the
frequencies of electromagnetic radiation emitted or absorbed
by a hydrogen atom must obey the rule
 1
e 4
1 
1 
15 1  1
 
  3.3 10 s  2 
 ,

3
2 3  2
2 
2 
4   0   n (n' ) 
 n ( n' ) 
for some n and n´. This formula, derived by Bohr 13 years
before Schrödinger, was soon verified via spectroscopy.
Energy levels for hydrogen, and
transitions among the levels
Energy levels for hydrogen, and
transitions among the levels
Corresponding spectral lines
Setup for emission spectroscopy:
Light from
source
The “magnetic” quantum number m:
When an atom is immersed in a uniform magnetic field, the
energies En split! Let B be the strength of the field and let point
up the z-axis. A state with quantum numbers n, l, m has energy
Enlm  
13 .6 eV
2
 μ BmB ,
n
where μB = eħ/2me is the Bohr magneton. How can we explain
this effect? An electron moving in a circular orbit of radius r, at
speed v, produces a current I = ev/2πr and a magnetic moment μz
= I(πr2) = evr/2 = eLz/2me. Since Lz = mħ, we have μz = μBm.
The corresponding extra potential term for the hydrogen atom is
e ˆ ˆ
ˆ
ˆ
VB  μ  B 
LB .
2me
The “magnetic” quantum number m:
But this is not the only magnetic effect. In a uniform magnetic
field, there is a torque on the atom (to make it anti-parallel to the
magnetic field). But in a non-uniform magnetic field, there is
also a force on the atom. Assume again that the field points up
the z-axis, so that VB = –μz B. Then if dB/dz ≠ 0, the force is
d
dB
Fz   VB   z
.
dz
dz
So, what happens if a beam of neutral atoms with –lħ ≤ Lz ≤ lħ
crosses a non-uniform magnetic field?
The “magnetic” quantum number m:
But this is not the only magnetic effect. In a uniform magnetic
field, there is a torque on the atom (to make it anti-parallel to the
magnetic field). But in a non-uniform magnetic field, there is
also a force on the atom. Assume again that the field points up
the z-axis, so that VB = –μz B. Then if dB/dz ≠ 0, the force is
d
dB
Fz   VB   z
.
dz
dz
So, what happens if a beam of neutral atoms with –lħ ≤ Lz ≤ lħ
crosses a non-uniform magnetic field?
We expect the beam to split into 2l+1 beams, one for each value
of μz. In some cases, the beam indeed splits into 2l+1 beams.
Stern-Gerlach and spin
But O. Stern and W. Gerlach saw a beam of silver atoms split
into two beams!
Stern-Gerlach and spin
But O. Stern and W. Gerlach saw a beam of silver atoms split
into two beams!
How can Lˆ z have an even number of eigenvalues?
G. Uhlenbeck and S. Goudsmit suggested that each electron has
its own intrinsic angular momentum – “spin” – with only two
eigenvalues.
But electron spin has odd features. For example, its magnitude
never changes, just its direction – and it has only two directions.
Stern-Gerlach and spin
Let’s try to understand spin better by reviewing the algebra of
ˆ 2 . Consider l = 1 and m = –1, 0, 1. The
Lˆ x , Lˆ y , Lˆ z , and L
matrix representation of Lˆ z in a basis of eigenstates of Lˆ z is
1 0 0 


ˆ
Lz   0 0 0  ,
 0 0  1


since the eigenvalues are 0 and ±ħ.
Stern-Gerlach and spin
What is Lˆ   1 ? We know it must equal a 0 ,but what is a?
We have 0 a * a 0   1 Lˆ  Lˆ   1
   
2
2

ˆ
ˆ
  1  Lx  L y  Lˆ z   1


 
2

2
ˆ
ˆ
  1 L  Lz  Lˆ z   1  2 2 .


Hence Lˆ   1  2  0 , up to an overall phase. Similarly, we
0

can show that Lˆ  0  2  1 , hence Lˆ   0

0

2
0
0
0 
2 .

0 

Stern-Gerlach and spin
 0

ˆ
Similarly, L   2
 0

0
0
2
0

0 . Now, since Lˆ  Lˆ  2Lˆ x,
0 
 0 1 0

 
ˆ
. Since Lˆ   Lˆ   2iLˆ y ,
we can write Lx   1 0 1 
 0 1 0 2


0  i 0 

 
ˆ
.
we can write L y   i 0  i 
2
0 i

0

Pauli matrices
It is straightforward to check that these matrix representations
have the correct commutation relations:
 Lˆ y , Lˆ z   iLˆ x ,  Lˆ z , Lˆ x   iLˆ y .
 Lˆ x , Lˆ y   iLˆ z ,
But Pauli discovered 2 × 2 matrices with the same commutation
relations:
0 1 
0  i 
1 0  
ˆ
ˆ
ˆ
 , S y  

 .
S x  
, S z  
1 0 2
i 0 2
 0  1 2
(The “Pauli matrices” are these matrices without the ħ/2 factors.)
These are the operators for the components of electron spin!
Pauli matrices
1 0  
ˆ

We can write the eigenstates of S z  
 0  1 2
1
as  or   for Sz = ħ/2
0
0
and as  or   for Sz = –ħ/2.
1
2
1
0




Since Sˆ x2  Sˆ y2  Sˆ z2  3
 s(s  1) 2 for s = ½,
0 1 4
we refer to electron spin as “spin-½”.
Stern-Gerlach and spin
More odd features of electron spin:
The eigenvalues of Sˆ x , Sˆ y and Sˆ z are ±ħ/2.
We can write an eigenstate of L̂ z with eigenvalue mħ as
1 im
e
but an analogous eigenstate of Ŝ z, namely
2
1
2
e i / 2 , would not be single-valued. Yet experiments show
that these electron spin eigenstates are not invariant under
rotation by 2π, but they are invariant under rotation by 4π!
This is reminiscent of a trick with a twisted ribbon….
Stern-Gerlach and spin
This is reminiscent of a trick with a twisted ribbon…one twist
cannot be undone, but two twists are equivalent to no twist.
Stern-Gerlach and spin
One more odd feature of electron spin:
For orbital angular momentum, we found that μz = eLz/2me.
For spin angular momentum, experiment shows that μz = eSz/me.
That is, electronic spin produces an anomalous “double”
magnetic moment.
Therefore, the total magnetic moment of an electron with orbital
angular momentum mħ and spin angular momentum ±ħ/2 is
e
m    .
z 
2me
Atomic quantum numbers (again)
We associated every eigenstate of the hydrogen atom with
three quantum numbers n, l and m. But now we have to
introduce a fourth quantum number, the spin: ms = ±½ .
The degeneracy of the energy eigenvalue En is therefore not
n2 but 2n2, since there are two spin states for every set of
quantum numbers n, l and m.
The nucleus, too, has spin angular momentum. But its
magnetic moment is relatively tiny because the mass of a
proton is about 2000 times the electron mass. In this course we
neglect the spin and magnetic moment of the nucleus.
Exercise: Show that the superposition of wave functions
1
5
ψ
1
2,1,1,
2

1
5
ψ
1
2,1,1,
2

3
5
ψ
2,1,1,
1
2
is normalized if each wave function ψ n,l ,m,m is, and calculate
s
ˆ
ˆ
Lz , S z and ΔLz.
Exercise: Show that the superposition of wave functions
1
ψ
5
1
2,1,1,
2

1
ψ
5
1
2,1,1,
2
3

5
ψ
2,1,1,
1
2
is normalized if each wave function ψ n,l ,m,m is, and calculate
s
ˆ
ˆ
Lz , S z and ΔLz.
Solution: Since the components have different eigenvalues,
they are orthonormal, and the normalization is obtained from
the absolute value of the squares of the coefficients:
 ψn,l,m,m
s
2
2
2
2
 1   1   3
 1 .
  
  
d x  


 5  5  5
3
Exercise: Show that the superposition of wave functions
1
5
ψ
1
2,1,1,
2

1
5
ψ
1
2,1,1,
2

3
5
ψ
2,1,1,
1
2
is normalized if each wave function ψ n,l ,m,m is, and calculate
s
ˆ
ˆ
Lz , S z and ΔLz.
Solution:
2
 1 
 1 
ˆ
  ()

Lz  ()
 5
 5
2
2
2
2
 3
3


 ( )
 
 5
5


2
  1 
  1 
  3 

ˆ
  ( )
  ( )
S z  ( )

2  5
2  5
2  5 
10
.
Exercise: Show that the superposition of wave functions
1
5
ψ
1
2,1,1,
2

1
5
ψ
1
2,1,1,
2

3
ψ
5
2,1,1,
1
2
is normalized if each wave function ψ n,l ,m,m is, and calculate
s
ˆ
ˆ
Lz , S z and ΔLz.
Solution:
2
2
2  1 
2  1 
ˆ
Lz  ( )
  ( )

 5
 5
Lz 
Lˆ2z  Lˆ z
2
2
2
2  3 
 ( )
 2
 5


2
4
3 
 2      .
5
5 
Exercise: What happens in a Stern-Gerlach experiment, if each
electron in an incident beam of hydrogen atoms has l = 1?
Exercise: What happens in a Stern-Gerlach experiment, if each
electron in an incident beam of hydrogen atoms has l = 1?
Solution: The magnetic moment of the electron depends on Lz
and Sz according to
e
e
L z  2 S z  
m  2ms  .
z 
2me
2me
Since m = –1, 0, 1 and, independently, ms = ±½, we get five
possible values of m +2ms : 2, 1, 0, –1, –2. We therefore expect
to see 5 separate spots on the screen.
Spin-orbit coupling
We discussed atomic magnetic moments in a magnetic field
that is uniform or non-uniform. But even without any external
magnetic field, an electron feels an effective field. Why?
Spin-orbit coupling
We discussed atomic magnetic moments in a magnetic field
that is uniform or non-uniform. But even without any external
magnetic field, an electron feels an effective field. Why?
The electron moves relative to the nucleus. Transforming the
Coulomb field to the electron’s rest frame yields a magnetic
field B' = –v × E/c2. Since E is radial, –v × E/c2 is –dV(r)/dr
times r × p /emec2r = L/emec2r. Since the electron’s magnetic
moment e Ŝ /me interacts with B', the spin-orbit interaction
ˆ . It enters the Hamiltonian as
contains also Sˆ  L
Hˆ SO 
1
dV (r ) ˆ ˆ
SL ,
2 2
2me c r dr
where V(r) is the Coulomb potential.
Spin-orbit coupling
ˆ , we must know how to add
To compute the eigenvalues of Sˆ  L
ˆ  Sˆ , we find that
angular momenta. Defining Jˆ  L
 Jˆ x , Jˆ y    Lˆx  Sˆx , Lˆ y  Sˆ y   iLˆz  iSˆz  iJˆ z
,
and so on, i.e. the components of Ĵ follow exactly the same
algebra as the components of L̂ and Ŝ . We immediately infer
that the eigenvalues of Ĵ 2 are j(j+1)ħ2 and that the eigenvalues
of Ĵ z are –jħ, (–j+1)ħ,…, (j–1)ħ, jħ.
Spin-orbit coupling
ˆ , we must know how to add
To compute the eigenvalues of Sˆ  L
ˆ  Sˆ , we find that
angular momenta. Defining Jˆ  L
 Jˆ x , Jˆ y    Lˆx  Sˆx , Lˆ y  Sˆ y   iLˆz  iSˆz  iJˆ z
,
and so on, i.e. the components of Ĵ follow exactly the same
algebra as the components of L̂ and Ŝ . We immediately infer
that the eigenvalues of Ĵ 2 are j(j+1)ħ2 and that the eigenvalues
of Ĵ z are –jħ, (–j+1)ħ,…, (j–1)ħ, jħ.
ˆ 2  Sˆ 2  2Sˆ  L
ˆ we derive
Now from Jˆ 2  L


2
1

ˆ  Jˆ 2  L
ˆ 2  Sˆ 2   j ( j  1)  l (l  1)  s(s  1)
Sˆ  L
2
2
.
Exercise: The spin-orbit coupling splits the degeneracy between
the hydrogen states ψ2,1,1, –½ and ψ2,1,1,½ by ΔE = 4.5 × 10-5 eV.
Estimate the magnetic field B' felt by the electron.
Exercise: The spin-orbit coupling splits the degeneracy between
the hydrogen states ψ2,1,1, –½ and ψ2,1,1,½ by ΔE = 4.5 × 10-5 eV.
Estimate the magnetic field B' felt by the electron.
Solution: The energy splitting is due to the interaction of the
electron’s magnetic moment with the effective magnetic field B'.
In the rest frame of the electron only the spin magnetic moment
contributes: μzB' = (eSz/me)B' = ±eB'ħ/2me, hence ΔE = eB'ħ/me
and
B' = meΔE/eħ
= ΔE/2μB
= (4.5 × 10-5 eV) / 2 × (5.79 × 10-5 eV/T)
= 0.39 T .