Physics 342 Lecture 24

The Hydrogen Atom

Lecture 24

Physics 342
Quantum Mechanics I

Monday, March 29th, 2010

We now begin our discussion of the Hydrogen atom. Operationally, this

is just another choice for spherically symmetric potential (i.e. Coulomb).
Morally, of course, this is one the great triumphs of our time (technically,
the time two before ours). We already know the angular solutions, the usual
Y`m (θ, φ), so all we need to do is establish the radial portion of the wave-
function, and put it all together. Notice that we are following Professor
Griffiths’ treatment here, and he uses a different initial dimensionless length
than you did for your homework. This is no problem, in the end, the spec-
trum has to be the same no matter which choice one makes. You will see a
slight difference in the recursion relation, and since the recursion relation in
this case is more directly related to the associated Laguerre definition, it is
somewhat easier to get the actual radial wavefunctions here.

24.1 Radial Wavefunction

The potential, in this case, represents the electrostatic field set up by the
nucleus of the Hydrogen atom, as felt by the electron:

e2
U (r) = − . (24.1)
4 π 0 r
This goes into the usual (with u(r) = r R(r) as before)

~2 d2 u ~2 ` (` + 1)
 
− + U (r) + − E u=0 (24.2)
2 m dr2 2m r2

where we are associating m with the mass of the electron. We just made a
pretty dramatic approximation. We know that the two-particle problem can

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24.1. RADIAL WAVEFUNCTION Lecture 24

be reduced to a stationary center, provided we use the reduced mass of the

system. On the one had, that is fine – but on the other: What do we mean
by a two-particle problem in quantum mechanics? For now, just imagine
the nucleus doesn’t have much “kinetic” energy, so that it remains pretty
much fixed (what about the energy associated with having it around at all?
Its relativistic “rest energy” is still there, but we are not doing relativistic
quantum mechanics yet).
If we write the above out, we have:

~2 d2 u e2 ~2 ` (` + 1)
 
− + − + − E u = 0. (24.3)
2 m dr2 4 π 0 r 2 m r2
√
As with the infinite square well, it makes sense to let κ = −2~m E (negative
inside the square root, now – bound states will have E < 0 and we want
to make κ real). We want to define a new “coordinate” ρ ≡ κ r. The
advantage is to render the coordinate variable itself unitless. Whenever we
want to consider limiting cases of an equation or more generally, a physical
setting, we need a point of comparison.
What does it mean to be “far away” from a distribution of charge, for
example? That clearly depends on how large the distribution itself is. By
re-parametrizing using a fundamental length in the problem, we have allowed
for easier classification of limits. For example, on the E&M side, suppose we
have a dipole moment with a certain length d. Then “far away” means that
a field point at a distance r from the origin is large compared to d: r  d.
Now suppose we wrote everything in our problem in terms of the new length
r̃ ≡ r/d. We have eliminated the explicit comparison with d and can refer
to “small r̃” unambiguously as r̃ ∼ 0, making for easier Taylor expansion,
etc.
The point is, κ has units of 1/length and involves the fundamental (and as
yet unknown) energy scale, it is a natural choice for constructing ρ = κ r, a
d d
unitless quantity. In the above, we just replace r −→ ρ/κ, and dr −→ κ dρ .
2m
Performing this simple change of variables, multiplying by ~2 in the process,
we have
d2 u(ρ) m e2
 
` (` + 1)
− + − + + 1 u(ρ) = 0. (24.4)
dρ2 2 π 0 ~2 κ ρ ρ
2
We have another scale defined by ρ0 ≡ 2 πm0e~2 κ (there are, evidently, two
energy scales of interest to us here, hence two lengths – we could have written

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24.1. RADIAL WAVEFUNCTION Lecture 24

ρ in terms of ρ = (ρ0 κ) r), and with this, we can write the final form:
d2 u
 
ρ0 ` (` + 1)
= 1− + u. (24.5)
dρ2 ρ ρ2

As for limiting cases, we can take ρ −→ ∞, which gives us growing and

decaying exponentials as solutions:
d2 u
= u −→ u(ρ) = A e−ρ (24.6)
dρ2
where we have thrown out the growing exponential, that will not be nor-
malizable.
On the other hand, when the barrier-term dominates, for small ρ, we have
(using ū to distinguish from the actual solution)
d2 ū ` (` + 1)
2
= ū, (24.7)
dρ ρ2
and we can solve this by consider a generic polynomial (always a good ansatz
for ODE’s of the above flavor): ū(ρ) = a ρp , then

a p (p − 1) ρp−2 = a (` (` + 1)) ρ−2 a ρp (24.8)

and then we have a solution for p (p − 1) = ` (` + 1), or p = −`, p = ` + 1.

The general solution is a linear combination:

ū(ρ) = a ρ−` + b ρ`+1 (24.9)

and we set a = 0, for ρ near zero, this will blow up.

Finally, we will use these two regimes to factor the full solution – take

u(ρ) = ρ`+1 e−ρ v(ρ), (24.10)

this is naturally dominated by the polynomial near ρ ∼ 0, and the exponen-

tial will help with integration at infinity. If we input this into our differential
equation, we get
d2 v dv
ρ 2
+ 2 (` + 1 − ρ) + (ρ0 − 2 (` + 1)) v = 0. (24.11)
dρ dρ
Let x ≡ 2 ρ, then in terms of x, the above is
d2 v
 
dv 1
x 2 + (2 (` + 1) − x) + ρ0 − (` + 1) v = 0. (24.12)
dx dx 2

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24.2. ASSOCIATED LAGUERRE POLYNOMIALS Lecture 24

Now, the differential equation:

d2 k dLkn (x)
x L (x) + (k + 1 − x) + n Lkn (x) = 0 (24.13)
dx2 n dx
has solutions Lkn (x), the “associated Laguerre polynomials”, for integer n.
This is almost the above, if we set k + 1 = 2(` + 1) and n = ( 21 ρ0 − (` + 1)) –
and we assume that n is an integer. In that case, the solution to our problem
is just:
v(x) = L21 `+1
ρ −(`+1)
(x), (24.14)
2 0

1
This pre-supposes that ρ0 ≡ n̄ is an integer, but we can return to that
2
later on. For now, this is the source of the quantization of energy, since we
have:
m e2 m e2 m e4
2 n̄ = ρ0 = = √ √ −→ E = − 2 ,
2 π 0 ~2 κ 2 2 0 ~ −E m π 32 0 ~2 π 2 n2
(24.15)
or in more standard form, labelled using n the “principal quantum number”:
 2 2 !
m e 1 E1
En̄ = − 2 2
≡ 2. (24.16)
2~ 4 π 0 n̄ n̄

This is the energy spectrum of Hydrogen – we shall return to it in a moment.

24.2 Associated Laguerre Polynomials

The associated Laguerre polynomials are defined as the solution to the above
differential equation (24.13). Most special functions arise as solutions to
“difficult” ODEs, meaning ones not solvable by exponentials or polynomials.
The solutions usually proceed by series expansion (Frobenius’ method), and
involve points at which we remove certain elements of the solution with
behaviour we do not want to allow. The associated Laguerre polynomials
have Rodrigues formula1
ex x−k dn  −x n+k 
Lkn (x) = e x (24.17)
n! dxn
and can be related to the Laguerre polynomials via
dk
Lkn (x) = (−1)k Ln+k (x) (24.18)
dxk
1
For a further compilation of properties, see Arfken and Weber, p. 832.

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24.2. ASSOCIATED LAGUERRE POLYNOMIALS Lecture 24

Because it is the “integerization” of ρ0 that quantizes the energy in the

Hydrogen atom, it is worthwhile to generate the series solution and see
how this appears effectively as a boundary condition (vanishing of the radial
wavefunction at infinity). We’ll return to the direct ODE,

d2 v dv
ρ 2
+ 2 (` + 1 − ρ) + (ρ0 − 2 (` + 1)) v = 0, (24.19)
dρ dρ
and make the series ansatz:
∞
X
v(ρ) = cj ρj
j=0
X∞
v 0 (ρ) = cj j ρj−1 (24.20)
j=0
X∞
v 00 (ρ) = cj j (j − 1) ρj−2 .
j=0

The relevant terms from the ODE are:

∞
X ∞
X
00 j−1
ρ v (ρ) = cj j (j − 1) ρ = ck+1 k (k + 1) ρk
j=0 k=−1
X∞
= ck+1 k (k + 1) ρk
k=0
∞
X ∞
X
0 j−1
2 (` + 1) v (ρ) = cj j 2 (` + 1) ρ = ck+1 (k + 1) 2 (` + 1) ρk
j=0 k=−1
∞
X
= ck+1 (k + 1) 2 (` + 1) ρk
k=0
X∞
−2 ρ v 0 (ρ) = ck k (−2) ρk
k=0
X∞
(ρ0 − 2 (` + 1)) v(ρ) = ck (ρ0 − 2 (` + 1)) ρk
k=0
(24.21)
where we have set k = j − 1 in the first two expressions, and noted that, for
each, the k = −1 term vanishes anyway.

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24.2. ASSOCIATED LAGUERRE POLYNOMIALS Lecture 24

Now we just read down the list to get the recursion relation:

ck+1 (k (k + 1) + 2 (k + 1) (` + 1)) − ck (2 k − ρ0 + 2 (` + 1)) = 0 (24.22)

holds for all ρk . Solving for ck+1 :

2 (k + ` + 1) − ρ0
ck+1 = ck . (24.23)
(k + 1) (k + 2 (` + 1))

This can clearly be used to find all the coefficients given just the first one,
c0 . But if we think about the large-k limit, large values of ρ dominate the
expansion (ρ100  ρ2 for ρ > 1, for example), then we have
2
ck+1 ∼ ck . (24.24)
k+1
Now, if this had been the recursion relation all along (meaning, for all or-
ders), we would have had:
∞ ∞
2k X X 2k
ck = c0 −→ ck ρk = c0 ρk = c0 e2 ρ , (24.25)
k! k!
k=0 k=0

where we recognize the Taylor expansion for the growing exponential. This is
a manifestation of a solution that lurks in (24.19) in the large-ρ limit, where
it is effectively just v 00 (ρ)−2 v 0 (ρ) = 0, meaning that v(ρ) = α e2 ρ +β. What
we want to do is kill the growing exponential term, which will overwhelm
the e−ρ we have already factored out of the radial u(r) solution. To accom-
plish this, we must truncate the series, eliminating, “by hand” the growing
solution. Truncation is simple, we just need ck̄+1 = 0 for some integer k̄,
then all successive coefficients are also zero by the recursion formula.
Define k̄ from (24.23):

2 (k̄ + ` + 1) − ρ0 = 0, (24.26)

and there we have it – ρ0 must be an integer if it is to kill the integer

2 (k̄ + ` + 1). This makes our ρ0 = 2 n critical value clear: n defines the
value of ρ0 which will cause the series to truncate:

n = k̄ + ` + 1. (24.27)

For a given n, now, which is the direction we are interested in, we can find
k̄ for a given ` from the above. What is interesting is that the energies

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24.3. HYDROGEN PHYSICS Lecture 24

themselves are indexed only by n, so we can ask: “How many states are
there with an energy En for given n?” The answer is simple – we have
k̄ = n − (` + 1), and the lower limit on the “cutoff” is k̄ = 0 so ` = n − 1.
The smallest value for ` is zero, so k̄ = n − 1 is the max. Then ` = 0 . . . n − 1
are the allowed values. But let’s not forget, in our state counting, the
m = −` . . . ` angular portion – there are 2` + 1 states with a given value of
`, so the total number of states with energy En is:
n−1
X
(2 ` + 1) = n2 (24.28)
`=0
this is the degeneracy of the Hydrogenic states.

24.3 Hydrogen Physics

What we have, then, is a quantized energy:

2 !
e2

E1 m
En = 2 E1 ≡ − (24.29)
n 2 ~2 4 π 0

shared by n2 states. Because we scaled the radial coordinate

√ r by the energy,
−2 m En
effectively, we have n-dependent scaling: ρ = κ r = ~ r and we can
write this naturally in terms of a characteristic length – that of the ground
state: √
−2 m E1 m e2 r r
κr = r= 2
≡ (24.30)
~n 4 π 0 ~ n an
2
with a = 4 πme02~ = 0.528 × 10−10 m, the “Bohr radius” of Hydrogen (half
an Angstrom). Using this for ρ, and normalizing the associated Laguerre
polynomials, we have the full time-independent solution:
s
2 3 (n − ` − 1)!  − r  2 r `  2`+1
   
ψn`m (r, θ, φ) = e a n Ln−`−1 (2 r/(a n)) Y`m (θ, φ).
an 2 n ((n + `)!)3 an
(24.31)
In particular, the ground state, with n = 1, ` = m = 0 has
1 r
ψ100 = √ e− a (24.32)
π a3
with energy:
 2 2 !
m e
E1 = − 2
= −13.6 MeV. (24.33)
2~ 4 π 0

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24.3. HYDROGEN PHYSICS Lecture 24

24.3.1 Electric field of the Ground State

One of the advantages of a quantum mechanical description of things like

the electron is that they tend to “smear out” point effects. For example, the
charge density of a point electron is, of course, ρ(r) = q δ(r), and its field is
singular “at” the electron, E = 4 π q0 r2 r̂. If we think of the “charge density”
of the electron in the ground state of Hydrogen, then
q −2r
ρ(r) = q |ψ100 |2 = e a. (24.34)
π a3
We take this with a grain of salt, of course – this is the probability per
unit volume of finding the charge at radius r, so while we can think of the
electron as a “cloud”, it’s not exactly the same as a known distribution of
charge. Still, we can calculate the electric field associated with this as if it
were a distribution of charge. Gauss’ law gives:
Z Z r
Qenc q −2 r/a 2
E · da = = 4π e r dr, (24.35)
0 0 π a3
and making the usual spherical ansatz, we have E = E(r) r̂, so that
q  2 −2 r/a

4 π r2 E = a − e (a2
+ 2 a r + 2 r 2
) (24.36)
0 a2
or
2 r 2 r2
  
q −2 r/a
E= 1−e 1+ + 2 r̂. (24.37)
4 π 0 r2 a a
This electric field is finite (zero) at the origin, and decays to zero at spatial
infinity. Much nicer physically, although a lot harder to work with. It’s also
not clear we should always consider an electron to be in the ground state of
Hydrogen . . .

24.3.2 Spectrum

Once an electron is in a stationary state of Hydrogen, it will remain there

(getting it into one is another issue). If there is some perturbation, and the
electron moves from one stationary state to another, then it will do so in a
quantized way – i.e. transitions between states occur with jumps in energy
related to the difference of integers (squared). For example, if we go from
some initial state ni (set of states) to a final state nf , the energy change is:
!
1 1
Ei − Ef = E1 − 2 . (24.38)
n2i nf

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24.3. HYDROGEN PHYSICS Lecture 24

Suppose the energy is released as radiation. We know, from the Planck

formula, that E = h ν for a photon, so
!
E1 1 1
ν= − 2 , (24.39)
h n2i nf

or, in terms of the wavelength of the emitted light λ = c/ν,

! !
1 E1 1 1 1 1
= − 2 = 1.097 × 107 m−1 − 2 (24.40)
λ h c n2f ni n2f ni
 2
e2
where the constant R = m
4πc~3 4π0 = 1.097 × 107 m−1 is the Rydberg
constant.

Homework

Reading: Griffiths, pp. 145–159.

Problem 24.1

Griffiths 4.13 (parts a. and b.). Practice calculating expectation values for
Hydrogen.

Problem 24.2

Griffiths 4.14. Most probable electron radius for the ground state of Hydro-
gen.

Problem 24.3

Griffiths 4.15. States that are not the ground state, and their time depen-
dence. Use (4.89) to write the full functional form of the wave function
Ψ(r, θ, φ, t).

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