Download Schrodinger equation in three dimensions

Schrodinger equation in three dimensions
•
The stationary Schrodinger equation in three dimensions is a partial dierential equation involving three
coordinates per particle.
The mathematical complexity behind such an equation can be intractable
by analytical means. However, there are certain high-symmetry cases when it is possible to separate
variables in some convenient coordinate system and reduce the Schrodinger equation to one-dimensional
problems.
•
The simplest example of variable separation is a particle in innitely deep three dimensional quantum
V (x, y, z) be zero inside a block with edges a1 , a2 , a3 and innite
0 , 0 < x < a 1 ∧ 0 < y < a2 ∧ 0 < z < a3
V (r) =
∞ ,
otherwise
well. Let the potential
outside:
We immediately note that this function can be written as a sum of three one-dimensional functions:
V (r) = V1 (x) + V2 (y) + V3 (z)
0 , 0 < xi < ai
Vi (xi ) =
∞ , otherwise
•
The Schrodinger equation
−
~2 ∇2
ψ(r) + V (r)ψ(r) = Eψ(r)
2m
can be expressed as a sum involving individual coordinates:
−
•
~2 ∂ 2 ψ(r)
~2 ∂ 2 ψ(r)
~2 ∂ 2 ψ(r)
−
−
+ V1 (x)ψ(r) + V2 (y)ψ(r) + V3 (z)ψ(r) = Eψ(r)
2m ∂x2
2m ∂y 2
2m ∂z 2
Whenever the operator acting on the unknown function can be expressed as a sum of operators involving
individual coordinates, the solution for the function has the form of a product:
ψ(r) = ψ1 (x)ψ2 (y)ψ3 (z)
Now, the one-variable functions
ψ1 , ψ2
and
ψ3
are unknown. We substitute
equation and note that the derivative with respect to
x
acts only on
ψ1 ,
ψ(r)
in the Schrodinger
etc.:
~2 ∂ 2 ψ1 (x)
+ V1 (x)ψ1 (x) ψ2 (y)ψ3 (z) +
−
2m ∂x2
~2 ∂ 2 ψ2 (y)
−
+ V2 (y)ψ2 (y) ψ3 (z)ψ1 (x) +
2m ∂y 2
~2 ∂ 2 ψ3 (z)
−
+ V3 (z)ψ3 (z) ψ1 (x)ψ2 (y) = (E1 + E2 + E3 )ψ1 (x)ψ2 (y)ψ3 (z)
2m ∂z 2
We have also expressed the total energy
E
as a sum of contributions from individual dimensions,
E = E1 + E2 + E3 .
•
Consider any point
r
at which
ψ(r) 6= 0.
Divide the whole equation by
~2 ∂ 2 ψ1 (x)
1
−
+ V1 (x)ψ1 (x)
ψ1 (x)
2m ∂x2
1
~2 ∂ 2 ψ2 (y)
−
+ V2 (y)ψ2 (y)
ψ2 (y)
2m ∂y 2
1
~2 ∂ 2 ψ3 (z)
−
+ V3 (z)ψ3 (z)
ψ3 (z)
2m ∂z 2
46
ψ(r) = ψ1 (x)ψ2 (y)ψ3 (z):
+
+
= E1 + E2 + E3
Each term on the left-hand side depends on only one coordinate and hence is completely independent
from other terms. The only way to satisfy this equation for any combination of
(x, y, z)
is to require
that three one-dimensional equations be satised:
−
~2 ∂ 2 ψ1 (x)
+ V1 (x)ψ1 (x) = E1 ψ1 (x)
2m ∂x2
−
~2 ∂ 2 ψ2 (y)
+ V2 (y)ψ2 (y) = E2 ψ2 (y)
2m ∂y
−
~2 ∂ 2 ψ3 (z)
+ V3 (z)ψ3 (z) = E3 ψ3 (z)
2m ∂z 2
One can now substitute these expressions into the full 3D Schrodinger equation and see that they solve
it even at the points
r
where
ψ(r) = 0.
Therefore, the solution of the 3D Schrodinger equation is
obtained by multiplying the solutions of the three 1D Schrodinger equations.
•
Now, in each dimension we have a simple one-dimensional innitely deep quantum well problem, which
we solved before:
π 2 ~2 2
n
2ma2i i
r
πni
2
sin
xi
ψi (xi ) =
ai
ai
Ei =
•
The full 3D solutions are characterized by three positive integer quantum numbers,
(nx , ny , nz ),
one
per dimension. The total energy is
π 2 ~2
E = E1 + E2 + E3 =
2m
n2y
n2x
n2
+ 2 + 2z
2
ax
ay
az
!
and the full wavefunction is:
r
ψ(r) = ψ1 (x)ψ2 (y)ψ3 (z) =
where
V = ax ay az
8
sin
V
πnx
πny
πnz
x sin
y sin
z
ax
ay
az
is the volume of the quantum well.
Hydrogen atom
•
Here we seek a proper quantum-mechanical description of a Hydrogen atom. We solve the stationary
Schrodinger equation to nd bound states of a proton and electron interacting via the Coulomb force.
•
rp be the proton position,
re the electron position. Then, the full wavefunction ψ(rp , re ) is a complex function of both rp
2
and re . We use Born's interpretation to give it a physical meaning: P (rp , re ) = |ψ(rp , re )| is the
probability density that the proton will be detected at the position rp , and that at the same time the
electron will be detected at re .
The full wavefunction must describe both the proton and the electron. Let
and
•
The full Schrodinger equation is constructed by analogy to classical mechanics.
The total classical
energy of the Hydrogen atom equals the sum of the proton kinetic energy, electron kinetic energy and
the Coulomb potential energy:
Eclassical =
p2p
p2
1
e2
+ e −
2mp
2me
4π0 |re − rp |
47
where
pp
and
pe
are proton and electron momenta respectively.
Hamiltonian operator, we replace
pp
and
pe
In order to obtain the quantum
by the appropriate operators:
∂
∂
∂
∂
≡ −i~ x̂
+ ŷ
+ ẑ
∂rp
∂xp
∂yp
∂zp
∂
∂
∂
∂
≡ −i~
≡ −i~ x̂
+ ŷ
+ ẑ
∂re
∂xe
∂ye
∂ze
pp = −i~∇rp ≡ −i~
pe = −i~∇re
The Schrodinger equation is
−
~2 ∇2rp
~2 ∇2re
1
e2
ψ(rp , re ) −
ψ(rp , re ) −
ψ(rp , re ) = Eψ(rp , re )
2me
2mp
4π0 |re − rp |
Relative motion
•
The Schrodinger equation is complicated because there are two sets of coordinates, one for proton and
one for electron. These coordinates are treated separately by kinetic energy terms, but appear as a
dierence
re − rp
in the potential energy term, in a non-linear fashion. We cannot make any progress
unless we express the equation in terms of some more convenient variables.
•
Let us dene the relative position
position
r
of the electron with respect to proton, and the center-of-mass
R:
Taking total derivatives relates small
r = re − rp
me re + mp rp
R=
me + mp
changes of re and rp to
small changes of
r
and
R:
dr = dre − drp
dR =
•
me dre + mp drp
me + mp
re and rp in terms of the partial derivatives with
ψ as a function of either (rp , re ) coordinates, or
equivalently as a function of (r, R) variables. A small change in the coordinates (drp , dre ) corresponds
to a particular small change of the alternative variables (dr, dR) according to the above equations, and
leads to a small change of the wavefunction ψ which equals:
Next, let us express partial derivatives with respect to
respect to
r
and
R.
We can regard the wavefunction
dψ =
∂ψ
∂ψ
∂ψ
∂ψ
dr +
dR
dre +
drp =
∂re
∂rp
∂r
∂R
drp = 0 and divide
dre (this is a short-hand notation; in reality, we have to treat x, y, z
set drp = 0, dye = dze = 0 and divide by dxe , than do the same for y, z
We use this equation to nd the relationship between partial derivatives. First, set
both sides of the
dψ
equation by
coordinates separately, that is
directions):
∂ψ
=
∂re
When
drp = 0
∂ψ dr
∂ψ dR
+
∂r dre
∂R dre
the equations from the previous bullet imply
drp =0
dr = dre
and
dR = dre me /(me + mp ).
We substitute this in the above equation and obtain:
∂ψ
∂ψ
me
∂ψ
=
+
∂re
∂r
me + mp ∂R
Similarly, if we were to set
dre = 0,
we would have
dr = −drp
and
dR = drp mp /(me + mp )
previous bullet, so that
∂ψ
=
∂rp
∂ψ dr
∂ψ dR
+
∂r drp
∂R drp
48
dre =0
=−
∂ψ
mp
∂ψ
+
∂r
me + mp ∂R
from the
•
We in fact need to calculate the Laplacians
we need to take one more derivative of
ψ.
∇2
which appear in the Schrodinger equation. Therefore,
We just re-apply the above rules for taking derivatives:
2
∂
∂2ψ
2me
∂2ψ
me
∂
∂2ψ
m2e
∂2ψ
=
ψ=
+
+
+
2
2
2
∂re
∂r me + mp ∂R
∂r
me + mp ∂r∂R (me + mp ) ∂R2
2
m2p
∂2ψ
∂
2mp
∂2ψ
mp
∂
∂2ψ
∂2ψ
2
∇r p ψ ≡
=
−
−
+
ψ
=
+
∂r2p
∂r me + mp ∂R
∂r2
me + mp ∂r∂R (me + mp )2 ∂R2
∇2re ψ ≡
•
Now, substitute these Laplacians in the kinetic energy terms of the Schrodinger equation:
~2 ∇2rp
~2 ∇2re
−
ψ−
ψ
2me
2mp
=
~2
−
2me
~2
−
2mp
=
where
M = me + mp
−
∇2r ψ
2me
m2e
∇2 ψ
+
∇r ∇R ψ +
me + mp
(me + mp )2 R
m2p
2mp
∇2r ψ −
∇2 ψ
∇r ∇R ψ +
me + mp
(me + mp )2 R
!
~2 ∇2r
~2 ∇2R
ψ−
ψ
2µ
2M
is the total atom mass, and
µ
is the reduced mass:
1
1
1
=
+
µ
me
mp
•
The Schrodinger equation can now be rewritten as:
−
~2 ∇2r
~2 ∇2R
1 e2
ψ(r, R) −
ψ(r, R) −
ψ(r, R) = Eψ(r, R)
2µ
2M
4π0 |r|
This is a tremendous simplication, which allows us to separate variables.
potential depends only on
r.
Note that the Coulomb
We write
ψ(r, R) = ψr (r)ψcm (R)
E = Er + Ecm
and using the same trick as before rewrite the Schrodinger equation as two equations involving only
one of the two sets of coordinates:
−
~2 ∇2r
1 e2
ψr (r) −
ψr (r) = Er ψr (r)
2µ
4π0 |r|
−
•
~2 ∇2R
ψcm (R) = Ecm ψcm (R)
2M
The second equation describes quantum mechanical motion of the Hydrogen atom as a whole. The
wavefunction
ψcm (R) depends only on the center-of-mass coordinates R and indeed captures the center-
of-mass motion. If there were external forces acting on the atom, they would aect the evolution of
ψcm (R), but in this case the atom propagates as a plane-wave with momentum any total momentum
P and energy Ecm = P2 /2M . If we knew nothing about the internal structure of the atom, we would
only write this center-of-mass equation. This is an example of a deeper principle. We don't know if an
electron has some constituent particles (we haven't found them yet), but one way or another we can
write and solve a Schrodinger equation for the electron.
•
The rst of the two separated equations describes the relative motion of the proton and electron. It
is satisfying to nd the reduced mass
electron mass, so that
µ ≈ me .
µ
in this equation. The proton mass is much larger than the
Therefore, this equation describes the quantum mechanical motion of
an electron around the heavy proton. The Coulomb potential is attractive and strong enough to bind
a low-energy electron to a nite distance from the proton. Therefore, we shall nd a discrete spectrum
of bound states
Er .
At high energies the electron is free, so there is also a continuum high-energy
portion of the spectrum
Er .
49
Bound states
•
Here we seek the bound states between an electron and proton. Therefore, we solve the Schrodinger
equation:
−
1 e2
~2 ∇2r
ψr (r) −
ψr (r) = Er ψr (r)
2µ
4π0 |r|
This is still a complicated equation, involving the coordinates
r = (x, y, z)
in a non trivial manner.
Specically, we know how to write the Laplacian in Cartesian coordinates:
∂2
∂2
∂2
+ 2+ 2
2
∂x
∂y
∂z
p
r ≡ |r| = x2 + y 2 + z 2 .
∇2r =
but the Coulomb potential is a function of
the
•
(x, y, z)
This is a hard problem because
coordinates are bundled together in a non-linear fashion.
Fortunately, the Coulomb potential is very symmetric. It is possible to separate variables using the
spherical coordinate system:
We want to express the position vector
the vector
the
•
r
and the
xy -plane.
z -axis,
x =
r sin θ cos φ
y
=
r sin θ sin φ
z
=
r cos θ
r using the radius r ∈ (0, ∞), the polar angle θ ∈ (0, π) between
φ ∈ (0, 2π) which species the projection of r on
depends only of r .
and the azimuthal angle
The Coulomb potential
Manipulating derivatives in the same manner as in the previous section, we can express the Laplacian
operator in spherical coordinates. This is a tedious procedure and will not be detailed here. Laplacian
in the spherical coordinates takes the form:
∇2r
1 ∂
= 2
r ∂r
1
∂
∂
1
∂2
2 ∂
r
+ 2
sin θ
+ 2 2
∂r
r sin θ ∂θ
∂θ
r sin θ ∂φ2
We can simplify notation by writing:
∇2r =
•
The operator
L̂ = r × (−i~∇)
1 ∂
r2 ∂r
∂
L̂2
r2
− 2 2
∂r
~ r
corresponds to angular momentum (L
=r×p
classically). It takes the
following form in Cartesian and spherical coordinates:
∂
∂
∂
∂
∂
∂
−z
x̂ − i~ z
−x
ŷ − i~ x
−y
ẑ
L̂ = −i~ y
∂z
∂y
∂x
∂z
∂y
∂x
∂
∂
∂
∂
∂
− cot θ cos φ
x̂ − i~ cos φ
− cot θ sin φ
ŷ − i~ ẑ
= −i~ − sin φ
∂θ
∂φ
∂θ
∂φ
∂φ
!
∂
θ̂ ∂
−
= −i~ φ̂
∂θ sin θ ∂φ
•
Let us rst hypothesize that the solution of the Schrodinger equation can be written as a product:
ψr (r) = ϕ(r)Y (θ, φ)
Substituting this in the Schrodinger equation, written in spherical coordinates, gives:
−
~2 1 ∂
2µ r2 ∂r
r2
∂ϕ(r)
∂r
Y (θ, φ) +
L̂2 Y (θ, φ)
1 e2
ϕ(r) −
ϕ(r)Y (θ, φ) = Er ϕ(r)Y (θ, φ)
2
2µr
4π0 r
50
Multiply this equation by
hand side and all
r2 ,
divide it by
(θ, φ)-dependent
−
1 ~2 ∂
ϕ(r) 2µ ∂r
R(r)Y (θ, φ),
(θ, φ)
r
−~2 l(l + 1)/2µ,
where
l
r,
(θ, φ).
Since the left-hand side does not
both sides must be equal to a constant.
l
must be a non-negative integer.
We end up with a set of two dierential equations:
−
~2 ∂
2µ ∂r
e2 r
∂ϕ(r)
~2
r2
−
ϕ(r) − Er r2 ϕ(r) = − l(l + 1)ϕ(r)
∂r
4π0
2µ
−
~2
L̂2 Y (θ, φ)
= − l(l + 1)Y (θ, φ)
2µ
2µ
The radial equation
•
r-dependent equation:
e2 r
~2 ∂
~2
2 ∂ϕ(r)
r
−
−
ϕ(r) − Er r2 ϕ(r) = − l(l + 1)ϕ(r)
2µ ∂r
∂r
4π0
2µ
Here we discuss the rst
Multiply by
−2µ/~2 :
∂
∂r
r
2 ∂ϕ(r)
∂r
Let us dene a new function,
+
2µe2
2µEr 2
rϕ(r) +
r ϕ(r) − l(l + 1)ϕ(r) = 0
2
4π0 ~
~2
R(r) = rϕ(r). Then:
∂ R
1 ∂R
∂ϕ
=
= 2
r−R
∂r
∂r r
r
∂r
∂
∂ ∂R
∂2R
2 ∂ϕ
r
=
r−R =
r
∂r
∂r
∂r ∂r
∂r2
The radial Schrodinger equation takes a simpler form in terms of
R(r):
∂ 2 R(r)
2µe2 R(r) 2µEr
l(l + 1)
+
+
R(r) −
R(r) = 0
∂r2
4π0 ~2 r
~2
r2
where we have also divided out one factor of
•
terms on the left-
is an unknown number ultimately determined
from angular momentum quantization. It turns out that
•
and
and the right-hand side does not depend on
We shall label that constant
r-dependent
∂ϕ(r)
e2 r
1
L̂2 Y (θ, φ)
r2
−
− Er r2 = −
∂r
4π0
Y (θ, φ)
2µ
This equation must be satised for any combination of
depend
and collect all
terms on the right-hand side:
r.
Note that
aB =
4π0 ~2
µe2
is the Bohr's radius for Hydrogen (0.53 Å). Also,
(H)
E0
~2
µ
=
=
2µa2B
2
is the binding energy of the Hydrogen atom (13.6 eV).
51
e2
4π0 ~
2
•
Let us change variables by dening dimensionless radius
(H)
Er /E0
.
Noting that
∂/∂r = a−1
B ∂/∂ρ
ρ = r/aB ,
and dimensionless energy
=
we obtain a the simplest form of the radial Schrodinger
equation:
l(l + 1)
∂ 2 R(ρ) 2
+ R(ρ) + R(ρ) −
R(ρ) = 0
2
∂ρ
ρ
ρ2
•
We are looking for bound-state solutions, corresponding to negative energies
write
= −α2 ,
d2 R
2 l(l + 1)
+ −α2 + −
2
dρ
ρ
ρ2
When
ρ→∞
< 0.
Therefore, let us
and
R=0
the equation reduces to
d2 R
= α2 R
dρ2
which has the solution
R(ρ) = Ae−αρ + Beαρ .
Assuming
the wavefunction nite and normalizable in the
ρ→∞
α > 0,
solution is not applicable. Therefore, we can seek a solution
ρ
to the exponential function in the large
we must select
B=0
ρ
limit. For nite values of
R(r)
in order to keep
the exponential
in the form of a sum, which reduces
limit:
R(ρ) = Ae
−αρ
nr
X
Bk ρδ+k
k=0
We anticipate that the sum must have a nite number of terms, because otherwise it is not certain
that
•
R(ρ)
could be normalized.
Substituting the hypothesized solution in the Schrodinger equation yields:
−2α
nr
X
Bk (δ + k)ρ
δ+k−1
+
nr
X
Bk (δ + k)(δ + k − 1)ρ
δ+k−2
+
k=0
k=0
This must hold for every
ρ.
2 l(l + 1)
−
ρ
ρ2
X
nr
k=0
Therefore, any net co-ecients multiplying any power of
The net co-ecient multiplying
ρδ+k
Bk ρδ+k = 0
ρ
must be zero.
is:
Bk+1 [−2α(δ + k + 1) + 2] + Bk+2 [(δ + k + 2)(δ + k + 1) − l(l + 1)] = 0
This gives us a recursion relationship for
Bk+1 =
•
We can determine
δ
in the limit
ρ→0
Bk
(take
k
reduced by one):
2α(δ + k) − 2
Bk
(δ + k)(δ + k + 1) − l(l + 1)
by substituting the hypothesized solution in the Schrodinger
equation and neglecting the smallest terms:
R(ρ) → AB0 ρδ
d2 R(ρ) l(l + 1)
−
R(ρ) = 0
dρ2
ρ2
We nd:
δ(δ − 1) − l(l + 1) = 0
which can be satised by setting
•
δ = l + 1.
Now we determine the last term in the sum for
R(ρ).
52
k = 0 and k = nr .
Bnr +1 = 0. The recursion
The sum runs between
We could interpret the sum as an innite sum in which the amplitude
relationship for
with
Bnr 6= 0,
Bk
would then automatically produce
Bk = 0 for
all
k > nr .
So, we require
Bnr +1 = 0
and this implies:
2α(δ + nr ) − 2
=0
(δ + nr )(δ + nr + 1) − l(l + 1)
Therefore,
α=
•
1
1
=
δ + nr
nr + l + 1
Finally, we recall that dimensionless energy is
= −α2 .
Therefore, the Hydrogen atom energy can
take values:
Er =
where
n = nr + l + 1
µ
=−
2
(H)
E0
e2
4π0 ~
2
1
n2
is a positive integer. This is precisely the form predicted phenomenologically by
Bohr's and Sommerfeld's models, which were designed to match experimental observations. However,
we have derived it microscopically from an equation of motion. This match (with experiments) was a
great conrmation of the Schrodinger's equation.
•
The bound-state energy levels are degenerate. There must be one quantum number for every dimension
(n,
l
and one more), but
Er
depends on a single integer
n.
This is a special feature of the Coulomb
potential.
•
Solutions for the wavefunction
R(ρ)
are given by associated Laguerre polynomials.
The angular equation
•
We now come back to the angular equation
L̂2 Y (θ, φ) = ~2 l(l + 1)Y (θ, φ)
2µ factors) reveals that we seek the eigenvalues
~2 l(l + 1). Expanding back L̂2 in spherical coordinates yields:
1 ∂
∂
1 ∂2
−
sin θ
+
Y (θ, φ) = l(l + 1)Y (θ, φ)
sin θ ∂θ
∂θ
sin2 θ ∂φ2
Written in this form (without the repeating
operator
•
L̂2
of the
in the form
We can anticipate a solution for
Y (θ, φ):
Y (θ, φ) = Θ(θ)eimφ
φ = 2π must be the same as at φ = 0, which
m must be an integer. Substituting this product form in the angular Schrodinger equation
2
2
2
eect of replacing ∂ /∂φ with −m :
d
m2
1 d
sin θ
−
+ l(l + 1) Θ(θ) = 0
sin θ dθ
dθ
sin2 θ
Since the wavefunction must be single-valued, its value at
implies that
has the
•
It is convenient to change variables from
θ
to
ζ = cos θ.
d
1 d
=−
d cos θ
sin θ dθ
⇒
We note that
d
d
= − sin θ
dθ
dζ
Θ(ζ) is:
d
m2
(1 − ζ)2
−
+
l(l
+
1)
Θ(ζ) = 0
dζ
1 − ζ2
Therefore, the dierential equation for
d
dζ
This is a known dierential equation in mathematics whose solutions are called associated Legendre
functions.
53
•
Solving this dierential equation is a tedious procedure which does not illuminate any new concepts.
One needs to guess the proper form of the expansion of
into the equation to obtain a recursion relation.
Θ(ζ)
in powers of
ζ,
and substitute it back
Using similar arguments as before, one then nds
restrictions on the values of l.
•
It turns out that
l
must be a non-negative integer, and the quantum number
the wavefunction) can take values
m
m ∈ {−l, −l + 1, −l + 2 . . . , l − 2, l − 1, l}.
(from the
eimφ
part of
Other combinations do
not produce normalizable wavefunctions. These conclusions are much easier to derive from algebraic
operator manipulations.
•
The eigenvalues of
L̂2
are
~2 l(l + 1).
These are the values of the angular momentum magnitude which
can be measured in an experiment. The eigenvalues of
axis) are
~m,
Lz
(angular momentum projection on the
z-
so these are the values which can be measured. In fact, by symmetry, measurements of
the angular momentum projections on any axis must yield integer multiples of
~.