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Schrödinger’s Equation
• The Schrödinger equation for a particle of mass equal to
(1)
ut = i∆u,
1
2
is
(x, t) ∈ Rn × R.
Here the units of length and time have been chosen so that Planck’s constant is
equal to 1. The particle is moving in the absence of external forces.
• In the traditional classification of PDEs, the Schördinger equation is neither
elliptic, parabolic, nor hyperbolic.
• The physical interpretation, in quantum mechanics, is that the square of the
modulus |u(t, x)|2 , is the probability density for finding the particle at time t
and place x. Precisely, for a Lebesgue measurable subset E ⊂ Rn ,
Z
(2)
Probability (particle is in E at time t) =
|u(x, t)|2 dx.
E
The probability density for the particle’s momentum is given by the Fourier
transform of u(·, t)
Z
(3)
Probability (momentum is in E at time t) =
|b
u(x, t)|2 dx.
E
• The probability interpretation requires that
Z
(4)
|u(x, t)|2 dx = 1,
∀t ≥ 0.
for physically relevant solutions.
– One is immediately led to think that L2 (Rn ) will play a distinguished role.
• Note that if (4) holds, then the Plancherel theorem shows that Fu also has
L2 -norm equal to 1, so the interpretation of |Fu|2 as a probability density is
consistent.
Proof of (4). Note that if a solution u is small enough as x → ∞, to justify
differentiation under the integral and neglecting terms at infinity from integrations
by parts, we have
Z
Z
Z
∂t |u|2 dx =∂t uu dx = uut + ut u dx
XZ
ui∂j2 u + (i∂j2 u)u dx
=
j
X Z
=
i ∂j u∂j u − ∂j u∂j u dx = 0.
j
R
This leads to the conclusion that (4) is satisfies if |u(x, 0)|2 dx = 1. R
• The physical interpretation of E |u|2 dx, as the probability of finding the particle
in E, suggests that we want solutions for which the quantity is continuous in t.
– We look for solutions such that t 7→ u(·, t) is continuous on R with values in
L2 (Rn ).
– This also forces related measurements of momenta to be continuous.
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1
2
Solutions of Schrödinger’s Equation with Data in F(Rn )
• The exponential solutions of the Schrödinger equation satisfy
τ = −i|ξ|2 .
The exponential solutions are
eiτ |ξ|
–
–
–
2
+ix·ξ
= eiξ·(x−ξt) ,
which is a function of x − ξt.
The exponential solution of frequency ξ evolves by a translation at velocity ξ.
The hyperplane of constant phase translate at speed |ξ| in the direction ξ/|ξ|.
Consequently, ξ is called the phase velocity at frequency ξ.
As we will see, the group velocity, equal to 2ξ for the Schrödinger equation, is a
more important quantity.
• Taking linear combinations of exponential solutions, yields the solutions
Z
2
a(ξ)e−it|ξ| +ix·ξ dξ.
To solve the initial value problem
(6)
ut = i∆u,
u(·, 0) = f (·),
set t = 0 and equate the resulting expression to f . This suggests a = (2π)−n/2 Ff ,
leading to the formula
Z
2
(7)
u(x, t) = (2π)−n/2 e−it|ξ| eixξ Ff (ξ)dξ.
• An alternate derivation of the same formula starts by the Fourier transform of
ut = i∆u with respect to x to obtain ∂t u
b = −i|ξ|2 u
b. This is an ODE in time
with ξ as parameter. Solving yields
2
u
b(ξ, t) ≡ e−it|ξ| fb(ξ).
Equivalently
(8)
2
u(ξ, t) ≡ F−1 (e−it|ξ| Ff ),
which is the same as (7).
• To show that (7)-(8) yields solutions of (6), observe that for f ∈ S, the
integral (7) can be differentiated arbitrarily often with respect to t and x.
– Indeed, the differentiation is justified by the rapid decrease of Ff since the t, x
derivatives of the integrand are expresses as a finite linear combination of terms
of the form
2
polynomial (ξ, t)e−it|ξ| eixξ Ff (ξ).
By the rapid decay of Ff , these are dominated by cN,T hξi−N on Rnx ×Rnξ ×[−T, T ].
– We find that u ∈ C ∞ (Rn × R) and satisfies the initial value problem (6).
• The operator f 7→ u(·, t) is called the propagator, and is denoted S(t).
It is defined for all t ∈ R.
– For each t, S(t) is a continuous linear map of S to itself given by the formula
2
S(t) = F−1 e−it|ξ| F.
• From the definition of S, it follows that S(t)∂xα f = ∂xα (S(t)f ).
3
Conservation of Momentum.
• Note that |Fu(ξ, t)|2 = |Ff (ξ)|2 is independent of t.
– Thus the probability density for momentum is independent of t,
a very strong form of conservation of momentum.
Integrating over ξ ∈ Rn shows that the L2 (Rn )-norm of u is independent of time.
– Thus the requirement (4) is satisfied as soon as itR is satisfied at t = 0.
– The conservation of total momentum states that ξ|Fu|2 dξ is independent of t.
This is a weaker version.
Example 1. Find the solution of the initial value problem (1) when f (x) =
2
e−a|x| /2 , a > 0, a Gaussian with “width” √1a and height 1.
The formula Ff (ξ) = a−n/2 e−|ξ|
2
/2a
yields
2
u
b(t) = e−it|ξ| Ff (ξ) = a−n/2 e−
((1/a)+2it)|ξ|2
2
,
again a Gaussian. As u
b is even, u = F−1 u
b = Fb
u, and
|x|2
1
u =a−n/2 ( + 2it)−n/2 e− 2((1/a)+2it)
a
2
a|x|
− 2(1+4a
2 t2 )
=(1 + 2ait)−n/2 e
ia2 t|x|2
e 1+4a2 t2 .
This is one of the few explicit solutions of the Schrödinger equation.
A good deal of intuitive content is hidden√in the long formula.
(a) As t → ∞, the width of u grows like t a. √
The physical interpretation is that (δx) ≈ t a.
√
(b) Similarly, the momentum√distribution is Gaussian √
with width δp ≈ a.
(c) The momentum of order a causes a spread like t a.
As t → ∞, the amplitude of u decays like (at)−n/2 .
√
– The geometric explanation is that the wave spreads over |x| ≤ t a.
2
Indeed,
√ if typical amplitude is M , then the square of the L -norm is like
M 2 (t a)n which must be independent of t, whence
Z
2
2 √ n
M (t a) ≈ e−a|x| /2 dx ∼ a−n/2 ,
which yields M 2 ∼ (at)−n .
Example 2. A closely related example illustrates the propagation of oscillatory
pulses. Take
2
f (x) = eix·η e−|x| /2 with η ∈ Rn .
Then
Ff (ξ) = e−|ξ−η|
2
/2
and
2
u
b(t) =e−it|ξ| e−|ξ−η|
=e−it(|ξ−η|
2
2
/2
+2(ξ−η)·η+|η|2 ) −(ξ−η)2 /2
e
≡ ψ(t, ξ − η).
4
Then
u = eix·η v, where vb ≡ ψ(t, ξ) = e−it(|ξ|
2
+2ξ·η+|η|2 ) −|ξ|2 /2
e
.
The factor e−i2tξ·η also corresponds to translation
2
v = τ2ηt w, where w
b = e−it|η| e−(1+2it)|ξ|
2
/2
.
b = Fw
b is given by
Since w
b is even, w = F−1 w
|x|2
2
w = e−it|η| (1 + 2it)−n/2 e− 2(1+2it) .
Then
|x−2ηt|2
u = eiη·(x−ηt) (1 + 2it)−n/2 e− 2(1+2it) .
Let g(x, t) be the solution with η = 0, so that g is an expanding Gaussian of
Example 1. Then
u = eiη·(x−ηt) g(x − 2ηt, t)
is the product of the plane wave solution with phase velocity η and an expanding
pulse translating with the group velocity 2η.
Note that the phase velocity is slower than the group velocity so the pulse appears
to overtake the plane wave.
Note that the speed of propagation depends on the frequency of the pulse.
– The parts of different frequencies move at different speeds which pulls the wave
apart.
The pulling apart of localized waves and separation according to frequency are
signatures of the phenomenon called dispersion.
Fundamental Solution of the Schrödinger Equation.
We proceed to find the (generalized) solution of the initial value problem
ut = i∆u,
u(·, 0) = δ.
By definition,
2
2
Fu(t) = e−it|ξ| Fδ = (2π)−n/2 e−it|ξ| .
The right-hand side is even in ξ, so F(rhs) = F∗ (rhs), and hence for t 6= 0,
(3)
u(t) = (2π)−n/2 (2it)−n/2 e
i|x|2
4t
= (4πit)−n/2 e
i|x|2
4t
.
– Note that u(0) = δ and u(t) ∈ C ∞ for all t 6= 0.
– The fact that the initial disturbance is localized at x = 0 and propagates immediately to a solution of amplitude 1 uniform in space reflects one aspect of
Heisenberg’s uncertainty principle.
This makes the instantaneous dispersion reasonable.