Download The Position Representation in Quantum Mechanics

The Position Representation in Quantum
Mechanics
Sungwook Lee
Department of Mathematics
University of Southern Mississippi
[email protected]
July 13, 2007
The x-coordinate of a particle is associated with an operator x̂. The
set of numbers that are eigenvalues of an operator is called the spectrum
of the operator. The spectrum of x̂ consists of real numbers −∞ < x <
∞. |xi is the state in which the particle is certainly at x and satisfies
the eigenvalue equation x̂|xi = x|xi. If the particle is in some state |ψi,
the amplitude to find the particle at x is the complex number hx|ψi. The
functional dependence of this number on x is what is called the wave function
ψ(x):
ψ(x) = hx|ψi.
Since the particle has to be somewhere, the sum of the probabilities |ψ(x)|2
has to be 1. Since the values of x is continuous,
Z
Z
1 = dx|ψ(x)|2 = dxhψ|xihx|ψi.
(1)
In discrete case, if |ψn i, n = 1, 2, · · · denote eigenkets, then
∞
X
|ψi =
So,
P∞
n=1 |ψn ihψn |
|ψn ihψn |ψi.
n=1
= I. Similarly, from (1) we obtain
Z
dx|xihx| = I.
By (2),
(2)
Z
0
hx |ψi =
dxhx0 |xihx|ψi.
1
(3)
Recall that the Dirac delta function is defined by
Z
0
ψ(x ) = dxδ(x − x0 )ψ(x).
Compare this with (3), we get
hx|x0 i = δ(x − x0 ).
(4)
If |φi and |ψi are two states, then1 by (2)
Z
hφ|ψi = dxhφ|xihx|ψi
Z
= dxφ∗ (x)ψ(x).
In the position representation, the functional form ψ(·) is identified with the
state |ψi and the complex number ψ(x) with the amplitude hx|ψi.
Now, let us calculate hx|x̂|ψi. First we assume that x̂ is hermitian, i.e.,
∗
(x̂) = x̂ as usual in quantum mechanics.
hx|x̂|ψi = h(x̂)∗ x|ψi
= hx̂x|ψi
= xhx|ψi
= xψ(x).
This calculation also can be done without assuming that x̂ is hermitian:
hx|x̂|ψi = hx|x̂|I|ψi
Z ∞
=
dxhx|x̂|x0 ihx0 |ψi
Z−∞
∞
=
dxhx|x̂x0 iψ(x0 )
Z−∞
∞
=
dxx0 δ(x − x0 )ψ(x0 )
Z−∞
∞
=
dxδ(x − x0 )x0 ψ(x0 )
−∞
= xψ(x).
1
Here, we consider only 1-dimensional case, so φ∗ (x) = φ̄(x).
2
Hence, the operator x̂ turns ψ that evaluates ψ(x) into the one that evaluates
xψ(x).
In the position representation, x-momentum p operates on wave functions by differentiation2 :
hx|p̂|ψi = hx|p̂ψi
= p̂ψ(x)
∂ψ
= −i~ .
∂x
h
~ = 2π
≈ 1.055 × 10−34 J · s is the Dirac constant or the reduced Planck
constant, where h ≈ 6.626 × 10−34 J · s is the Planck constant. Let up (x) :=
hx|pi. Then
hx|p̂|pi = hx|p̂pi
∂up
= −i~
∂x
= pup (x),
since p̂|pi = p|pi. That is, we obtain the differential equation
−i~
∂up
= pup .
∂x
The differential equation has a general solution up (x) = Aeipx/~ . The unknown constant A can be determined as follows:
δ(p − p0 ) = hp0 |pi
Z
= dxhp0 |xihx|pi
Z
= dxu∗p0 (x)up (x)
Z
0
2
=A
dxei(p−p )x/~
= 2π~A2 δ(p − p0 ).
Z ∞
1
Here, we used the familiar integral
dteixt = δ(x) in Fourier trans2π −∞
forms. Recall that this integral came from the sequence of functions
Z n
1
sin nx
=
dteixt , n = 1, 2, · · ·
δn (x) =
πx
2π −n
2
For now, we will take it as a definition, but it will be proved later.
3
which approximates the Dirac delta function δ(x), i.e.,
Z ∞
1
δ(x) ≈ lim δn (x) =
dteixt .
n→∞
2π −∞
1
. Hence,
Since A > 0, A = √
2π~
Figure 1: The sequence δn (x)
1
up (x) = √
eipx/~ .
2π~
This is the wave function of a state of momentum.
In physics, the hamiltonian H plays an important role in the study of
dynamics. It is given by
H = T + V,
where T and V are the kinetic energy and the potential energy, respectively.
The hamiltonian H represents the energy of a physical system. The kinetic
p2
energy is given by T = 2m
where p is the momentum. So, the hamiltonian
can be written as
p2
+ V (x).
H(p, x) =
2m
4
A typical classical example is the hamiltonian for a harmonic oscillator:
H(p, x) =
p2
1
+ mω 2 x2 .
2m 2
In the position representation, the hamiltonian for a particle acts by a combination of differentiation and multiplications:
hx|H|ψi = hx|Hψi
= Hψ(x)
µ
¶
~ ∂2
= −
+ V (x) ψ(x).
2m ∂x2
For a single particle moving along a real line R, there are two important
observables: position and momentum. As we discussed earlier, in the quantum mechanical description of such a particle, the position operator x̂ and
momentum operator p̂ are respectively given by
hx|x̂|ψi = xψ(x),
∂ψ(x)
.
hx|p̂|ψi = −i~
∂x
(5)
(6)
Now, we prove (6). This is what Larry Mead taught me3 . Let H be
a Hilbert space of states. Suppose that O is a hermitian operator4 on H
such that the unitary operator g(a) = eiOa/~ , −∞ < a < ∞ satisfies the
property:
g(a)ψ(x) = ψ(x + a).
(7)
That is, g(a) gives rise to a translation of ψ(x) along the real line. The
unitary operator g(a) = eiOa/~ acts on ψ(x) as5
µ
¶
ia
a2 2
eiOa/~ ψ(x) = 1 + O −
O
+
·
·
·
ψ(x)
~
2!~2
a2 2
ia
= ψ(x) + Oψ(x) −
O ψ(x) + · · · .
~
2!~2
3
If there is any misleading in the following argument, it is solely due to my misunderstanding on the material.
4
Let G be the group of unitary operators on a Hilbert space H and O a hermitian
operator on H. Then {eiOa/~ : a ∈ R} is a one-parameter (sub)group of G.
5
Here O2 means the function composition O ◦ O.
5
Let x denote the position operator and δa denote infinitesimal change of a,
so that (δa)n can be neglected for n ≥ 2. For any wave function ψ(x),
eiOδa/~ xe−iOδa/~ ψ(x) = eiOδa/~ xψ(x − δa)
= (x + δa)ψ(x).
Thus,
eiOδa/~ xe−iOδa/~ = x + δa.
On the other hand,
e
iOδa/~
−iOδa/~
xe
µ
¶ µ
¶
iδa
iδa
= 1+
O x 1−
O
~
~
iδa
=x−
[x, O],
~
(8)
(9)
where [x, O] = xO − Ox is the canonical commutator. From (8) and (9), we
obtain
iδa
x + δa = x −
[x, O].
~
This implies
[x, O] = i~.
(10)
Let a ∈ R. From (7) one obtains
|ψ(x)|2 = |ψ(x + a)|2 .
(11)
So, one may consider the complex-valued function ψ(x) itself as a periodic
function, and a simple candidate for ψ(x) satisfying (11) can be ψ(x) = eikx .
Note that k is the wave number k = 2π
λ , where λ is the wave length. Using
h
the de Broglie’s formula p = λ , one obtains p = ~k or k = ~p . Hence,
ψ(x) = eipx/~ .
(12)
The wave function in (12) may be regarded as hx|pi, i.e., ψ(x) = hx|pi and
ψ ∗ (x) = hp|xi. Since p̂|pi = p|pi,
hx|p̂|pi = phx|pi = pψ(x).
(13)
hx|p̂|pi = hxp̂|I|pi
Z
= dx0 hx|p̂|x0 ihx0 |pi
Z
= dx0 h|p̂|x0 iψ(x0 ).
(14)
On the other hand,
6
From (13) and (14), one obtains
Z
0
dx0 hx|p̂|x0 ieipx /~ = peipx/~ .
By Fourier transform6 ,
Z
0
dp −ipx0 /~ ipx/~
e
pe
2π
Z
dp ip(x−x0 )/~
∂
e
.
= −i~
∂x
2π
∂
= −i~ δ(x − x0 ).
∂x
hx|p̂|x i =
This implies that
hψ|p̂|ψi = hψ|I|p̂|I|ψi
Z
Z
= dx dx0 hψ|xihx|p̂|x0 ihx0 |ψi
¸
·
Z
Z
∂
0
0 ∗
= dx dx ψ (x) −i~ δ(x − x ) ψ(x0 )
∂x
µ
¶Z
Z
∂
= dxψ ∗ (x) −i~
dx0 δ(x − x0 )ψ(x0 )
∂x
µ
¶
Z
∂
∗
= dxψ (x) −i~
ψ(x).
∂x
Z
Since hψ|p̂|ψi = dxψ ∗ (x)p̂ψ(x), we see that
p̂ = −i~
∂
.
∂x
(15)
Let us write the momentum operator p̂ simply as p. Then for any wave
6
Recall that if
then
1
φ(k) = √
2π
Z
1
ψ(x) = √
2π
dxe−ikx ψ(x),
Z
dkeikx φ(k)
and vice versa.
7
function ψ(x),
[x, p]ψ(x) = xpψ(x) − pxψ(x)
µ
¶
∂xψ(x)
∂ψ(x)
= x −i~
+ i~
∂x
∂x
= i~ψ(x).
Hence,
[x, p] = i~.
(16)
That is, the momentum operator p satisfies the commutation relation (10).
Clearly,
eipa/~ ψ(x) = eipa/~ eipx/~ = eip(x+a)/~ = ψ(x + a).
Therefore, the momentum operator p = −i~
8
∂
is the desired operator O.
∂x