Elements of Quantum Mechanics
3.2 The One-Dimensional Schrödinger Equation
Chapter 3
represents the wavenumber. Further, as established by Einstein’s explanation of the photoelectric effect, the energy E of the particle is related to the frequencyωby the following equation:
E=ω (3.4)
The simplest type of a wave is a one-dimensional plane wave described by the wave function
(x, t)=A exp [i(kx−ωt)] (3.5)
where A is the amplitude of the wave and the propagation is assumed to be in the +x direction. If we now use Eqs. (3.2) and (3.4), we would obtain
=exp i
(px−Et)
(3.6) Elementary differentiation will give us
i∂
∂t =E (3.7)
−i∂
∂x =p (3.8)
which suggests that, at least for a free particle, the energy and momentum can be represented by differential operators given by
E→i∂
∂t, p→ −i∂
∂x (3.9)
Further, if we again differentiate Eq. (3.8), we would obtain
− 2 2m
∂2
∂x2 = p2
2m (3.10)
For a free particle, the energy and momentum are related by the equation E= p2
2m (3.11)
Thus the right-hand sides of Eqs. (3.7) and (3.10) are equal and we obtain
i∂
∂t = −2 2m
∂2
∂x2 (3.12)
which is the one-dimensional Schrödinger equation for a free particle. If we use the operator representations of E and p [see Eq. (3.9)], we may write the above
3.2 The One-Dimensional Schrödinger Equation 35 equation as
E= p2
2m (3.13)
We next consider the particle to be in a force field characterized by the potential energy V(x); thus, classically, the total energy of the system is given by
E= p2
2m +V(x) (3.14)
If we now assume p and E to be represented by the differential operators, the equation
E = p2
2m +V(x)
(3.15)
would assume the form i∂
∂t =
−2 2m
∂2
∂x2+V(x)
(3.16)
which represents the one-dimensional time-dependent Schrödinger equation. The above equation can be written in the form
i∂
∂t =H (3.17)
where
H= p2
2m+V(x)= −2 2m
∂2
∂x2 +V(x) (3.18)
is an operator and represents the Hamiltonian of the system. Equations (3.16) and (3.17) represent the one-dimensional time-dependent Schrödinger equation.
The above is a very heuristic derivation of the Schrödinger equation and lacks rigor. Strictly speaking Schrödinger equation cannot be derived. To quote Richard Feynman
Where did we get that equation from ? Nowhere. It is not possible to derive it from anything you know. It came out of the mind of Schrödinger.
Although we have obtained Eq. (3.16) starting from an expression for a plane wave, the Schrödinger equation as described by Eq. (3.16) is more general in the sense that ψ (r, t)called the wave function contains all information that is knowable about the system. As will be discussed in Section 3.4, ψ∗(r, t)ψ(r, t)dτ represents the probability of finding the particle in a volume element dτ. Note also that observables such as momentum energy are represented by operators [see Eq. (3.9)].
When the Hamiltonian, H, is independent of time1, Eq. (3.16) can be solved by using the method of separation of variables:
(x, t)=ψ(x) T(t) (3.19)
Substituting in Eq. (3.16) and dividing by, we obtain i 1
T(t) dT
dt = 1 ψ
−2 2m
d2ψ
dx2 +V(x)ψ
=E (3.20)
where E is a constant (and now a number). Thus dT
dt + i
ET(t)=0 giving
T(t)∼exp
−i
Et (3.21)
Further, Eq. (3.20) gives us
− 2 2m
d2ψ
dx2 +Vψ=Eψ (3.22)
or
Hψ=Eψ (3.23)
which is essentially an eigenvalue equation. Forψto be “well behaved,” the quan- tity E takes some particular values (see, e.g., Examples 3.1 and 3.2), these are known as the energy eigenvalues and the corresponding forms ofψ are known as eigenfunctions; by “well-behaved” we imply functions which are single valued and square-integrable (i.e.,
|ψ|2dτ should exist).
In Section3.4we will interpret the wave functionψas the probability amplitude;
thereforeψshould be single valued and|ψ(x)|2dx has to be finite for finite values of dx. Thus
dxlim→0|ψ|2dx=0
In practice this is satisfied by demanding thatψ be finite everywhere. We also have the following theorems:
1Whenever we are considering bound states of a system (like those of the hydrogen atom or that of the harmonic oscillator) the Hamiltonian is independent of time; however, for problems such as the interaction of an atom with radiation field, the Hamiltonian is not independent of time (see, e.g., Section 4.7).
3.2 The One-Dimensional Schrödinger Equation 37 Theorem 1 The derivative of the wave function dψ/dx is always continuous as long as the potential energy V(x) is finite, whether or not it is continuous2.
Proof We integrate the Schrödinger equation [Eq. (3.22)] from x–εto x+εto obtain
x+ε
x−ε
d2ψ
dx2dx= −2m 2
x+ε
x−ε
[E−V(x)]ψ(x)dx or
ψ(x+ε)−ψ(x−ε)= −2m 2
x+ε
x−ε
[E−V(x)]ψ(x)dx
Since V(x) is assumed to be finite (it could, however, be discontinuous), the RHS tends to zero asε→0. Thusψ is continuous at any value of x. It is obvious that ψhas to be necessarily continuous everywhere. Alternatively one may argue that if dψ/dx is discontinuous then d2ψ/dx2must become infinite; this will be inconsistent with Eq. (3.22) as long as V(x) does not become infinite.
Theorem 2 If the potential energy function V(x) is infinite anywhere, the proper boundary condition is obtained by assuming V(x) to be finite at that point and car- rying out a limiting process making V(x) tend to infinity. Such a limiting process makes the wave function vanish at a point where V(x)=∞.
Example 3.1Particle in a one-dimensional infinitely deep potential well
We will determine the energy levels and the corresponding eigenfunctions of a particle of massμin a one-dimensional infinitely deep potential well characterized by the following potential energy variation (see Fig.3.1):
+∞ +∞
0 a
Fig. 3.1 Particle in a one-dimensional box
2It may be mentioned that in many texts the continuity ofψand dψ/dx are taken to be axioms.
This is not correct because it follows from the fact thatψ(x) satisfies a second-order differential equation [see Eq. (3.22)]. Indeed, when V(x) becomes infinite, dψ/dx is not continuous.
V(x)=0 for 0<x<a
= ∞ for x<0 and for x>a (3.24)
For 0 < x < a, the one-dimensional Schrödinger equation becomes d2ψ
dx2 +k2ψ(x)=0 (3.25)
where
k2=2μE
2 (3.26)
The general solution of Eq. (3.25) is
ψ (x)=A sin kx+B cos kx (3.27)
Since the boundary condition at a surface at which there is an infinite potential step is thatψis zero (see Theorem 2), we must have
ψ (x=0)=ψ (x=a)=0 (3.28)
Using the boundary condition given by the above equation, we get
ψ (x=0)=B=0 (3.29)
and
ψ (x=a)=A sin ka=0 Thus, either A=0 or
ka=nπ, n=1, 2,. . . (3.30)
The condition A=0 leads to the trivial solution ofψvanishing everywhere; the same is the case for n=0.
If we now use Eq. (3.26), the allowed energy values are therefore given by En= n2π22
2μa2 , n=1, 2, 3,. . . (3.31)
The corresponding eigenfunctions are ψn=
2 a sinnπ
a x 0<x<a
=0 x<0 and x>a
(3.32)
where the factor√2/a is such that the wave functions form an orthonormal set a
0
ψm∗(x)ψn(x)dx=δmn (3.33)
and δkn=1 if k=n
=0 if k=n (3.34)
is known as the Kronecker delta function. It may be noted that whereasψn(x) is continuous everywhere, dψn(x)/dx is discontinuous at x=0 and at x=a. This is because of V(x) becoming infinite at x=0 and at x=a (see Theorem 1). Figure3.2gives a plot of the first three eigenfunctions and one can see that the eigenfunctions are either symmetric or antisymmetric about the line x=a/2; this follows from the fact that V(x) is symmetric about x=a/2 (see Problem 3.2).
3.2 The One-Dimensional Schrödinger Equation 39
ψ1(x) ψ2(x)
ψ3(x)
+∞ +∞
Fig. 3.2 The energy eigenvalues and
eigenfunctions for a particle in an infinitely deep potential well. Notice that the eigenfunctions are either symmetric or antisymmetric about x=a/2
The following points are also to be noted
(i) E cannot be negative because if we assume E to be negative then the boundary conditions at x=0 and x=a cannot be simultaneously satisfied.
(ii) The eigenvalues form a discrete set.
(iii) The eigenfunctions given by Eq. (3.32) form a complete set, i.e., an arbitrary (well-behaved) function f(x) (in the domain 0 < x < a) can be expanded in terms of the eigenfunctions of H:
f (x)= n
cnψn(x)= 2
a ∞ n=1,2,...
cnsinnπ a x
(3.35)
where cnare constants which can be determined by multiplying both sides of the above equation byψm∗(x) and integrating from 0 to a to obtain
a 0
ψm∗(x)f (x)dx= n
cn a 0
ψm∗(x)ψn(x)dx= n
cnδmn=cm (3.36)
where we have used the orthonormality condition given by Eq. (3.33).
(iv) The most general solution of the time-dependent Schrödinger equation i∂
∂t =H= −2 2μ
∂2
∂x2 +V(x)(x, t) (3.37)
with V(x) given by Eq. (3.24) will be (x, t)=
∞ n=1,2,...
cnψn(x)e−iEnt/= ∞ n=1,2,...
cnψn(x)e−in2τ (3.38)
where
τ= t t0
; t0=2μL2
π2 (3.39)
Substituting forψn(x) and Enwe get (x, t)= ∞
n=1,2,...
cn 2
asinnπx a
exp
−in2π2 2μa2 t
(3.40)
Since
(x, 0)= n
cnψn(x) (3.41)
the coefficients cncan be determined from the initial form of the wave function:
cn= a 0
ψn∗(x)(x, 0)dx (3.42)
Thus, the recipe for determining(x, t) is as follows:
If we know(x, 0) we can determine cnfrom Eq. (3.42), we substitute these values in Eq. (3.38), and sum the series to obtain(x, t).
(v) We assume(x, 0) to be normalized:
a 0
| (x, 0)|2dx=1 (3.43)
This would imply
1= a 0
n
c∗nψn∗(x) m
cmψm(x)dx= n
n
c∗ncm a 0
ψn∗(x)ψm(x) dx
= n
m
c∗ncmδmn = n
|cn|2
(3.44)
where we have used the orthonormality condition given by Eq. (3.33). Further, a
0
| (x, t)|2dx= n
m
c∗ncmei(En−Em)t
a 0
ψn∗(x)ψm(x)dx
= n
m
c∗ncmei(En−Em)t
δmn= n
|cn|2=1
Thus, if the wave function is normalized at t=0, then it will remain normalized at all times.
Further, we can interpret Eq. (3.44) by saying that|cn|2represents the probability of finding the system in the nth eigenstate which remains the same at all times. Thus there are no transitions.
Indeed, whenever the potential energy function is time-independent, we obtain what are known as stationary states and there is no transition between states.
(vi) As a simple example, let us assume that the particle is described by the following wave function (at t=0):
(x, 0)= 1
6 ψ1(x)+ i
√2ψ2(x)+ 1
3ψ4(x)
Notice that
n
|cn|2=1
so that the wave function is normalized. Thus, if we carry out a measurement of energy, the probabilities of obtaining the values E1, E2, and E4 would be 1/6,1/2, and 1/3, respectively.
How will such a state evolve with time? Well, we just multiply each term by the appropriate time-dependent factor to obtain(x, t) [see Eq. (3.38)].
3.2 The One-Dimensional Schrödinger Equation 41 (x, t)=
1
6ψ1(x)e−it/t0+ i
√2ψ2(x)e−4it/t0+ 1
3ψ3(x)e−16it/t0 (3.45) where t0is given by Eq. (3.39). Obviously
a 0
|(x, t)|2dx=1 (3.46)
for all values of t. The quantity
P(x, t)= |(x, t)|2 (3.47)
would represent the time evolution of the probability distribution function. However, at all values of time, the probability of finding the system in a particular state remains the same. Further, the average value of the energy is given by
E =1 6E1+1
2E2+1 3 E4
= 1
6 +2+16 3
π22 2μa2 = 15
2 π22
2μa2 (3.48)
Thus, if one carries out a large number of measurements (of energy) on identically prepared systems characterized by the same wave function as given by Eq. (3.45), then the average value of the energy would be given by Eq. (3.48).
(vii) What happens to the wave function if E=En, i.e., if E is not one of the eigenvalues? For such a case the boundary conditions cannot be satisfied and therefore it cannot be an allowed value of energy. For example, if
E= 0.81π22 2μa2 then the wave function cannot be zero for both x=0 and x=a.
Example 3.2The linear harmonic oscillator
We next consider the linear harmonic oscillator problem where the potential energy function is given by V(x) = 1
2μ ω2x2 (3.49)
and the Schrödinger equation [Eq. (3.22)] can be written in the form d2ψ
dξ2 + −ξ2
ψ=0 (3.50)
whereξ= αx and we have chosen
α= μω
(3.51)
so that
= 2E ω
For the wave function not to blow up at x=± ∞(which represents the boundary condition),must be equal to an odd integer (see Appendix A), i.e.,
= 2E
ω=(2m+1) ; m=0, 1, 2, 3,. . .. (3.52) The above equation would give us the following expression for the discrete energy eigenvalues:
E=Em=
m+1
2 ω, m=0, 1, 2, 3,. . .. (3.53) The corresponding normalized wave functions are the Hermite–Gauss functions (see Appendix A):
ψm(ξ)=NmHm(ξ) exp
−1
2ξ2 , m=0, 1, 2, 3,. . .. (3.54) where
Nm= α
π1/22mm! 1/2
(3.55) represents the normalization constant. The first few Hermite polynomials are
H0(ξ)=1, H1(ξ)=2ξ
H2(ξ)=4ξ2−2, H3(ξ)=8ξ3−12ξ,. . . (3.56) The wave functions form a complete set of orthonormal functions:
+∞
−∞
ψk∗ψndx=δkn (3.57)
The most general solution of the time-dependent Schrödinger equation [Eq. (3.16) with V(x) given by Eq. (3.49)] will be
(x, t)= ∞ n=0,1,2,...
cnψn(x)e−iEnt/
= ∞ n=0,1,2,...
cnψn(x)e−i n+12
ωt (3.58)
The values of cnwill be determined by the initial state of the oscillator.