3.1 Uncertainties dened

Let us now dene the uncertainty precisely.

If we have a wave of amplitudeA(x), then average of function F (x) over the associated probabilitydistribution A2(x) (i.e. the intensity) is given by

F x =dxF (x)A2(x)

dxA2(x) (15)

and the root mean square (RMS) uncertainty is given by

F =F 2x F 2x (16)

3.1.1 Gaussian wavepacket

Let us take a concrete example, of a Gaussian wave packet where(k) = ( a

)1/4eak2/2 with the prefactor so that

dk (k) = 1.

Here k0 = 0, (k) is real, so x0 = 0 as well.

A(x) = 12

(a)1/4

dk eak

2/2 eikx

Doing this integral, we see that

A(x) = ( 1a

)1/4ex

2/2a

Here notice an interesting feature, if we put a = 1/a, then A(x) = (a)1/4eax2/2, i.e. duality of

gaussian functions. Now let us calculate

kk =dk k 2(k)dk 2(k) = (a/)

1/2

dk k eak2 = 0

k2k =dk k2 2(k) = (a/)1/2

dk k2 eak2 = (a

)1/2

4a3/2 2 =12a

Thus k = 12a .

Similarly xx = 0, x2x = 1/(2a) = a/2 and x =

a2

and kx = 1/2.

A mathematical theorem (not proved here) gives that this is the minimum product of uncertaintiespossible, so in general kx 1/2

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4 The Schrodinger Wave Equation

Let us start with the DeBroglie relations:

p = h/ for photons Apply to electronsThus, p = h/ for electrons, giving the de Broglie wavelength for electrons as = h/p.

So how do we now treat an electron? We dene a wavefunction (x, t) which has certain prop-erties. This wavefunction is the solution to the Schrodinger equation, which is the equation ofmotion of our wave-particle. Usually, we might also say that the particle is the state (x, t).

4.1 The Schrodinger equation

We have seen that a wave can be written as (a plane wave exhibiting interference)

(x, t) ei(kxt)

Let us use E = h, thus = E/h, andp = hk, thus k = p/h

(x, t) e ih (pxEt)

(Remember that we can make different wave functions by superimposing different waves of dif-ferent p and E).

Let us differentiate w.r.t time t and position x.

t= i

hE

= ihp and thus 2 = p

2

h2

For a non-relativistic particle, E = p2/(2m), So dividing the 2 side by 2m and equating thetwo RHS

h22m

2(x, t) = ih t(x, t) (17)

This is the free-particle Schrodinger equation.

If the particle is moving in some potential V (x), then classically we have E = p2/2m+ V (x) andthus we can write h2

2m2 + V (x)

(x, t) = ih

t(x, t) (18)

This Schrodinger equation cannot be derived. It is a basic postulate of quantum mechanics thatthe wave function of a particle satises this equation. Although we motivated this equation usingplane waves, the solutions of this can be completely different from plane wave. Our goal will be

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Figure 4: Figure shows a wavefunction squared (in red) and the area between two points is inter-preted as the probability of nding particle between those points at given time.

to study this equation for various potentials, and try and nd explicit solutions to it, thus giving usthe wavefunctions for particles in that particular potential.

4.2 The wavefunction

Let us come back to look at what this (x, t) is. There is no physical meaning to (x, t). Insteadthe accepted interpretation is that 2 describes the probability distribution for the particle, and (seeFig: 4) b

a|(x, t)|2 dx = Probability of nding particle between a and b, at time t (19)

If this is the interpretation, and we know that the particle exists, then the probability of nding itsomewhere must be one.

|(x, t)|2 dx = 1But wait a minute, nowwe are starting to impose general conditions on the solution of the Schrodingerequation. How do we ensure that a solution also obeys this probability business? Well it turns outthat since Schrodinger eqn is linear, so any multiple of is also a solution. So we take an unnor-malized solution un and turn it into a normalized solution.

norm(x, t) = Nun(x, t)and

|N |2 d3x|un(x, t)|2 = 1Thus,

|N | = 1d3x|un(x, t)|2

(20)

In other words, once we have correctly determined N , we can then use the normalized wavefunc-tion norm and we are assured that the probability condition is met, i.e.

|(x, t)|2 dx = 1.

In the tutorial we shall see that as time progresses, the wavefunction does not lose its normalization.

d

dt

|(x, t)|2 dx =

t|(x, t)|2 dx = 0 (21)

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We shall see d/dt of the probability is zero: if is normalized at some time, it stays normalized atsome other time.

4.3 The average of x,p

Since |(x, t)|2 is the probability distribution, we can write down the average,

x =

x|(x, t)|2dx (22)

Is this the average position of the particle? NO.Rather it is the average of all measurements (of position) made on many many identically preparedsystems.

We write p = mv = mdxdt, so we can evaluatemdx

dt.

dxdt

= ddt

x|(x, t)|2dx =

x

t|2|dx (23)

In the tutorial we shall see that we get

dxdt

= ihm

xdx (24)

that isp = ih

xdx (25)

Customarily, this is how we write things

x =

(x, t)x(x, t)dx (26)

p =

(x, t)ih

x

(x, t)dx (27)

4.4 Position and momentum bases

Recall that we learned earlier, for any wavefunction (x, t) we can calculate its Fourier transform,which is another wavefunction (p, t) as follows

(p, t) = 1(2) 32

d3x(x, t) e ih px (28)

It turns out that if (x, t) normalized, then (p, t) is also normalized. Thus it is natural to interpret|(p, t)|2 as the probability of the particle having momentum p at time t. In analogy to x, we canwrite p as follows

p =d3p p |(p, t)|2 (29)

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which is based on our generalised statement

g(p) =dp g(p) |(p, t)|2 (30)

We have already written

p =

(x, t)ih

x

(x, t)dx (31)

which we arrived at by consideringmdxdt

. Now let us consider the other denition

p =d3p p |(p, t)|2 (32)

which in one dimension can be written as

p =dp p |(p, t)|2 =

dp(p, t) p(p, t) (33)

Examining these two equations, and since for a given system p is the same, we can see theconnection between two representations of p. And moreover (x, t) and (p, t) are related in aspecial way (by a Fourier transform). Let us introduce some terminology to describe this. We referto (x, t) as the wavefunction in the x-basis and (p, t) as the wavefunction in the p-basis.Then, we may say that in the x-basis, the momentum p is represented by the differential operatorih (i.e. in one dimension it is ih

x). In fact, there is also a converse statement that one can

prove. In the p-basis, the position x is represented by the differential operator ihp. Here p isthe gradient with respect to the momentum:

(p)i = pi

, i = 1, 2, 3 (34)

We can choose the basis to work in at our own convenience. We can work permanently in the x-basis, in which the position is just x, but momentum is given byih. At the general level, both xand p are to be thought of as operators. In different bases, their representations are different. Thusin the x-basis, it just happens that position is just a multiplication and momentum a differentiation.

In QM, all observables are operators. And the converse is also true, all Hermitian operators areobservables (We shall dene Hermiticity later).

4.5 The Hamiltonian Operator

Let us now look at a very important operator, which is the QM-analogue of energy. We call it theenergy operator or for historical reasons, the Hamiltonian operator. It is dened as follows

H = p2

2m + V (x)

In the x-basis, we have p ih, and thus p2 = h22,

H = h2

2m2 + V (x)

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and thus the Schrodinger equation looks like

H = ih t

The above discussion allows us to determine the expectation value of the energy of the particle instate . It is just

H =

(x, t)H (x, t) d3x

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