Chapter 2

Informal Introduction to QM:

Remember that in case of , the of nding a at a location is given by the of the square of electric eld at that point. And if there are no sources present in the region, the components of the electric eld are governed by the equation (1D case only) ∂2u 1 ∂2u − =0 (2.1) ∂x2 c2 ∂t2 Note the features of the solutions of this dierential equation:

1. The simplest solutions are harmonic, that is

u ∼ exp [i (kx − ωt)]

where ω = c |k|. This represents the of with ω and k.

2. holds, that is if u1 = exp [i (k1x − ω1t)] and u2 = exp [i (k2x − ω2t)] are two solutions of equation 2.1 then c1u1 + c2u2 is also a solution of the equation 2.1. 3. A general solution of the equation 2.1 is given by

ˆ ∞ u = A(k) exp [i (kx − ωt)] dk. −∞

Now, by analogy, the rules for matter particles may be found. The functions representing matter will be called wave functions.

£ First, the ψ(x, t)=A exp [i(px − Et)/]

8 represents a particle with momentum p and energy E = p2/2m. Then, the probability density function P (x, t) for nding the particle at x at time t is given by

P (x, t)=|ψ(x, t)|2 = |A|2 .

Note that the function is independent of both x and t.

£ Assume that superposition of the waves hold. Then the wave function

ψ(x, t)=A exp [i(p1x − E1t)/]+B exp [i(p2x − E2t)/]

2 2 represents a particle with momentum either p1 or p2 with propabilities |A| and |B| respectively. Extending this to the wave function

N ψ(x, t)= An exp [i(pix − Eit)/] n=1 2 represents a particle with momentum pi with probability |Ai| . £ In general, since p is a continuous variable, ˆ 1 ∞ ψ(x, t)=√ A(p) exp [i(px − Et)/] dp 2π −∞

represents a particle with momentum p with probability density function |A (p)|2. Example 1. Let 1 1 ψ(x, t)=√ exp [i(px − Et)/]+√ exp [i(−px − Et)/] . 2 2 The pdf for nding particle at x at time t is

P (x, t)=ψ∗(x, t)ψ(x, t)   2px = 1 + cos .  Now this pdf is not what one expects in classical . There are some points, in the vicinity of which, the probability of nding particle is 0!

Now the wavefunction of a particle with denite momentum presents a problem. The proba- bility density function P (x, t) is not integrable. Thus the net probability of nding a particle somewhere is innite. One way to look at this is to say that P (x, t) represents the relative probability and not the absolute. Really speaking, one does not nd harmonic waves in nature. What one encounters are wave trains or wave pulses. Think of water ripples when a stone is dropped in it. Think of particles

9 which make tracks in bubble chambers. It is always known that particles is in some region of space. This can be put in mathematical terms as

ˆ ∞ P (x, t)dx < ∞ −∞ or ˆ ∞ |ψ(x, t)|2 dx < ∞ −∞ Here are the rules for the wave function of a free particle.

1. A free particle will be described by a square integrable function called as wave function or probability amplitude. The absolute square of the wave function is proportional to the probability of nding the particle at a location at an instant.

2. The wave function ˆ 1 ∞ ψ(x, t)=√ A(p) exp [i(px − E(p)t)/] dp 2π −∞ where A(p) is a square integrable function, represents a free particle with momentum p with probability density function |A(p)|2and energy E(p)=p2/2m.

Wave Packets

Thus, the particles with reasonably sharply dened momentum may be described by a pulse like wave functions. Consider a particle with momentum p0 with uncertainty of ∆p0 ( p0). If the wave function of the particle is ˆ 1 ∞ ψ(x, t)=√ A(p) exp [i(px − E(p)t)/] dp 2π −∞ then the form of A(p) must be a sharply peaked function about p0 (see gure). (For present argument, assume that A(p) is real.)

p p0

Claim 2. If A(p) is a sharply peaked function about p0 with uncertainty ∆p0, then the wave function ψ is also localized. (such wave functions will be called wave packets).

10 To justify the claim, rst, let t =0. It is the behaviour of exp (ipx/) that must be investigated as a function of p near p0. Now period of exp (ipx/) is h/|x|.

h £ Case 1: When |x| , then ∆p0  h/|x|. That is ∆p0 is very large compared to ∆p0 the period of exp (ipx/). Thus this function has highly oscillatory behaviour near p0 and hence ˆ 1 ∞ ψ(x, 0) = √ A(p) exp [i(px)/] dp ≈ 0 2π −∞

h £ Case 2: When |x| , then ∆p0  h/|x|. That is ∆p0 is very small compared to ∆p0 the period of exp (ipx/). Thus this function is nearly constant near p0 and hence ˆ 1 ∞ ψ(x, 0) = √ A(p) exp [i(px)/] dp 2π −∞ ˆ 1 ∞ = √ exp [i(p0x)/] A(p)dp 2π −∞

Thus ψ is signicantly nonzero only for |x|

p p p0 p0

p p p0 p0

p p p0 p0

Figure 2.1: The three graphs in the rst column are that of A(p), cos(px/) and the product of the two when x  h/∆p0. The second column contains the same graphs, but for x  h/∆p0.

11  Example 3. Let A(p)=1/ ∆p if |p − p0| < ∆p/2. Then the wave functin is given by ˆ 1 ∞ ψ(x, t)=√ A(p) exp [i(px)/] dp 2π  −∞ 2 ∆ sin (∆ x/) = √ p p 2π ∆px/ The plots are shown in the gure.

Ap Ψx

x

p

To analyze the wavefunction ψ(x, t), when t =0, remember E is a function of p. Let β = (px − Et)/ be the phase of the exponential function. If β changes rapidly, then exponential function is oscillatory. Alternatively if β is stationary, exponential function is nearly constant. Thus, the peak of the ψ is at x0 if ∂β (p )=0 ∂p 0 ∂E when x − (p ) t =0 0 ∂p 0 This means that the peak of the wave packet is moving with speed ∂E v = (p ) . g ∂p 0 This is called as as against the of the wave which is dened as E/p0. In macroscopic limit, the velocity of the particle is p0/m, this suggests that E(p)= p2/2m. This ts nicely. Now,

E(p)=E(p0)+vg(p − p0)+··· Thus if t is suciently small, then ˆ 1 ∞ i(p0x−E(p0)t)/ ψ(x, t)=√ e A(p) exp [i(p − p0)(x − vgt)/] dp 2π −∞

And compare this with ˆ 1 ∞ i(p0x/) ψ(x, 0) = √ e A(p) exp [i(p − p0)x/] dp 2π −∞ the wave packet just moves without changing shape with speed vg.

12 Gaussian Wave Packets

Now let     1 2 1 2 (p − p ) A(p)= √ exp − 0 . π∆ 2∆2 p p Note that this function is normalized to 1, that is, ˆ ∞ |A (p) |2dp =1. −∞ The wave packet is     1 ˆ ∞ 2 1 1 2 (p − p ) px ψ(x, 0) = √ √ exp − 0 + i dp π∆ 2∆2  2π p −∞ p   1   ∆ 2 p x ∆ x2 = √ p exp i 0 exp − p π  22 (See footnote1) The following gures show the plot of Gaussian wavepacket.

Re Ψ Ψ

x x

Figure 2.2: The left gure shows Reψ and right hand gure shows |ψ|2.

Dierential Equation (Schrödinger Equation)

If the wave function at time t =0is known, that is ψ(x, 0) is known for all x, how does one nd ψ(x, t)? Remeber that em waves are governed by a dierential equation called as . What is the dierential equation for matter waves? Here one borrows the classical energy expression E = p2/2m. The wave function ψ(x, t) of a free particle is ˆ 1 ∞ ψ(x, t)=√ A(p) exp [i(px − E(p)t)/] dp. 2π −∞ Then   ˆ ∂ 1 ∞ i ψ(x, t)=√ EA(p) exp [i(px − E(p)t)/] dp ∂t 2π −∞

1 ˆ ∞ 2 π 1/2 2 e−αu e−βudu = eβ /4α −∞ α

13 and   ˆ   2 ∂2 1 ∞ p2 − ψ(x, t)=√ A(p) exp [i(px − E(p)t)/] dp 2m ∂x2 2π 2m ˆ−∞ 1 ∞ = √ EA(p) exp [i(px − E(p)t)/] dp 2π −∞ since p2/2m = E. Thus,     ∂ 1 ∂ 2 i ψ(x, t)= −i ψ(x, t). ∂t 2m ∂x This is called Schrödinger equation. Note this rst ordered dierential equation in t.Thus knowledge of ψ(x, 0) is enough obtain ψ(x, t).2 It was implicitly assumed that A(p) is time-independent. In classical case, momentum of a free particle does not change in time. Simillary, in QM, the pdf of momentum does not change in time and so does A(p). Then there is another way to nd ψ(x, t) from ψ(x, 0):

1. Given ψ(x, 0), compute momentum space wave function: ˆ 1 ∞ A(p)=√ ψ(x, 0) exp [−i(px)/] dx 2π −∞

2. Then ˆ 1 ∞ ψ(x, t)=√ A(p) exp [i(px − Et)/] dp 2π −∞ Example 4. A wave function of a free particle at t =0,is

B |x|

B is a real constant. Find ψ(x, t). First normalize the wave function. ˆ a |ψ(x, 0)|2 dx =1 −a √ This implies that B =1/ 2a. Now the momemtum space wave function is ˆ 1 ∞ A(p)=√ ψ(x, t) exp [−i(px)/] dx 2π  −∞ a pa = sinc . π 

2Of course, one also must know ψ(±∞,t), since its a second ordered dierential equation in x. For free particle, the square integrability requires that ψ(±∞,t)=0.

14 Now  ˆ   1 a ∞ pa p2 ψ(x, t)=ψ(x, t)=√ sinc exp i px − t / dp 2π π −∞  2m But this is not easy to evaluate in terms of simple functions. But it can be numer- ically evaluated and plotted. Clearly the average momentum is zero, thus the peak of the wavefunction remains at x =0.|

Expectation Values

Thus, if the wave packet (or wave function)of a particle is known then all other information can be obtained. Let ψ(x, t) is the wave function of a particle.

£ Probability density function for nding the particle at x is

P (x, t)=|ψ(x, t)|2

£ The average of the particle or expectation value of x is

ˆ ∞ x(t) = xP (x, t)dx −∞ ˆ ∞ = ψ ∗ (x, t)[xψ(x, t)] dx −∞

£ Let ˆ 1 ∞ φ(p)=√ ψ(x, t) exp [−i(px − E(p)t)/] dx. 2π −∞ Then probability density function for nding the particle with momentum p is given by

P (p)=|φ(p)|2

£ The expectation value of p is

ˆ ∞ p = p |φ(p)|2 dp ∞ ˆ ∞ = φ∗(p)[pφ(p)] dp ∞ Now, notice that ˆ ∂ 1 ∞ −i ψ(x, t)=√ [pφ(p)] exp [i(px − E(p)t)/] dp ∂x 2π −∞

15 Then ˆ ˆ ˆ 1 ∂   p = dx dxψ∗(x,t) −i ψ(x, t) dp exp[i p(x − x)/ 2π ∂x ˆ ˆ 1 ∂   = dx dxψ∗(x,t) −i ψ(x, t) 2πδ(x − x) 2π ∂x ˆ   ∞ ∂ = ψ(x, t) −i ψ(x, t) dx −∞ ∂x Exercise 5. Show that ˆ   ∞ ∂ E = ψ(x, t) i ψ(x, t) dx. −∞ ∂t

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