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Informal Introduction to QM: Free Particle

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Informal Introduction to QM: Free Particle Chapter 2 Informal Introduction to QM: Free Particle Remember that in case of light, the probability of nding a photon at a location is given by the square 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 wave 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 function represents the probability amplitude of photons with energy ω and momentum k. 2. Superposition principle 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 waves will be called wave functions. £ First, the wave function 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 probability distribution 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 mechanics. 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| <h/∆p0.Thus∆x∆p0 ∼ h/2. 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 wave packet 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 group velocity as against the phase velocity 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 wave equation. 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.
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