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Gaussian Wave Packets The Free Particle Gaussian Wave Packets The Gaussian wave packet initial state is one of the few states for which both the {|x i} and {|p i} basis representations are simple analytic functions and for which the time evolution in either representation can be calculated in closed analytic form. It thus serves as an excellent example to get some intuition about the Schr¨odinger equation. We define the {|x i} representation of the initial state to be 2 „ «1/4 − x 1 i p x 2 ψ (x, t = 0) = hx |ψ(0) i = e 0 e 4 σx (5.10) x 2 ~ 2 π σx √ The relation between our σx and Shankar’s ∆x is ∆x = σx 2. As we shall see, we 2 2 choose to write in terms of σx because h(∆X ) i = σx . Section 5.1 Simple One-Dimensional Problems: The Free Particle Page 292 The Free Particle (cont.) Before doing the time evolution, let’s better understand the initial state. First, the symmetry of hx |ψ(0) i in x implies hX it=0 = 0, as follows: Z ∞ hX it=0 = hψ(0) |X |ψ(0) i = dx hψ(0) |X |x ihx |ψ(0) i −∞ Z ∞ = dx hψ(0) |x i x hx |ψ(0) i −∞ Z ∞ „ «1/2 x2 1 − 2 = dx x e 2 σx = 0 (5.11) 2 −∞ 2 π σx because the integrand is odd. 2 Second, we can calculate the initial variance h(∆X ) it=0: 2 Z ∞ „ «1/2 − x 2 2 2 1 2 2 h(∆X ) i = dx `x − hX i ´ e 2 σx = σ (5.12) t=0 t=0 2 x −∞ 2 π σx where we have skipped a few steps that are similar to what we did above for hX it=0 and we did the final step using the Gaussian integral formulae from Shankar and the fact that hX it=0 = 0. Section 5.1 Simple One-Dimensional Problems: The Free Particle Page 293 The Free Particle (cont.) We calculate the {|p i}-basis representation of |ψ(0) i so that calculation of hPit=0 2 and h(∆P) it=0 are easy (by contrast, Shankar Example 4.2.4 does this in the {|x i} basis): Z ∞ ψp(p, t = 0) = hp |ψ(0) i = dx hp |x ihx |ψ(0) i −∞ 2 Z ∞ „ «1/4 − x 1 − i p x 1 i p x 2 = √ dx e e 0 e 4 σx ~ 2 ~ 2 π ~ −∞ 2 π σx 2 !1/4 (p−p0) 1 − 2 = e 4 σp (5.13) 2 2 π σp 1 1 where the √ comes from the normalization |p i = √ |k i, where σp ≡ ~ , and the ~ ~ 2 σx final step is done by completing the square in the argument of the exponential and R ∞ −u2 √ using the usual Gaussian integral −∞ du e = π. With the above form for the 2 {|p i}-space representation of |ψ(0) i, the calculation of hPit=0 and h(∆P) it=0 are 2 calculationally equivalent to what we already did for hX it=0 and h(∆X ) it=0, yielding 2 2 hPit=0 = p0 h(∆P) it=0 = σp (5.14) Section 5.1 Simple One-Dimensional Problems: The Free Particle Page 294 The Free Particle (cont.) We may now calculate |ψ(t) i. Shankar does this only in the {|x i} basis, but we do it in the {|p i} basis too to illustrate how simple it is in the eigenbasis of H. The result is of course 2 !1/4 (p−p0) − p2 1 4 σ2 − i t ψ (p, t) = hp |ψ(t) i = e p e 2 m (5.15) p 2 ~ 2 π σp That is, each {|p i} picks up a complex exponential factor for its time evolution. It is immediately clear that hPi and h(∆P)2i are independent of time. Calculationally, this occurs because P, and (∆P)2 simplify to multiplication by numbers when acting on |p i states and the time-evolution complex-exponential factor cancels out because the two expectation values involve hψ | and |ψ i. Physically, this occurs because the P operator commutes with H; later, we shall derive a general result about conservation of expectation values of operators that commute with the Hamiltonian. Either way one looks at it, one has 2 2 2 hPit = hPit=0 = p0 h(∆P) it = h(∆P) it=0 = σp (5.16) Section 5.1 Simple One-Dimensional Problems: The Free Particle Page 295 The Free Particle (cont.) Let’s also calculate the {|x i} representation of |ψ(t) i. Here, we can just use our propagator formula, Equation 5.6, which tells us Z ∞ 0 0 ψx (x, t) = hx |ψ(t) i = dx [U(t)]xx 0 hx |ψ(0) i −∞ (x 0)2 Z ∞ r m (x−x 0)2 „ «1/4 − 0 m i 1 i p x 0 2 = dx e 2t e 0 e 4 σx ~ 2 ~ −∞ 2 π ~ i t 2 π σx 2 3 ` p ´2 »q „ i t «–−1/2 x − 0 t = 2 π σ2 1 + ~ exp 6− m 7 x 2 “ ” 2 m σ 4 2 i ~ t 5 x 4 σx 1 + 2 2 m σx „ « „ 2 « i i p0 exp p0 x exp − t (5.17) ~ ~ 2 m where we do the integral in the usual fashion, by completing the square and using the Gaussian definite integral. Section 5.1 Simple One-Dimensional Problems: The Free Particle Page 296 The Free Particle (cont.) The probability density in the {|x i} basis is P(x) = |hx |ψ(t) i|2 2 3 −1/2 " „ «2!# `x − p0 t´2 2 ~ t 6 m 7 = 2 π σ 1 + exp 6− 7 (5.18) x 2 m σ2 „ “ ”2« x 4 2 ~ t 5 2 σx 1 + 2 2 m σx p0 Because the probability density is symmetric about x = m t, it is easy to see that p0 p0 hX it = t = hX it=0 + t (5.19) m m i.e., the particle’s effective position moves with speed p0/m, which is what one expects for a free particle with initial momentum p0 and mass m. Section 5.1 Simple One-Dimensional Problems: The Free Particle Page 297 The Free Particle (cont.) The variance of the position is given by the denominator of the argument of the Gaussian exponential (one could verify this by calculation of the necessary integral), " # " # „ t «2 „ t «2 h(∆X )2i = σ2 1 + ~ = h(∆X )2i 1 + ~ (5.20) t x 2 t=0 2 2 m σx 2 m σx The position uncertainty grows with time because of the initial momentum uncertainty of the particle – one can think of the {|p i} modes with p > p0 as propagating faster than p0/m and those with p < p0 propagating more slowly, so the initial wavefunction 2 spreads out over time. In the limit of large time (t 2 m σx /~), the uncertainty q 2 h(∆X ) it grows linearly with time. The “large time” condition can be rewritten in a more intuitive form: 2 σx σx σx t t0 = 2 m = m = (5.21) ~ σp σv where σv = σp/m is the velocity uncertainty derived from the momentum uncertainty. So, t0 is just the time needed for the state with typical velocity to move the width of the initial state. We should have expected this kind of condition because σx and ~ are the only physical quantities in the problem. Such simple formulae can frequently be used in quantum mechanics to get quick estimates of such physical phenomena; we shall make such use in the particle in a box problem. Section 5.1 Simple One-Dimensional Problems: The Free Particle Page 298 The Free Particle (cont.) Position-Momentum Uncertainty Relation Before leaving the free particle, we note an interesting relationship that appeared along the way. Recall that, because the position and momentum operators do not commute, [X , P] = i ~, no state is an eigenstate of both. If there is no uncertainty in one quantity because the system is in an eigenstate of it, then the uncertainty in the other quantity is in fact infinite. For example, a perfect position eigenstate has a delta-function position-space representation, but it then, by the alternative representation of the delta function, Equation 3.146, we see that it is a linear combination of all position eigenstates with equal weight. The momentum uncertainty will be infinite. Conversely, if a state is a position eigenstate, then its position-space representation has equal modulus everywhere and thus the position uncertainty will be infinite. Section 5.1 Simple One-Dimensional Problems: The Free Particle Page 299 The Free Particle (cont.) When we considered the Gaussian wave packet, which is neither an eigenstate of X nor of P, we found that the t = 0 position and momentum uncertainties were 2 h(∆X )2i = σ2 h(∆P)2i = σ2 = ~ (5.22) t=0 x t=0 p 2 4 σx Hence, at t = 0, we have the uncertainty relation q q 2 2 ~ h(∆X ) it=0 h(∆P) it=0 = (5.23) 2 We saw that, for t > 0, the position uncertainty grows while the momentum uncertainty is unchanged, so in general we have q q h(∆X )2i h(∆P)2i ≥ ~ (5.24) 2 We will later make a general proof of this uncertainty relationship between noncommuting observables. Section 5.1 Simple One-Dimensional Problems: The Free Particle Page 300 The Particle in a Box The Hamiltonian A “box” consists of a region of vanishing potential energy surrounded by a region of infinite potential energy: L 0 |x| ≤ 2 V (x) = lim L (5.25) V →∞ 0 V0 |x| > 2 It is necessarily to include the limiting procedure so that we can make mathematical sense of the infinite value of the potential when we write the Hamiltonian.
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