Problem: Evolution of a free Gaussian wavepacket Consider a free particle which is described at t = 0 by the normalized Gaussian wave function µ ¶1=4 2a 2 ª(x; 0) = e¡ax : ¼ The normalization factor is easy to obtain. We wish to ¯nd its time evolution. How does one do this ? Remember the recipe in quantum mechanics. We expand the given wave func- tion in terms of the energy eigenfunctions and we know how individual energy eigenfunctions evolve. We then reconstruct the wave function at a later time t by superposing the parts with appropriate phase factors. For a free particle H = p2=(2m) and therefore, momentum eigenfunctions are also energy eigenfunctions. So it is the mathematics of Fourier transforms! It is straightforward to ¯nd ©(k; 0) the Fourier transform of ª(x; 0): Z 1 dk ª(x; 0) = p ©(k; 0) eikx : ¡1 2¼ Recall that the Fourier transform of a Gaussian is a Gaussian. Z 1 dx Á(k) ´ ©(k; 0) = p ª(x; 0) e¡ikx (1.1) ¡1 2¼ µ ¶ µ ¶ 1=4 Z 1 1=4 2 1 2a ¡ikx ¡ax2 1 ¡ k = p dx e e = e 4a (1.2) 2¼ ¼ ¡1 2¼a What this says is that the Gaussian spatial wave function is a superposition of di®erent momenta with the probability of ¯nding the momentum between k1 and k1 + dk being pro- 2 portional to exp(¡k1=(2a)) dk. So the initial wave function is a superposition of di®erent plane waves with di®erent coef- ¯cients (usually called amplitudes). Note that exp ikx for each real k is also an eigenfunction ¹h2k2 of the Hamiltonian with eigenvalue Ek = 2m . Therefore, we know that exp(ikx) evolves in time with a time dependence given by e¡iEkt=¹h . So the time-evolved state, ª(x; t), is given by Z 1 dk ª(x; t) = p Á(k) eikx e¡iEkt=¹h (1.3) ¡1 2¼ µ ¶ 1=4 Z 1 2 1 dk ¡ k ¡iE t=¹h ikx = p e 4a e k e (1.4) 2¼a ¡1 2¼ where we know Ek and this integral has to be done. The exponential has an argument given by µ ¶ k2 2i¹hat bk2 1 2i¹hat ¡ [ 1 + ] = ¡ where b = 1 + : 4a m 2 2a m 1 Thus the integral to be done is µ ¶1=4 Z 1 1 dk 2 p e¡bk =2 eikx 2¼a ¡1 2¼ and this can be done in the usual way in spite of the fact that b is complex since the integral is well-de¯ned. We obtain, using r Z 1 2 ¡bk2=2 ikx 2¼ ¡ x dk e e = e 2b ; (1.5) ¡1 b µ ¶ r 1=4 2 1 1 ¡ x the result ª(x; t) = e 2b (1.6) 2¼a b µ ¶ 2a 1=4 1 ax2 = q exp[ ¡ ] : (1.7) ¼ 2i¹hat (1 + 2i¹hat ) 1 + m m The probability density is given by (after a wee bit of algebra) r 2 ¡ 2ax 2 2a 1 2¹hat 2 jª(x; t)j = q e 1+( m ) (1.8) ¼ 2¹hat 2 1 + ( m ) from which we can read o® " µ ¶ # 1 2¹hat 2 hx2i = 1 + : (1.9) 4a m 1 Note that the initial width 4a increases quadratically as a function of time. To understand this result it is useful to rewrite it. We note that since < x >= 0 we have 2 2 the initial < x > is 1=(4a). Let us call this σ0 = 1=(4a), in a great notational leap forward. 2 So the width at time t is given by (chasing the factors of a and replacing them by σ0=2) 2 2 2 2 2 ¹h t hx it ´ σx(t) = σ0 + 2 2 : (1.10) 4m σ0 Now we note that the uncertainty in the momentum can be read o® from the wave function in k-space. Since Á(k) is Gaussian we ¯nd 2 2 2 2 ¹h hk i = a and hence σp = ¹h a = 2 4σ0 1 where we have rewritten a in terms of σ0 and used the de Broglie relation. σ2t2 Therefore, σ2(t) = σ2 + p : (1.11) x 0 m2 1Why can the initial wave function be called a minimum uncertainty wave packet? How does the momentum uncertainty change as a function of time ? Why ? 2 So the initial uncertainty in the position increases with an extra part which can be thought of as follows: σpt=m is the distance traveled by a classical particle with momentum equal to the uncertainty σp in a time t. So for t << σ0=(σp=m) the width remains unchanged. Given an arbitrary initial wave function ª(x; 0) for a free particle we can ¯nd ª(x; t). First ¯nd its decomposition in terms of Á(k): 2 Z 1 dx ª(x; 0) = p Á(k) eikx : ¡1 2¼ Then each plane wave, an eigenfunction of the free particle of the Hamiltonian with eigenvalue ¡iE t=¹h Ek, evolves in time and at time t picks up a phase factor e k . Therefore we have Z 1 dk ª(x; t) = p Á(k) eikx e¡iEkt=¹h (1.12) ¡1 2¼ Z Z 1 dk 1 dy = p eikx e¡iEkt=¹h p e¡ikyª(y; 0) replacing Á(k) (1.13) ¡1 2¼ ¡1 2¼ Z ·Z ¸ 1 1 dk = dy eik(x¡y) e¡iEkt=¹h ª[y; 0] (1.14) ¡1 ¡1 2¼ Z 1 = dy G(x; t; y; 0) ª[y; 0] (1.15) ¡1 where we have de¯ned the corresponding Green's function for the SchrÄodingerequation: Z ·Z ¸ 1 1 dk G(x; t; y; 0) ´ dy eik(x¡y) e¡iEkt=¹h : ¡1 ¡1 2¼ Note that given the function ª(y; 0) (y is just a dummy variable), G tells you how to propagate the system in time and obtain the function ª(x; t) at time t. G is known as the free-particle propagator. It is also called the Feynman propagator for a non-relativistic free particle. Performing the integral we ¯nd the explicit expression r 2 m i m(x¡y) G(x; t; y; 0) = e 2¹ht 2¼i¹ht for t > 0. As t ! 0 this goes over to a delta function as one can check. Finding the Green's function for other problems such as the quantum harmonic oscillator is somewhat tedious. (See R. P. Feynman and A. R. Hibbs, \Quantum Mechanics and Path Integrals", unfortunately, poorly proof-read.) 2Á(k) can be found as usual by the inverse Fourier transform: Z 1 dy Á(k) = p ª(y; 0) e¡iky : ¡1 2¼ 3.