SJP QM 3220 Ch. 2 part 3. Page 1 (S. Pollock, taken from M. Dubson) with thanks to J. Anderson for typesetting. Fall 2008 The Free Particle: Consider V(x) = 0. No PE, just a free particle. We talked about this at the start, (de Broglie's ideas). Now we can tackle it with the Schrödinger Equation. There is some funny business here, but clearly this too is an important physics case, we need to describe quantum systems that aren't bound! )()('' xuExum⋅=−22 That looks easy enough! Define )()('' xukxumEk2so ,2−==  I can solve that 2nd order ODE by inspection… ikxikxkBeAexu−+=)( (The label "k" identifies u, different k's  different u's) There is no boundary condition, so k (thus E) is not quantized. Free particles can have any energy! And Ψk(x,t ) = (Aeikx+ Be−ikx)e−iEt / (Let's define ω= E / , as usual) )/ mkE 2(with 22= Rewriting: Ψk(x,t ) = Aeik (x −ωkt )this is a fn of x-vtright-moving wave!speed v=ω/k  + Be−ik(x +ωkt )this is a fn of x+vtleft moving wave!speed ω/k   (The speed of a simple plane wave is also called "phase velocity". Go back to our chapter 1 notes on classical waves to see why v=ω/k) Here, v =ωk=E / k= 2k2/ 2mk=k2m Is that right? What did we expect? I'm thinking p = k* so v =pm= km * [ For the function Ψk(x,t ) = Aei (kx −k22 mt ), you get ˆpΨk=i∂∂xΨk= kΨk So this Ψk is an eigenfunction of ˆpwith eigenvalue k. Just exactly as our "de Broglie intuition" says . (The Ψk at the top of the page, (with k > 0 only) is a mix of +k and -k, which means a mix of right moving and left moving plane waves….)] But wait, What's with that funny factor of 2, then? Why did I get v =k2m? We're going to need to consider this more carefully, coming soon!SJP QM 3220 Ch. 2 part 3. Page 2 (S. Pollock, taken from M. Dubson) with thanks to J. Anderson for typesetting. Fall 2008 Summary So we have solution Ψk(x,t ) = Aei (kx −ωt ) where k = ±2mE (Note: 2 values of k for each E, one plus, one minus.) These waves have λ=2πk , because Ψ(x ±λ) = Ψ(x) The speed (phase = k2m=h2mλvelocity) (Small λ  faster speed) There's that funny "2" we saw above. Classically, I expected v =pm=km=2Emwithout the factor of 2. Funny v (factor of 2 curiosity) is one issue we need to understand. Also, here's a wave function which is not renormalizable! € Ψk(x,t)2dx = A21⋅ dx−∞∞∫−∞∞∫  yikes! Conclusion from this: This Ψk is not representing a physical object! There are no free (quantum) particles with definite energy! But … not to worry, these Ψk's are useful, even essential! We can form linear combos that are physical! k > 0 or k < 0 Re _ (k) k > 0 or k < 0 ψkSJP QM 3220 Ch. 2 part 3. Page 3 (S. Pollock, taken from M. Dubson) with thanks to J. Anderson for typesetting. Fall 2008 Since k is a continuous variable, combining Ψk's requires not a sum over k, but an integral over k! Consider € Ψgeneralfree(x,t) =12πφ(k)Ψk(x,t)dk−∞∞∫ This is the continuous analogue of our old familiar "Fourier" formula: € Ψgeneralsquare well(x,t) = cnn∑Ψnsquare well(x,t) , where we basically replace € cn→φ(k)dk  This Ψ has many k's (momenta, energies) superposed.  If I give you € Ψgeneral(x,0) =12πφ(k)Ψk(x,0)dk−∞∞∫ then you can figure out φ(k) (like Fourier's trick!) and given φ(k), you then know, from equation at top of page, Ψ at all future times. So we need to review "Fourier's trick" for continuous k! Note Ψk(x,0) = eikx  simple enough … Suppose I give you f(x), and ask "what φ(k) is needed" so that€ f (x) =12πφ(k)eikxdx−∞∞∫ This is our problem: Given Ψgen(x, t=0), find φ(k). Because once you've got it, you know Ψ at all times. stuck this in for later convenience This plays the role of cn before, it's a function of k Sum over all k's, + or - .SJP QM 3220 Ch. 2 part 3. Page 4 (S. Pollock, taken from M. Dubson) with thanks to J. Anderson for typesetting. Fall 2008 The answer is Plancherel's Theorem, also known as the "Fourier Transform". € φ(k) =12πf (x)e−ikxdx−∞∞∫ (Note the minus sign in the exponential here!) Recall that if € f (x) = cn2asinnπxan∑ ,then Fourier's trick gave € cn= f (x)∫2asinnπxadx It's rather similar; the coefficients are integrals of the desired function with our "orthogonal basis" functions.  φ(k) is the "Fourier transform" of f(x) here.  If f(x) (here Ψ(x, t=0) is normalizable to start with, Ψ(x,t) will be too, as will φ(k).  You can "build" almost ANY FUNCTION f(x) you want this way!! Digression – Some examples of Fourier transforms. Griffiths (p. 62) has "width" is € ≈ 2a. then € φ(k) =12πf (x)e−ikxdx−∞∞∫is easy, you get € φ(k) =1πasin kak (Do it, try it yourself!) If a → 0, Need many momenta to build a narrow wave packet If a → ∞, One (sharp) momentum  broad wave packet, where is it? -a -a f(x) x π/a is "width", roughly f(x) localized φ(k) ~ constant spread out f(x) spread outSJP QM 3220 Ch. 2 part 3. Page 5 (S. Pollock, taken from M. Dubson) with thanks to J. Anderson for typesetting. Fall 2008 Digression continued: Let's do another: € f (x) = Ae−x2/ 4α Here € 12πf (x)e−ikxdx−∞∞∫ can be done analytically (guess what's on the next homework?) € φ(k) =A2πe−x24α+ikx )      dx =−∞∞∫A2πe−k2αe−k2αe−x2α − ikα      2dx−∞∞∫ (That little trick on the right is called "completing the square", you need to DO that algebra yourself to see exactly how it works, it's a common and useful trick) Basically, letting € x' =x2α− ikα ; dx'=dx2α ; e-x' 2dx'=π-∞∞∫ you then get € φ(k) =2α2Ae−k2α Once again, narrow in f  wide in φ, and vice versa. ▪ If f is centered around x0, € f (x) = Ae−(x − x0)2/ 4α work it out… This will add (well, really multiply!) a phase € eikx0 to φ(k). So the phase of φ(k) does carry some important information (but not about momentum itself!) (Puzzle for you: what does a phase € eik0x multiplying f(x) do? …) width ~ 2α width ~ 1/√αSJP QM 3220 Ch. 2 part 3. Page 6 (S. Pollock, taken from M.