Solution to a Second-Order Ordinary Differential Equation in Quantum Mechanics - Prof. Ste, Study notes of Quantum Mechanics

Download Solution to a Second-Order Ordinary Differential Equation in Quantum Mechanics - Prof. Ste and more Study notes Quantum Mechanics in PDF only on Docsity! 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! )()('' xuExu m ⋅=− 2 2  That looks easy enough! Define )()('' xukxumEk 2so ,2 −==  I can solve that 2nd order ODE by inspection… ikxikx k BeAexu −+=)( (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) = (Ae ikx + Be− ikx )e− iEt / (Let's define ω = E /  , as usual) )/ mkE 2(with 22 = Rewriting: Ψk (x,t) = Ae ik (x−ω k t ) this is a fn of x-vt right-moving wave! speed v=ω /k   + Be − ik (x+ω k t ) this is a fn of x+vt left 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 = ω k = E /  k =  2k2 / 2m k = k 2m Is that right? What did we expect? I'm thinking p = k* so v = p m = k m * [ For the function Ψk (x,t) = Ae i(kx− k 2 2m t ) , you get p̂Ψk =  i ∂ ∂x Ψk = kΨk So this Ψk is an eigenfunction of p̂with 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 = k 2m ? 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) = Ae i(kx−ω t ) where k = ± 2mE  (Note: 2 values of k for each E, one plus, one minus.) and € ω = E / = k 2 /2m comes from the usual free particle energy (basically, p^2/2m) These waves have λ = 2π k , because Ψ(x ± λ) = Ψ(x) The speed (phase =  k 2m = h 2mλ velocity) (Small λ  faster speed) There's that funny "2" we saw above, again. Classically, I expected v = p m = k m = 2E m without the factor of 2. So the 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 = A2 1⋅ 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. So these “plane wave solutions” are eigenfunctions of momentum, they are eigenfunctions of the free-particle Hamiltonian (i.e. if V=0), they are easy to write down, they have a simple form. That’s all “good news”. They are not normalizable, that’s “bad news”. I would consider them to be idealizations of physical systems, a kind of limit (which cannot be actualized, but can be approximated) of an ideal free particle with definite momentum. And as we will see, by the methods of Fourier transforms, they provide the mathematical basis for constructing perfectly physical free-particle states. k > 0 or k < 0 Re _ (k) k > 0 or k < 0 ψk SJP 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−x 2 / 4α Here € 1 2π f (x)e− ikxdx −∞ ∞ ∫ can be done analytically (guess what's on the next homework?) € φ(k) = A 2π e − x 2 4α + ikx )       dx = −∞ ∞ ∫ A 2π e−k 2αe−k 2α e − x 2 α − ik α      2 dx −∞ ∞ ∫ (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'= x 2 α − ik α ; dx '= dx 2 α ; e-x ' 2 dx '= π -∞ ∞ ∫ you then get € φ(k) = 2 α 2 Ae−k 2α 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/√α SJP QM 3220 Ch. 2 part 3. Page 6 (S. Pollock, taken from M. Dubson) with thanks to J. Anderson for typesetting. Fall 2008 So here's the central idea: instead of ψ(x) = Pure eikx, (which is not normalizable, and thus not a physical state) we will start with a more physical ψ(x), a "wave packet". Any ψ(x) can be "built" like this, the math of Fourier transforms yields the following, in general:  φ(k) may very well turn out complex, (but might also be real.)  If ψ(x) is "mostly sinusoidal" with wavelength ~ λ0 (like above) then φ(k) will be centered around k = 2π / λ0  If ψ(x) is localized with size Δx, then φ(k) is localized with size Δk ≈ 1 /Δx .  Given Ψ( x , t=0)  φ(k)  Ψ( x, t ) is determined for all times Claim: If we start with a simple wave packet, = As time goes by, the ripples in the ψ(x) wave packet will move with phase velocity ω/k, but the envelope itself (which we interpret as "where the particle is located"!) moves with a different velocity, "group velocity" € = dω dk . λ0 which has a Fourier transform: Δk k0=2π/λ0 envelope Δx SJP QM 3220 Ch. 2 part 3. Page 7 (S. Pollock, taken from M. Dubson) with thanks to J. Anderson for typesetting. Fall 2008 DIGRESSION on “phase and group velocity”, and the resolution of the “speed” puzzle. Let’s look more at this last statement, about phase and group velocity. In our case (a free particle), € ω = E  =  2m k 2 , so dω dk = k m , no funny factor of 2 present! This resolves both our "free particle issues": 1) The packet moves with € υclassical = k m 2) The Ψ( x, t ) is perfectly normalizable. ▪ And something else happens: different k's move with different speeds, so the ripples tend to spread out the packet as time goes by, So → → So Δx grows with time! Let me go back, and make a casual proof of the "claim" above about the time development of a wave packet: € Ψgeneral (x, t) = 1 2π φ(k)ei(kx−ωt )dk −∞ ∞ ∫ Recall ω is itself dependent on k, € ω = k 2 2m in this case of a free particle, it's just a consequence of E = p^2/2m. Let's consider a wave packet where φ(k) is peaked near k0. So we have a "reasonably well-defined momentum, (this is what you think of when you have a real particle like an electron in a beam …) ( If you had many k's, each k travels at different speeds  packet spreads, not really a "particle" …) So this is a special case, but a common/practical one … So if k's are localized, consider Taylor expanding ω(k) ω( k ) = ω( k0 ) + ω'(k0 ) ( k - k0 ) + . . . This should be a fine approximation for the ω( k ) in that integral we need to do at the top of the page! Let's define k' ≡ k - k0 as a better integration variable, so dk' = dk, but ω ≈ ω( k0 ) + ω'( k0 ) k' + . . . higher k's move up front SJP QM 3220 Ch. 2 part 3. Page 10 (S. Pollock, taken from M. Dubson) with thanks to J. Anderson for typesetting. Fall 2008 But I can DO that integral over x: just use our cute integral (1) to simplify. Let me first take integral (1) and rewrite it, flipping x’s and k’s: € δ(k) = 1 2π e+ikxdx −∞ ∞ ∫ (1’) (Convince yourself, treat x and k as pure symbols!) But this is precisely what I have in my triple integral, except instead of k I have (k’-k). Substitute this in (getting rid of the dx integral, which “eats up” the Sqrt[2π] terms too!) € δ(k'−k)φ * (k)φ(k ')dk dk '∫∫ =1 (It’s easy to “glaze over” this kind of math. Pull out a piece of paper and DO it, the steps are not hard, and you’ll use these tricks many times. Check that you see what happened to the 2π‘s and the integral over x) There’s still a double integral over k and k’, but with the δ(k’-k) it’s easy enough to do, I love integrating over δ’s! Let’s integrate over k’, and collapse it with the δ, giving € φ * (k)φ(k)dk −∞ ∞ ∫ =1 Ah – this is important. We started with a normalized wave function Ψ(x), and what I just showed is that the Fourier transform, φ(k) , is ALSO normalized. But there’s more! Consider now € p = Ψ* (x) −∞ ∞ ∫ i ∂ ∂x Ψ(x)dx . Play the exact same game as before, but with that derivative (d/dx) inside. All it will do will pull out an ik’ (convince yourself), everything else looks like what I just did, and we get € p = (k)φ * (k)φ(k)dk −∞ ∞ ∫ . Now, deBroglie identifies € k with p. So just stare at this until you realize what it’s saying. I would expect (from week 1!) that € p = p Prob density(p)dp −∞ ∞ ∫ . So here it is, apparently (other than annoying factors of € , since p and k are essentially the same thing, p= € k) we know what the “Probability density for momentum” is, it’s just € φ * (k)φ(k) = φ(k) 2 . To get the € ’s right, I would define € Φ(p) = 1  φ(k) , again with p= € k . I’ll let you convince yourself that Φ(p) is properly normalized too, € 1= Φ* (p)Φ(p)dp −∞ ∞ ∫ , and € p = pΦ* (p) −∞ ∞ ∫ Φ(p)dp We’ve just done a lot of math, so let’s pull this all together and summarize SJP QM 3220 Ch. 2 part 3. Page 11 (S. Pollock, taken from M. Dubson) with thanks to J. Anderson for typesetting. Fall 2008 Bottom line: We’d gotten used to thinking about Ψ(x) as “the wave function”. When you square it, you learn the probability density to find a particle at position x. Expectation values tell us about OTHER things (any other operator/measurable you want.) Ψ(x) contains ALL information about a quantum particle. Now we have the Fourier transform, φ(k), or equivalently € Φ(p) = 1  φ(k) . This function depends on “p”. It contains the SAME information as Ψ(x)! It just lives in “momentum space” instead of “position space”. It’s a function of p. • Φ(p) is normalized, just like Ψ(x) was. • |Φ(p)|^2 tells the “probability density of finding the particle with momentum p” • Φ(p) is the wave function too. There’s nothing special about x! Φ(p) is called the “momentum space wave function”. • We can compute expectation values in momentum space to learn about anything, just like we could before. Let’s now consider this in more detail: We already know € p = pΦ* (p) −∞ ∞ ∫ Φ(p)dp In fact, it shouldn’t be hard to convince yourself that for any function of momentum f(p), € f (p) = f (p)Φ* (p) −∞ ∞ ∫ Φ(p)dp. How about <x>? Here again, start with the usual old formula in terms of Ψ(x) and play the same math game as on the previous pages: expand Ψ(x) as a Fourier transform, you get a triple integral, do the “δ-function” trick, and (try this, it’s not so hard!) you find € x = Φ* (p) i ∂ ∂p−∞ ∞ ∫ Φ(p)dp. Look at that for a second. There’s a lovely symmetry here: In position space, we’ve used € ˆ x = x and € ˆ p = −i ∂ ∂x . Now we’re seeing that in momentum space, € ˆ x = i ∂ ∂p , and € ˆ p = p . It’s almost perfect “symmetry” (except for a minus sign, but don’t forget, the inverse Fourier transform has an extra minus sign in front of the “ikx”, which is the source) We will discover there are other “spaces” we might live in. Energy-space comes to mind, but pretty much any-operator-you-like-space can be useful too! The momentum-space example will be the first of many! But for now we move back to our central problem, solving the Schrodinger equation. We’ve done two “bound states”, and the free particle. Next we will consider particles with positive energy (so, they’re still “free”) but where V is not just simply 0 everywhere. This is still a lot LIKE the free particle, it’s a “free particle interacting with things”. This is getting to be real physics, like the beam of protons at LHC interacting with nuclei! We’ll still stick to one dimension (your beam can only go forward and backward, you can only transmit or reflect – no “scattering off at 27 degrees” just yet!) And, at the same time, we will tackle other “bound state” problems too - to keep the math simple, we’ll consider examples with “piecewise constant” V(x), like in the picture on the next page: SJP QM 3220 Ch. 2 part 3. Page 12 (S. Pollock, taken from M. Dubson) with thanks to J. Anderson for typesetting. Fall 2008 Back to our problem, in general, of solving the TISE, Schrodinger‘s equation: € ˆ H un (x) = Enun (x), or writing it out : −  2 2m d2 dx 2 u(x) + V (x)u(x) = Eu(x) € u' '(x) = − 2m  2 (E −V (x))u(x).  So u and u' need to be continuous, at least if V is finite. If E > V(x), that means u'' = - (k2) u Classically, KE > 0 € k 2 ≡ 2m(E −V )  2 , or E −V =  2k 2 2m De Broglie = p2 2m as usual! u'' = - k2 u => curvature is towards axis. This is a wiggly function! If k2 is constant, u = A sin kx + B cos kx or = A' eikx + B' e-ikx Big K.E.  Big k2  more wiggly (smaller λ) Classically Big K.E.  faster speed  less likely to be found there. This is very different, it corresponds in QM not to λ, but amplitude. ← (well, usually, but there are exceptions) So as a specfic example: V(x) x L slow fast V(x) I expect === E smaller k  longer λ Big k  small λ L x low prob  small | ψ| higher prob  big |ψ| V(x) x V(x) E<v (KE) E>V E SJP QM 3220 Ch. 2 part 3. Page 15 (S. Pollock, taken from M. Dubson) with thanks to J. Anderson for typesetting. Fall 2008 Let's start with even solutions for u(x): Region Region Region Boundary conditions: Continuity of u at x=+a: € uIII(x) |x=+a= uII(x) |x=a Continuity of u' at x=+a: € u'III (x) |x=+a= u'II (x) |x=a (Conditions at x = - a add no new info, since we've already used fact that u is even!) coefficients We have 3 unknowns: E and (A and C) The energy eigenvalue. We have 2 boundary conditions and normalization, so we can solve for 3 unknowns! Continuity of u says € Ce−Ka = Acoska Continuity of u' says € −KCe-Ka = −Ak sinka Dividing these  € −K = −k tanka € tan 2mE  a       − K k = V0 − E E (A nasty, transcendental equation with one variable, E) E V(x) -a a x I II III Region We've solved for u(x) in each region already I No good, blows up II No good, looking for even solution right now III Same C, 'cause looking for even solution! SJP QM 3220 Ch. 2 part 3. Page 16 (S. Pollock, taken from M. Dubson) with thanks to J. Anderson for typesetting. Fall 2008 Can solve that transcendental equation numerically, or graphically, € Let z ≡ ka = 2mE  2 a Let z0 = 2mV0  2 a        so € V0 E = z0 z     2 So we want € tan z = (z0 /z) 2 −1 I see there are solutions for z (which in turn means E, remember, z and E are closely related, by definition, if you know one you know the other) They are discrete. There is a finite number of them. No matter what z0 is, there's always at least solution. (Look at the graph!) If V0 → ∞, solutions are near π/2, 3 π/2, … Giving € E =  2 2ma2 Z 2 =  2π 2 2m(2a)2 (12, or 32, …) which is exactly the even solutions for our ∞ well with width 2a. Knowing z  E  k and K , our B.C.'s give A in terms of C, and normalization fixes C. I leave the odd solutions to you. The story is similar and you will get, as your transcendental equation, € cot(ka) = −K k This time, it's not always the case there's even a single solution. (The well needs to be deep enough to get a 1st excited state, but even a shallow well has one bound state.) SJP QM 3220 Ch. 2 part 3. Page 17 (S. Pollock, taken from M. Dubson) with thanks to J. Anderson for typesetting. Fall 2008 What about E > V0? This is the continuous region, also called the "scattering states", because u will be nonzero off to ∞, it's like a particle comes in, interacts with our potential well, and is affected (scattered). Lots of physics in this example (- like "shooting electrons" at some object it interacts with.) Griffiths treats the square well, I'll do a simpler example: We could build a wave packet, send it in from left with E > V0 , and solve for time dependence. That's the right way to make this physically realistic. But our approach will be simpler. Look for u(x) which solves S.E. and which is physically interpreted as "something coming in from the left". € uI(x) = Ae ikx + Be− ikx with k = 2mE  2 uII(x) = Ce ik'x + De−ik'x with k'= 2m(E −V0)  2 Remember our interpretation of free particle solutions from awhile back, the Aeikx term  incoming wave, traveling rightwards with € p = k Be-ikx represents a left moving wave, but it's in region I, so it must physically represent reflected wave (in steady state, remember!) Ceik'x represents "transmitted wave", going right in region II. We set D = 0, because otherwise we'd have "incoming wave from the right", and our physics condition was "waves coming from left". Boundary Conditions: Continuity of u and u' at the origin gives: € uI(x) |x=0= uII(x) |x=0 u'I (x) |x=0= u'II (x) |x=0 E V0 Region x I II 0 0 Region The STEP POTENTIAL. Think of wire, w/junction: Higher voltage on right, lower ˝ ˝ left. SJP QM 3220 Ch. 2 part 3. Page 20 (S. Pollock, taken from M. Dubson) with thanks to J. Anderson for typesetting. Fall 2008 So back to our scattering problem, with particles coming in from the left and hitting a "step up" potential, but with E>V: If J = "probability current", then R= reflection coefficient = € Jref Jinc € which is just B 2 /A 2 , do you see that from the formula for J we got on the previous page? T= Transmission coefficient = € Jtrans Jinc , which is NOT € C 2 /A 2 , it is € Jtrans Jinc = C 2 A 2 k' k (Think a little about the presence of the k'/k term there - I might not have expected it at first, thinking maybe T = C2/A2, but when you look at our probability current formula, you realize it makes sense, the waves have a different speed over there on the right side, and probability current depends not JUST on amplitude, but ALSO on speed!) We solved for B/A and C/A, so here € R = 1− k 'k 1+ k 'k       2 T = 4 1 + k 'k( ) 2 ⋅ k ' k          convince yourself then, that R + T =1 (!) Convince yourself Note: If V0 → 0 , € k' k = E −V0 E →1 (Convince yourself!) and then check, from the formulas above, that R → 0 and T → 1 . This makes sense to me - there's no barrier in this limit! On the other hand, if E → V0, I claim € k' k → 0 (again, do you see why?), and in THIS limit, the formulas above tell us that R→1 and T→0. Again, this makes sense - in this limit, it is "forbidden" to go past the barrier. (Classically, if E > V0, R = 0. QM gives something new here.) E0 V0 Jref C Jtransmitted Jinc B A SJP QM 3220 Ch. 2 part 3. Page 21 (S. Pollock, taken from M. Dubson) with thanks to J. Anderson for typesetting. Fall 2008 Other examples: Pain, but can compute R, T. E < V0: T is not 0, even though classically it would be. This is tunneling. E > V0: Classically, T=1, but QM  some reflection. E V0 V