Schrödinger s equation.

Transcription

1 Physics 342 Lecture 5 Schröinger s Equation Lecture 5 Physics 342 Quantum Mechanics I Wenesay, February 3r, 2010 Toay we iscuss Schröinger s equation an show that it supports the basic interpretation of the funamental object of stuy in quantum mechanics: the wave function. The operational proceures of quantum mechanics begin with this PDE, an its associate bounary an initial conitions. So while it is easy to state the basic axioms, an even motivate the expressions, the shift in focus is ramatic. In classical mechanics, our goal is always to fin or escribe the salient features of a curve x(t), the trajectory of a particle uner some impresse force. The relation of force to curve is provie by Newton s secon law: F = m ẍ(t), an the physically inspire initial (or bounary) conitions. The job is straightforwar, if complicate at times. The interpretation is simple: The particle moves along the curve. In quantum mechanics, our initial goal will be to fin the probability ensity Ψ(x, t) 2 for a particular system, from which we can calculate expectation values (the outcomes of experiment). The relation of a classical potential to the ensity (or equivalently, Ψ(x, t), the wave function) is provie by Schröinger s equation. There are a few obvious questions that come up as we exploit the probabilistic interpretation of the wave function. We know that if we have a probability ensity ρ(x, t), we can compute the outcome of experiments that rely on position, since f(x) = f(x) ρ(x, t) x. What, then, shoul we make of experiments that measure velocity (or momentum)? A relate question is why, given that it is ρ(x, t) that we want, oes Schröinger s equation govern Ψ(x, t)? Shouln t we have a PDE that tells us about the evolution of ρ(x, t) itself? 1 of 9

2 5.1. WITHOUT FURTHER ADO Lecture Without Further Ao Here it is for a complex fiel Ψ(x, t), Schröinger s equation says that the ynamical evolution of Ψ is given by (in one spatial imension): i Ψ t = 2 2 Ψ + V (x) Ψ (5.1) x2 for a potential V (x). This is the quantum mechanical analogue of the classical Newton s secon law: the ynamical evolution of a particle is given by its trajectory x(t) governe by mẍ = V. Our statistical interpretation is provie by viewing Ψ(x, t) Ψ(x, t) as a ensity: ρ(x, t) = Ψ(x, t) 2 so that P (a, b) = b a ρ(x, t) x = b a Ψ(x, t) 2 x (5.2) is the probability a particle is foun between x = a an x = b. We are talking about the probability ensity associate with a particle, that explains the m in Schröinger s equation. The constant tells us this is not a classical system. If we turn off, we get Ψ = 0, which seems reasonable classically, the probability ensity woul be given by: ρ(x, t) CM = δ(x x(t)), so that with probability 1, the particle is on its classical trajectory x(t). Without some notion of x(t), there is no probability ensity at all, so Ψ = 0 is as goo a solution as any. Of course, is oing more than setting units, but we will see that later on. No statistical interpretation exists in Schröinger s equation alone for example, the equation itself oes not eman that Ψ(x, t) 2 x = 1 for all time. But, it oes support the statistical interpretation in the sense that if we solve the above, an require that, at time t = 0, Ψ(x, 0) 2 x = 1, then (as we ll show), Ψ(x, t) 2 x = 1 for all time so once a probability ensity, always a probability ensity. We can see this by taking the time erivative of Ψ(x, t) 2 x, an showing that, as long as Ψ(x, t) satisfies (5.1), the time erivative is zero, so that the normalization remains intact. It is, then, up to us to ensure that Ψ(x, t) is appropriately normalize at some time. Take: t Ψ(x, t) 2 x = t (Ψ (x, t) Ψ(x, t)) x, (5.3) 2 of 9

3 5.1. WITHOUT FURTHER ADO Lecture 5 then we just nee the temporal erivative of the norm (square) of Ψ(x, t): t (Ψ (x, t) Ψ(x, t)) = Ψ Ψ Ψ + Ψ t t, (5.4) which is fine, but this is uner an integral w.r.t. x, so we like a mechanism for turning temporal erivatives into spatial ones (preparatory to integration by parts). Schröinger s equation, an its complex conjugate provie precisely the esire connection: an we can write the above as: Ψ t = i 2 Ψ x 2 i V (x) Ψ Ψ = i 2 Ψ t x 2 + i V (x) Ψ, t (Ψ (x, t) Ψ(x, t)) = i ( 2 Ψ x 2 (5.5) ) Ψ + Ψ 2 Ψ x 2. (5.6) Now, uner the integral, we can use integration by parts 1 to simplify the above consier the first term: 2 Ψ ( ) Ψ x 2 Ψ x = x Ψ Ψ Ψ x. (5.9) x x We can get ri of the bounary term by assuming (an aitional requirement) that Ψ(x, t) 0 as x. This is a reasonable assumption, an a common one when ealing with ensities (charge istributions o not exten to infinity except in artificial cases, for example). Then, going back to the original time erivative of the ensity: (Ψ (x, t) Ψ(x, t)) = i [( 2 Ψ )] t x 2 Ψ + Ψ 2 Ψ x 2 x = i = 0, 1 For two functions u(x) an v(x), we have Z b [ Ψ an using the prouct rule on the left gives: Z b» u(x) v(x) v(x) + u(x) x x a a x Ψ x Ψ x Ψ x ] x (5.10) x (u(x) v(x)) x = u(x) v(x) b a, (5.7) from which the usual form of integration by parts follows. x = u(x) v(x) b a, (5.8) 3 of 9

4 5.1. WITHOUT FURTHER ADO Lecture 5 an we have the result: The time erivative of the probability that a particle exists somewhere spatially is zero, hence we can normalize Ψ(x, t) so that the probability the particle exists somewhere is = 1, an it will remain 1 for all time. Notice that the statement: t ρ(x, t) x = 0 (5.11) is reminiscent of charge conservation. This is a probability conservation sentiment. By analogy with charge, though, we shoul be able to fin a probability current associate with the conservation. This must be an object whose x-erivative is equal to the right-han-sie of (5.10) (the top line, before we integrate by parts) easy to fin in this case. We want: i J(x, t) = [ 2 Ψ ] x x 2 Ψ + Ψ 2 Ψ x 2, (5.12) (the minus sign is convention, an we will use it in a moment to make contact with other conservation laws) an it is pretty clear that: has the esire property. J(x, t) = i [ Ψ Ψ ] x Ψ x Ψ (5.13) Interesting then we have another view of the conservation statement we woul write it, in one imension, as t ρ(x, t) x = J(x, t) x. (5.14) x Here, the right-han sie is clearly zero by the funamental theorem of calculus, an the requirement that Ψ vanish at spatial infinity. We have basically rewritten integration by parts, but this rewrite allows us to make a ifferential continuity statement: In three imensions, this woul rea; ρ t = J x. (5.15) ρ t = J J = i [Ψ Ψ Ψ Ψ ]. (5.16) 4 of 9

5 Now we see the analogy: In E&M, the rate of change of charge in a volume is etermine by the charge flowing in through a surface. We have V ρ τ as the charge in the volume, an J a is the surface integral for the current then the statement is ρ t = J ρ τ = J a, (5.17) t V where we get the integral form by integrating both sies of the ifferential one an applying the ivergence theorem. The same conservation law hols for mass flow in a flui. Eviently, in quantum mechanics, the local probability of fining a particle in a volume can change over time accoring to a probability flux that can be use to account for probability flowing into an out of the volume through its bounary. V 5.2 Expectation Values Let s take stock: We have a PDE that governs the temporal evolution of a wavefunction Ψ(x, t) given a classical potential V (x), Schröinger s equation. This PDE supports a probabilistic interpretation for the absolute value (square) of Ψ(x, t), i.e. Ψ 2 is the probability ensity for fining the particle at location x at time t. provie Ψ(x, t) vanishes at spatial infinity. That gives us a bounary conition for Ψ(x, t), but we are missing the initial conition what is the spatial function Ψ(x, t = 0)? We ll come back to this, but one thing is clear, it must have the property: Ψ(x, t = 0) 2 x = 1 for the probabilistic interpretation to hol. We have, so far, a complex Ψ(x, t) to be foun subject to: i Ψ t = 2 Ψ(x ±, t) = 0 Ψ(x, 0) 2 x = 1. 2 Ψ x 2 + V (x) Ψ (5.18) If we ha the wavefunction, all manner of interesting physical properties coul be calculate. We can start with the simplest the expectation value of position: x. From the ensity, we know that x = x ρ(x, t) x = 5 of 9 Ψ x Ψ x (5.19)

6 which is reasonable. We have put x in between Ψ an its complex conjugate, that oesn t matter here, just notation. One can efine the expectation of velocity to be the time-erivative of x, i.e. if we know how the expecte position changes in time, we might call that the expecte velocity. Since expectation values procee by integration over all positions, the notion of time-epenence for x itself is not particularly useful. The objects with time-epenence are expectation values, an these inherit their time-epenence irectly from the wavefunction s, so we are sort of force into viewing particle velocity as the time-epenence of the particle s expecte position. At that point, we have x t = = = = (ρ(x, t) x) x t ρ t x x J x x x J x, (5.20) where we have use the local form of conservation: ρ t = J x, an integration by parts. This makes some sense when we think of the current ensity from E&M, or the flui flux ensity in hyroynamics in either case, with appropriate interpretation of ρ, we have J = ρ v in our present setting, the ρ is probability ensity, an the v then represents some sort of velocity with which the probability is flowing the natural expectation value here woul be v = ρ v τ in one imension. But that s what the last line of (5.20) is. There is a sleight-of-han going on here, we just associate v with probability flux, an yet, we like to unerstan the expectation value of v in terms of a particle s velocity on t take the above argument as anything more than motivation at this point. Pressing on, we can write the integral itself in a more illuminating form using the efinition of J from (5.13): [ ( ) ( ) ] i i v = J x = Ψ Ψ Ψ Ψ x 2m x 2m x = Ψ ( ) (5.21) i Ψx m x 6 of 9

7 where we use integration by parts to combine the two terms. In the en, we can move the mass over to the right, an efine the expectation value of momentum (of the particle, now): ( ) p = m v = Ψ Ψ x. (5.22) i x Looking at the expectation value of position again, x = Ψ x Ψ x (5.23) we are tempte to efine operators x an p (whose expectation values are given by the above) via: x x p i x. (5.24) These are operators in the sense (obvious for p) that they act on the wave function. This seems like a strange move, an it is a literal translation of what the expectation values are telling us, so there s no reason to believe any of it just yet we will let the correctness of the preictions of this theory o the convincing. For now it is the case that with a sensible efinition of expectation value, an the peculiar factoring of probability ensity provie by the wave function, we are le to the above association. With this in han, we can efine any classical quantity s expectation value by replacing x x an p i x, an sanwhiching it between the wave function an its complex conjugate. For a classical function Q(x, p), we have ( Q = Ψ Q x, ) Ψ x. (5.25) i x Things are beginning to look suspiciously like α Q α with an inner prouct efine by integration, which is no accient. What is also no accient is the fact that for the classical Hamiltonian, a function of x an p: H = p2 + V (x), our prescription for the expectation value is: ( H = Ψ 2 2 ) x 2 + V (x) Ψ x, (5.26) an if you cover up the integral an Ψ, what you have is one sie of Schröinger s equation. 7 of 9

8 Homework Reaing: Griffiths, pp Problem 5.1 Griffiths 1.4. In part c., you are aske where the particle is most likely to be foun technically, what we want is the maximum of ρ(x) here that is, the location of the particle with the largest ρ(x) x. Problem 5.2 Griffiths 1.7. This problem is about (an example of) Ehrenfest s theorem it provies goo practice with integration by parts, in aition to the computational utility of the result. Problem 5.3 Working from the separate form of Schröinger s equation: For Ψ(x, t) = ψ(x) φ(t), we ha 2 2 ψ(x) x 2 + V (x) ψ(x) = E ψ(x) (5.27) φ(t) = e i E t. Consier a physical configuration in which we have V (x) = 0 an require that ψ(0) = ψ() = 0. This represents a particle confine to a oneimensional box of with. The wavefunction (an hence the probability ensity) is zero outsie the interval, an continuity requires that the wavefunction must then vanish at the enpoints of the interval. a. Solve the upper equation for ψ(x), with bounary conitions. This shoul look familiar by now, an there will be an integer n that inexes your solutions. Write the energy E as a function of n, an form the full wavefunction: Ψ n (x, t). This shoul have an unetermine overall constant in it. b. Suppose we have a wavefunction Ψ(x, t) = Ψ n (x, t) in orer to interpret ρ(x, t) = Ψ (x, t) Ψ(x, t) as a probability ensity, we must have 8 of 9

9 ρ(x, t) x = 1. Fix your overall constant in Ψ n(x, t) by normalizing the wavefunction. c. Again, using Ψ(x, t) = Ψ n (x, t), calculate: x an p. (5.28) 9 of 9