J. Broida UCSD Fall 2009 Phys 130B QM II Identical Particles 1 Two-Particle Systems Suppose we have two particles that interact under a mutual force with potential energy V (x x ), and are also moving in an external potential V (x ). Then the 1 − 2 i Hamiltonian is H = T1 +T2 +V +V (x1)+V (x2) so the time-dependent Schr¨odinger equatione becomes e ~2 ~2 ∇2 ∇2 + V (x x )+ V (x )+ V (x ) Ψ(x , x ,t) −2m 1 − 2m 2 1 − 2 2 2 1 2 1 2 ∂ e = i~ Ψ(x , x ,t) (1) ∂t 1 2 ∇2 where i refers to the laplacian with respect to the coordinates of particle i. To solve this equation, we first separate time and spatial variables in the usual manner by writing Ψ(x1, x2,t)= ψ(x1, x2)T (t) and substituting this into (1). Dividing both sides by ψT yields 1 ~2 ~2 i~ ∂T ∇2 ∇2 + V (x x )+ V (x )+ V (x ) ψ = . ψ −2m 1 − 2m 2 1 − 2 2 2 T ∂t 1 2 Since the left side is a function ofe spatial variables only and the right side is a function of time only, both sides must be equal to a constant which we call E. Then the time equation has the solution (up to normalization) ~ T (t)= e−iEt/ (2a) and the spatial equation becomes ~2 ~2 ∇2 ∇2 1 2 + V (x1 x2)+ V (x2)+ V (x2) ψ(x1, x2)= Eψ(x1, x2) −2m1 − 2m2 − (2b) e If the solutions to (2b) are En and ψn(x1, x2), then the general solution to (1) is of the form −iEnt/~ Ψ(x1, x2,t)= cnψn(x1, x2)e . n 2 X Now we interpret Ψ(x1, x2,t) as the probability density for particle i to be in the 3 | | volume d xi, and the constants cn are chosen so as to normalize the wave function to Ψ(x , x ,t) 2 d3x d3x =1 . | 1 2 | 1 2 Z 1 Let us now specialize to the case where the external potentials vanish. In this case, (2b) becomes ~2 ~2 ∇2 ∇2 + V (x x ) ψ(x , x )= Eψ(x , x ) (3) −2m 1 − 2m 2 1 − 2 1 2 1 2 1 2 where we now drop the tilde on V . The presence of the term x x suggests that 1 − 2 we change variables from x1 and x2 to e r = x x (4a) 1 − 2 R = αx1 + βx2 . (4b) We will see that these turn out to be the usual center-of-mass coordinates. Using the chain rule, we have ∂ ∂r ∂ ∂R ∂ ∂ ∂ = j + j = + α ∂x1i ∂x1i ∂rj ∂x1i ∂Rj ∂ri ∂Ri ∂ ∂ ∂ = + β ∂x2i −∂ri ∂Ri so that ∇2 = (∂ + α∂ )(∂ + α∂ )= ∇2 +2α∇ ∇ + α2∇2 1 ri Ri ri Ri r r · R R ∇2 = ( ∂ + β∂ )( ∂ + β∂ )= ∇2 2β∇ ∇ + β2∇2 2 − ri Ri − ri Ri r − r · R R (where ∂ri = ∂/∂ri etc). Therefore 1 1 1 α β α2 β2 ∇2 + ∇2 = ∇2 +2 ∇ ∇ + + ∇2 m 1 m 2 µ r m − m r · R m m R 1 2 1 2 1 2 where m m µ = 1 2 m1 + m2 is called the reduced mass. We can eliminate the cross-term by choosing m β = 2 α (5) m1 which results in 2 1 ∇2 1 ∇2 1 ∇2 α ∇2 1 + 2 = r + M 2 R m1 m2 µ m1 with M = m1 + m2. Our choice of alpha is still arbitrary, and it is most convenient to choose it so that 3 3 3 3 d x1 d x2 = d r d R Z Z 2 which means the Jacobian of the transformation must be unity. In other words, letting xij be the jth component of xi we have ∂r1/∂x11 ∂r1/∂x12 ∂r1/∂x13 ∂r1/∂x21 ∂r1/∂x22 ∂r1/∂x23 ∂r2/∂x11 ∂r2/∂x12 ∂r2/∂x13 ∂r2/∂x21 ∂r2/∂x22 ∂r2/∂x23 ∂r3/∂x11 ∂r3/∂x12 ∂r3/∂x13 ∂r3/∂x21 ∂r3/∂x22 ∂r3/∂x23 =1 . ∂R1/∂x11 ∂R1/∂x12 ∂R1/∂x13 ∂R1/∂x21 ∂R1/∂x22 ∂R1/∂x23 ∂R /∂x ∂R /∂x ∂R /∂x ∂R /∂x ∂R /∂x ∂R /∂x 2 11 2 12 2 13 2 21 2 22 2 23 ∂R /∂x ∂R /∂x ∂R /∂x ∂R /∂x ∂R /∂x ∂R /∂x 3 11 3 12 3 13 3 21 3 22 3 23 I leave it as a good exercise (and not as hard as you might think) for you to show that substituting equations (4) and (5) in this and evaluating the determinant yields m m α = 1 = 1 . m1 + m2 M Then our change of variables becomes r = x x (6a) 1 − 2 m x + m x R = 1 1 2 2 (6b) M which are the familiar CM coordinates as claimed. Putting this all together we have 1 ∇2 1 ∇2 1 ∇2 1 ∇2 1 + 2 = r + R m1 m2 µ M and hence the Schr¨odinger equation (3) becomes ~2 ~2 ∇2 ∇2 + V (r) ψ(r, R)= Eψ(r, R) . (7) −2µ r − 2M R This is clearly separable by letting (in a somewhat sloppy notation) ψ(r, R) = ψ(r)φ(R) which results in 1 ~2 1 ~2 ∇2 + V (r) ψ(r)= E ∇2 φ(R) . ψ(r) −2µ r − φ(R) −2M R Again, the left and right sides of this equation depend solely on different variables and so must be constant. This yields the two equations ~2 ∇2 + V (r) ψ(r)= E ψ(r) (8a) −2µ r r and ~2 ∇2 φ(R)= E φ(R) (8b) −2M R R 3 with E = Er + ER . This expresses the total energy as the sum of the translational center-of-mass energy plus the energy of relative motion. We can always choose ER = 0 as defining our zero of energy. Equation (8a) is the starting point for the solution to the hydrogen atom problem you solved in 130A, although you unknowingly may have approximated the reduced mass µ by the electron mass me. This is a pretty good approximation since the proton mass mp is roughly 2000 times the electron mass, so if me mp, then µ = m m /(m + m ) m . ≪ e p e p ≈ e 2 Identical Particles and Exchange Degeneracy In classical mechanics, it is possible to distinguish between identical particles by following their distinct trajectories without disturbing them in any way. As a result, exchanging the particles results in a physically distinct configuration. However, in quantum mechanics there is no such thing as a continuous trajectory, and hence there is no physical basis for distinguishing between identical particles. This means that two systems that differ only by the exchange of identical particles must be described by the same state vector. So, by identical particles, we mean particles which are completely and fun- damentally indistinguishable. There is no physical significance attached to the labeling of such particles, and there is no observable effect if any two of them are interchanged. As a consequence, the coordinates of such particles must enter into the Hamiltonian in exactly the same way. (In fact, Feynman once said that the rea- son electrons were indistinguishable was that there was really only one electron in the universe. By allowing this one electron to travel forward and backward in time, a single time slice at a given instant would show the existence of many identical electrons at different locations.) For example, consider two noninteracting identical particles moving under the influence of some external force. Since each particle feels the same potential, the Hamiltonian must be of the form p 2 p 2 H = 1 + 1 + V (x )+ V (x ) . 2m 2m 1 2 The stationary state solutions are then ψkl(x1, x2)= ψk(x1)ψl(x2) (9) and the corresponding energy is Ekl = Ek + El where Hspψk = Ekψk and Hsp is the single particle Hamiltonian p2 H = + V (x) sp 2m 4 common to both particles. If k and l are different, the state ψkl is degenerate with respect to the interchange of the particles, and hence the state ψkl = ψk(x1)ψl(x2) must have the same energy as the state ψlk = ψl(x1)ψk(x2). This degeneracy is called exchange degeneracy, and is a consequence of the invariance of H under the exchange of the coordinates of the two particles. Let us show that this exchange degeneracy is a property of the solutions to Schr¨odinger’s equation for any system of identical particles, independently of the number of particles present or the forces they may be moving under. To do this, we define the exchange operator Pij = Pji which exchanges the coordinates of particles i and j when acting on any function of those coordinates: Pij f(x1,...,xi,...,xj ,...,xN )= f(x1,...,xj ,...,xi,...,xN ) . In particular, note for example, that P12P13f(x1, x2, x3,...,xN )= P12f(x3, x2, x1,...,xN ) = f(x3, x1, x2,...,xN ) . It is very important to note that this example also shows that P12 and P13 do not commute, as we easily see from P13P12f(x1, x2, x3,...,xN )= P13f(x2, x1, x3,...,xN ) = f(x2, x3, x1,...,xN ) . And in general, for N 3 we find that P and P will not commute. (However, ≥ ij ik it may very well be that Pij and Pkl will commute if i = j = k = l.) What this example also shows is that the subscript on a coordinate6 labels6 the6 particle to which the coordinate refers, and not the order in which a coordinate appears in the function.