Multi-Particle Wave functions

Transcription

1 Multi-Particle Wave functions Multiparticle Schroedinger equation (N particles, all of mass m): 2 2m ( N ) Multiparticle wave function, + U( r 1, r 2,..., r N )=i (~r 1,~r 2,...,~r N,t) t (r 1,..., r N ) 2 dv 1 dv 2... dv N is probability particle 1 is found at r 1 while particle 2 is found at r 2 and particle N is found at r N (within volume range for each particle) at time t.

2 apple Consider 2 distinguishable non-interacting particles in a onedimensional problem. ~ U Since the particles are non-interacting we can express U as U(x 1,x 2 )=U(x 1 )+U(x 2 ) ~ 2 2m Let s use separation of variables (x 1,x 2,t)= a (x 1 ) b (x 2 ) exp( iet/ ) Substituting into Schroedinger equation: 1 a(x 1 2 a (x apple + U(x 1 ) + ~ 2 2m 1 b(x 2 2 b (x U(x 2 ) = E = E a + E b ) ~2 2m ) ~2 2 a (x b (x U(x 1 ) a (x 1 )=E a a (x 1 ) + U(x 2 ) b (x 2 )=E b b (x 2 )

3 Solve individual 1-particle equations and then combine. Example: 1d-inifinte square well: a = 2 L sin(n x 1/L) b = 2 L sin(m x 2/L) n,m = 2 L sin(n x 1/L) sin(m x 2 /L) exp( ie n,m t/ ) Suppose particle a is in the n=4 state and particle b is in the m=3 state. 4,3 = 2 L sin(4 x 1/L)sin(3 x 2 /L)

4 Energy of the two particles: E n,m = 2 2 2mL 2 (n2 + m 2 ) If potential energy is not separable, then separation of variables would not work. Example: if U depended on distance between particle 1 and particle 2, U( x 1 -x 2 )

5 Identical Particles Mathematically, if we switch particle 1 and particle 2, no observational consequence should occur. Thus, P (x 1,x 2 )=P (x 2,x 1 ) Let s revisit our example of two particles in a 1D well, where n=4 and m=3 4,3 = 2 L sin(4 x 1/L)sin(3 x 2 /L) Does P(x 1,x 2 ) = P(x 2,x 1 )? P (x 1,x 2 ) = P (x 2,x 1 ) What do we do? Use superposi*on to force a symmetric probability distribu*on (x 1,x 2 )=C[ n (x 1 ) m (x 2 ) ± m (x 1 ) n (x 2 )]

6 We call the state with the + the symmetric state and we call the state with the the antisymmetric state. Symmetric: S(x 1,x 2 )=C S [ n (x 1 ) m (x 2 )+ m (x 1 ) n (x 2 )] Note this means the wave func*on is symmetric under par*cle exchange: S(x 1,x 2 )= S (x 2,x 1 ) Antisymmetric: A(x 1,x 2 )=C A [ n (x 1 ) m (x 2 ) m (x 1 ) n (x 2 )] This wave func*on is an*- symmetric under par*cle exchange: A(x 1,x 2 )= A (x 2,x 1 ) The probability distribution for either of these states is symmetric: (x 1,x 2 )= (x 1,x 2 ) 2 = C 2 [ 2 n(x 1 ) 2 m(x 2 )+ 2 m(x 1 ) 2 n(x 2 ) ± 2 n (x 1 ) m (x 2 ) m (x 1 ) n (x 2 )]

7 (x 1,x 2 )=C[ n (x 1 ) m (x 2 ) ± m (x 1 ) n (x 2 )] Rather than saying particle 1 is in n state and particle 2 is in m state, we say one particle is in n state and one particle is in the m state. Which one should we use???

8 Fermions and Bosons Bosons - Particles with integer spin (s=0, 1, 2, ) Examples include photons, pions, and alpha particles. Fermions Particles with half-integer spin (s=1/2, 3/2, ) Examples include electrons, protons, neutrons, neutrinos, and quarks. Exchange Symmetry Principle The multiparticle state of identical bosons is symmetric The multiparticle state of identical fermions is antisymmetric Consider a system of two fermions occupying individual states n and m (here n and m is shorthand for all quantum numbers required to specify the individual particle s state. For an electron in hydrogen, for example, n!{n, l, m l,m s } ) A(x 1,x 2 )=C A [ n (x 1 ) m (x 2 ) m (x 1 ) n (x 2 )]

9 A(x 1,x 2 )=C A [ n (x 1 ) m (x 2 ) m (x 1 ) n (x 2 )] What happens if n=m? A(x 1,x 2 )=C A [ n (x 1 ) n (x 2 ) n (x 1 ) n (x 2 )] = 0! We conclude that two fermions cannot have the same quantum numbers. This is Pauli s Exclusion Principle

10 Pauli Exclusion Principle No two indistinguishable fermions my occupy the same individual quantum state. Note: This principle was formulated by Wolfgang Pauli in 1925 (before the advent of wave functions) in order to account for the structure of the periodic table (using quantum numbers). This also led Pauli to predict a fourth quantum number that can have two possible values (which we now know is m s )

11 Example: Two fermions in 1D Box Let s revisit the case of two (non-interacting) identical fermions in the 1-dimensional infinite potential well. The individual quantum state is specified using two quantum numbers: n and m s. The spatial state is given by n(x) = p 2/L sin(n x/l) The spin state is specified by m s. One common notation is using m s (χ + and χ - for m s =1/2 and m s =-1/2, respectively) Thus the overall quantum state can be expressed as either n(x) + or n (x) For two identical particles, the two-particle quantum state is the product of a spatial state times the spin state: = (x 1,x 2 ) (1, 2)

12 Two Fermions in a Box = (x 1,x 2 ) (1, 2) For the case of two identical fermions, the overall quantum state must be anti-symmetric. Thus either the spatial state is anti-symmetric and the spin state is symmetric, or the spatial state is symmetric and the spin state is anti-symmetric. SA = S (x 1,x 2 ) A (1, 2) AS = A (x 1,x 2 ) S (1, 2)

13 Clicker Question For the ground state of two (non-interacting) identical fermions in the 1D box, what can you say about the spin state? 1. It must be symmetric 2. It must be anti-symmetric 3. It can be either symmetric or anti-symmetric. 4. Huh????

14 For the ground state, both fermions are in the n=1 state, so the spatial portion of the quantum state is symmetric. S (x 1,x 2 )= 1 (x 1 ) 1 (x 2 ) This means the spin state must be anti-symmetric A (1, 2) = 1 p 2 [ +(1) (2) (1) + (2)] We can think of this as one particle having spin up, and the other particle having spin down. In this case, the total spin of the system is S=0. We call this the singlet state. The spatial state can hold two fermions

15 Symmetric Spin State Now consider a symmetric spin state: Three ways to make a symmetric spin-state. What are they? S (1, 2) = + (1) + (2) S (1, 2) = 1 p 2 [ +(1) (2) + (1) + (2)] S (1, 2) = (1) (2) For all cases, the total spin of the system is S=1 (but the component of the spin in the z-direction is ħ, 0, -ħ, respectively) We call the S=1 state a triplet state.

16 What is the quantum state for the first excited state, and what is the degeneracy?

17 What is the quantum state for the first excited state, and what is the degeneracy? 3 symmetric spin states to go along with the anti-symmetric spatial state 1 anti-symmetric spin state to go along with the symmetric spatial state. Degeneracy = 4 (four quantum configurations that have the same energy).

18 Exchange Force Identical (non-interacting) particles with a symmetric/antisymmetric spatial wave-funcition behave as if an attractive/ repulsive force acts between them, as they tend to be more/less likely to be closer together than expected classically. Consider two non-interacting particles in the 1-D inf. square box. What is probability both particles are found in the left half of the box? P = L/2 L/2 (x 1,x 2 ) 2 dx 1 dx 2 0 0

19 P = L/2 L/2 0 0 If particles are distinguishable: (x 1,x 2 ) 2 dx 1 dx 2 = n (x 1 ) m (x 2 ), i (x) = 2 L sin(i x/l) P = 0 L/2 L/2 2 n(x 1 ) dx m(x 2 ) dx 2 = (1/2)(1/2) = 1/4 25% chance both on left, 25% chance both on right, and 50% chance one is on left and one is on right half.

20 Now consider indistinguishable particles with a symmetric spatial state (with particles in different spatial states): (x 1,x 2 )= 1 2 [ m(x 1 ) n (x 2 )+ n (x 1 ) m (x 2 )] P = 1 2 L/2 L/2 0 0 P = m(x 1 ) 2 n(x 2 )+ 2 n(x 1 ) 2 m(x 2 )+2 m (x 1 ) n (x 2 ) n (x 1 ) m (x 2 ) dx 1 dx K L/2 L/2 K = m (x 1 ) n (x 2 ) n (x 1 ) m (x 2 ) dx 1 dx 2 K = L/2 0 m(x) n (x) dx 2

21 For symmetric spatial states (singlet spin state), the probability both are are in the same half is increased by an amount of K/2. Particles are closer together on average than what we expect classically. For anti-symmetric spatial states (triplet spin state), probability is decreased by an amount of K/2. Particles are farther apart from each other on average than what we expect classically. Note: The particles are not interacting! This is solely due to them being indistinguishable Very important for understanding ground-state configurations of multi-electron atoms (Hund s rule, ferromagnetism, etc), as well as understanding covalent bonds in molecules