Lecture 01

Lecture Lecture 1: Identica Identicall particles, introduction (9/20/2005) The main unifying theme of this course will be time-dependence (and dynamics), but we will start with identical particles. In classical classical physics, physics, particles particles are distinguis distinguishable hable.. Even Even if two electrons electrons happen happen to have have exactly exactly the same mass, mass, charge, harge, and other quant quantiti ities es,, you may trace them and distinguish which electron is Peter and which electron is Paul. However, particles in quantum mechanics don’t have any ID. If two electrons scatter, it is impossible (even in principle) to decide which of them is Peter Peter and which which is Paul. Paul. Actually Actually,, both possibi p ossibilitie litiess contribu contribute te to the final wavefunction that decides about the probabilities of various results. Draw a picture of two possible histories of electrons that scatter elastically; one can’t say which of them occured.

Recomme Recommended nded reading: reading: Griffiths Griffiths ch. 5.1, Gasioro Gasiorowicz wicz ch. 20.

Symmetrizing the multi-particle wavefunctions Recall that in quantum mechanics, two particles are not generally described by two independent “waves” ψ(x) and ψ and  ψ((y ) but rather by a wavefunction in six dimensions ψ(r1, r2) that determines the probability for the first one to be in the volume dV 1 around  around r1  and the second to be in a small volume dV 2  around   around  r2  to be dP  = dV 1 dV 2 ψ(r1 , r2) 2.

|

|

Similar statements apply for many particles, too. The wavefunction becomes a function of 3k 3k  coordinates for k  particles.  particles. Imagine Imagine a quantum quantum mechanical mechanical system for two electrons in an external field. The Hamiltonian is gonna be p21 p21  +  + V   +  V  (  (r1 ) + V  + V  (  (r2 ) + V  +  V rel H  = r1 rel ( 2m 2m 1

− r ) 2

where p2i = h ¯ 2 2i   enters the kinetic energy, V  (r) is the potential energy between an electron and the nucleus, and  V rel (r1 r2 ) is the mutual interactions of the electrons. Obviously it looks nicely symmetric in both electrons. ˆ12  that What do we mean mathematically? Define the exchange operator P  exchanges the electrons. It means that it acts as follows:

 − ∇

−

 ˆ12 ψ)(r1, r2 ) = ψ(r2, r1). (P  If the electrons’ wavefunction depended on other quantities, such as their spin, the spins would have to be exchanged on the RHS, too. Note that ˆ 2  = 1; it squares to the identity operator: P  12  ˆ 2 ψ)(r1, r2 ) = ( P   ˆ12 (P   ˆ12ψ))(r1 , r2) = (P   ˆ12ψ)(r2 , r1) = 1 ψ(r1, r2). (P  12

·

Now it’s important to realize that the exchange operator commutes with the Hamiltonian:  ˆ12, H ] P  ˆ12H  H  ˆ [P  P 12  = 0.

≡

 −

It does not matter whether you first act with the Hamiltonian and then you switch the dependence of the resulting wavefunction on r1 , r2 , or you first exchange the electrons and then act with the Hamiltonian. You’ll get the same results. Fine. So now, because these two operators commute, we can simultaneously diagonalize them. Because one of the two operators is the ˆ12  will be Hamiltonian that determines time evolution, the eigenvalues of  P  conserved. ˆ12 . Because we’ve More concretely, it is useful to consider eigenstates of  P  proved that the exchange operator squares to 1, the same thing holds for the eigenvalues. If we find an eigenfunction (let’s start to omit some awkward parentheses) i.e. if  ˆ12 ψ(r1, r2 ) = pψ(r1 , r2), P  then also

ˆ 2 ψ(r1 , r2) = p 2 ψ(r1, r2 ). P  12

But we know that the latter equals ψ(r1, r2 ) which means that p2 = 1; the eigenvalues must satisfy the same algebraic equation as the operator. Obviously this is solved by p = 1. How do the eigenfunctions look like?

±

 p = +1 :  p =

−1 :

ψ(r1 , r2) = +ψ(r2, r1 ), ψ(r1 , r2) =

−ψ(r , r ), 2

1

We say that the wavefunctions are symmetric in the first case, and antisymmetric in the second case. 2

Bosons and fermions Now imagine 3, 4, or more electrons, and prepare them in the eigenstate of  ˆ12 , P  ˆ13 , and other exchange operators. If the electrons are indistinguishable, P  all these eigenstates must be equal. So for any kind of particles, the eigenvalues of all exchange operators must be universally either +1 – this is when we call the particles bosons - or they must equal 1 – which is the case of  fermions. Now it is natural to assume that the wavefunction of bosons must be symmetric – and the wavefunction of identical fermions must be antisymmetric - with respect to any exchange of a pair of particles. More general wavefunctions, especially antisymmetric wavefunctions for bosons and symmetric for fermions, are simply forbidden by mother Nature. This assertion actually follows from quantum field theory that describes multiparticle states as action of “creation operators” for single-particle states acting on the “vacuum state” called 0 . For example, in quantum field theory, a state where one electron is at place X  and another one is at Y  is written as ˆb† (X )ˆb† (Y  ) 0  = ˆb† (Y  )ˆb† (X ) 0 which follows from  ˆb† (X )ˆb† (Y  ) = ˆb† (Y  )ˆb† (X ). (In the real life, the dagger usually kills. But in quantum field theory, it creates things.) Translated to the language of ordinary wavefunctions this means our usual

 −

|

|   −  −

ψ(r2 , r1) =

|

 −ψ(r , r ). 1

2

For bosons there would be a plus sign. Because these creation operators commute for bosons – and anticommute for fermions – the required (anti)symmetry of the wavefunction follows. Also, one can follow Wolfgang Pauli and prove the so-called  spin-statistics relation  using quantum field theory:

•  the bosons, i.e.

the particles with symmetric wavefunctions, p = +1, must have integral spin j Z , such as 0, 1, 2, . . .; examples include photons, W,Z-bosons, Higgs bosons, and bound states containing an even number of fermions

∈

•   the fermions, i.e. the particles with antisymmetric wavefunctions, p = −1, must have half-integral spin j ∈ Z + 1/2 such as 0.5, 1.5, 2.5 and so on; this includes electrons (and neutrinos), quarks, protons, neutrons, and bound states containing an odd number of other electrons 3

These relations between the spin and the symmetry of the wavefunction follow from the assumption that the energy is bounded from below and the norm of the states (and thus the probabilities) must be non-negative. The wavefunction of two Helium atoms is easily seen to be symmetric ( p = +1) because the exchange of two Helium atoms ( He42) is nothing else than the exchange of two pairs of electrons, two pairs of neutrons, and two pairs of protons. And p2e p2n p p2  = +1 regardless of the individual signs. But more generally, one needs quantum field theory to prove the spin-statistics relation above.

Statistics, the Pauli exclusion principle, and the distribution You know from high school chemistry that two electrons can’t occupy exactly the same state of an atom. This Pauli exclusion principle can easily proved from antisymmetry of the fermionic wavefunctions: ψ(MA,MA) =

 −ψ(MA,MA) ⇒

ψ(MA,MA) = 0.

We chose Massachusetts but it holds for any states; the wavefunction is always antisymmetric whatever basis we choose for the Hilbert space. It is antisymmetric in the coordinate representation; it is also antisymmetric in the momentum representation (which can be trivially seen from the Fourier transform); ˜  p1, p 2 ) = ψ(  ˜  p2 , p 1 ) ψ( 

 −

it is also antisymmetric if the states such as 1s(up), 1s(down), 2s(up), 2s(down) etc. are chosen as the basis. The antisymmetry of the fermions’ wavefunctions is crucial for the diversity of atoms and the existence of things such as life. Bosons behave very differently than the fermions. They not only have no difficulties to live in the same state, but they actually like it – which is the driving force behind the so-called Bose-Einstein condensation and behind the coherence of LASERs (photons are bosons). In classical physics, the thermal distribution of particles treats all of them independently. The probability that a given particle appears in the state of  energy E   decreases as the Maxwell-Boltzmann factor  p =

1 1 1  exp( E/kT ) = Z  Z  exp(E/kT ) + 0.

−

4

where Z  is a normalization factor that must be chosen in such a way that the probabilities add up to one. In quantum statistical physics, the particles are indistinguishable and we must look at particular states, not particles. In a given state, the number N f  of fermions of the same kind may be either 0 or 1. In another state, the number of bosons  N b  of the same kind may be 0, 1, 2, 3, etc. If the energy of the state is  E f , the energy of  N f  fermions in this state is N f E f ; similarly if you replace f  by b. Different configurations described by the number of bosons and fermions in various states are weighted again by the Maxwell-Boltzmann factor.

Fermi-Dirac distribution Consider the fermions. The number  N f  for a given state is either 0 or 1. The probability that there is exactly one fermion there must be exp( E f /kT ) times the probability that there is no fermion. Because the probabilities add up to one, it implies that

−

 p(N f  = 0) =

1 , exp( E f /kT ) + 1

p(N f  = 1) =

−

  exp( E f /kT ) . exp( E f /kT ) + 1

−

−

What is the   average   number of particles in that state? It is obviously equal to the probability that there is one particle: 1 N   = exp(E  /kT ) +1 f 

f 

In Maxwell-Boltzmann statistics, the +1 term in the denominator was missing; the number of particles in a different state was simply proportional to the probability distribution of each individual distinguishable particle.

Bose-Einstein distribution Consider the bosons. The number N b  for a given state is either 0 or 1 or 2 or any non-negative integer. The probability that there are  N b  bosons must depend on N b  like exp( N b E b /kT ). But the sum of these numbers must add up to one. You can see that the unique solution is

−

 p(N b  = nb ) = exp(nb E b /kT ) 5

− exp((n  + 1)E  /kT ) b

b

What is the average number of bosons in the state? ∞

1 exp(E b /kT )

N   =   p(N   = n )n  = b

b

b

b

nb =0

−1

The series can be calculated as a derivative of a geometric series with respect to an extra parameter. The result may be more important: note that it has a similar form like the Maxwell-Boltzmann or the Fermi-Dirac distribution, but the relative sign in the denominator differs from the fermions. When we add some volumes of the phase space (the density of states which gives a power law factor ω 3 ) to the Bose-Einstein formula above, we obtain the famous blackbody formula due to Planck (1900). Note that for E f , E b kT , the 1 term in the denominator may be neglected (the exponential wins) and the quantum distributions reduce to the classical distribution. However, for states whose energy  E f , E b  is comparable to kT  (or smaller), quantum mechanics and indistinguishability is always important.



 ±

The Slater determinant Let us return to the wavefunctions. Imagine that you want to construct a multi-particle wavefunction for n  identical particles that are as independent as possible and that are described by the wavefunctions ψ1 (r),

ψ2(r),

ψ3 (r)   . . .

ψn (r).

Our first guess is the product ψ(r1 , r2 , . . . rn ) =  ψ1 (r1)ψ2 (r2) . . . ψn (rn ). But because ψi are different functions, this does not have the right (anti)symmetry. We must (anti)symmetrize the wavefunction to obtain permutations



ψ(r1 , r2 , . . . rn) = A

ψ p (r1)ψ p (r2 ) . . . ψ pn (rn ) 1

2

 p

for the bosons or permutations

ψ(r1, r2 , . . . rn ) = A

p

 (−1)  p

6

ψ p (r1 )ψ p (r2) . . . ψ pn (rn) 1

2

where  p  is the sign of the permutation for fermions. The latter wavefunction can also be written as the Slater determinant:

 ψ (r )  ψ (r )  . A det ψ (r ) = A   ..  ψ (r ) i

i,j

1

1

1

2

1

n

ψ2 (r1 ) ψn (r1 ) ψ2 (r2 ) ψn (r2 ) .. .. ... . . ψ2 (rn ) ψn(rn)

··· ···

 j

    

···

The normalization constant A  may be determined easily if all one-particle wavefunctions ψ i (r) are orthogonal, i.e. ψ¯ j (r)ψi (r) = 0 for i =  j. Then the norm of the multiparticle wavefunction is simply a sum of  n! identical terms (the mixed terms cancel by orthogonality) which tells us that



  

√ 1n! .

A =

Note that for two-particle states, the (anti)symmetrized wavefunction is as simple as ψ1 (r1 )ψ2(r2 ) ψ2 (r1 )ψ1 (r2 ) ψ(r1, r2) = . 2

± √ 

Exchange forces Because two fermions can’t occupy the same state, you may think that they don’t even want to be too close. And you would be right. Even non-interacting fermions tend to be further apart than if they were distinguishable. And bosons tend to be closer together than non-interacting distinguishable particles would be. To see it, start with distinguishable particles whose wavefunction is ψ(x1, x2 ) = ψa (x1 )ψa (x2 ). Calculate 2

2 1

2 2

(x − x )  = x  − 2x x  + x  1

2

1

2

It is easy to see that the result is simply . . . = x2 a + x2

    − 2x x . b

a

b

For example, 2 1

x  =

 

2 1 1

dx x ψa (x1 )

|

2

|

7

 

dx2 ψb (x2 ) 2 = x2

|

|   

a

because the second integral in the product equals one (normalization). The same thing applies to x2 b . Also, the integral calculating 2 x1 x2 factorizes as indicated. For identical particles we must change a couple of things, starting from the wavefunction. The sign below refers to bosons or fermions, respectively.

 

ψ±

−



± 1 (x , x ) = √  [ψ (x )ψ (x ) ± ψ (x )ψ (x )]. 2 1

2

=

 

a

b

1

b

2

a

1

2

In this case, 2 1

x 

dx1 dx2 x21 ψ± (x1, x2 ) 2

|

1 = 2 1 + 2 1 2 1 2

(1)

|

    dx x |ψ (x )| dx |ψ (x )|     dx x |ψ (x )| dx |ψ (x )|   ¯    ¯ ± dx x ψ (x )ψ (x ) dx ψ (x )ψ (x )   ¯    ¯ 2 1 1

a

1

2 1 1

b

1

2 1 1

a

2

2

b

1

2

b

2

2

a

2

1

(2)

2

(3)

b

2

dx1 x21 ψb (x1 )ψa (x1 )

±

2

a

2

(4)

2

dx2 ψa (x2 )ψb (x2 )

(5)

The last two terms (among the four) vanish because their latter integral factor equals zero by orthogonality of  ψa and ψb . The first two terms give you 1 . . . = ( x2 a  + x2 b ) 2 Obviously, the result for x22  is the same thing because of an  a-b symmetry. The last term in the expectation value of the distance, namely x1x2 , requires an extra calculation:

   

  

1 = 2 1 + 2 1 2 1 2 = x





    x x  dx x |ψ (x )| dx x |ψ (x )|     dx x |ψ (x )| dx x |ψ (x )|    ¯    ¯ ± dx x ψ (x )ψ (x ) dx x ψ (x )ψ (x )    ¯    ¯ 1

2

1 1

a

1

1 1

b

1

1 1

a

2

2

b

1

1

dx1 x1 ψb (x1 )ψa (x1 )

±

  x ± |x | a

b

ab

8

2

2

b

2

2 2

a

2

2 2

b

2

2

(6)

2

(7)

2

a

1

dx2 x2 ψa (x2 )ψb (x1)

(8) (9) (10)

where we defined the so-called “overlap integral”

x

ab

 =

 

dx x ψ¯a (x)ψb (x)

as the first factor in the third term (with x1 x, and it also appears in the conjugated form, both in the third and the fourth term). This overlap integral is only nonzero if the particles overlap in space; only in this case, their indistinguishability affects the expectation value (and thus the force). Otherwise we get back the previous result. The qualitative conclusion of  the calculated expectation values is, of course, that bosons tend to be closer together while the fermions tend to be further apart.

→

Spins and binding energies of some nuclei and atoms Hydrogen molecules and helium Even a qualitative analysis of this result shows that the indistinguishability of  the particles matters and it can be shown to be responsible for the differences between various superficially similar systems in atomic (electrons moving around nuclei), molecular (electrons moving among many nuclei) and nuclear (protons and neutrons moving inside nuclei) physics. For example: Hydrogen is a gas that likes to form the  H 2   molecules. But the helium and the hydrogen bind neither to He2 nor HeH . What causes the difference? The molecule H 2   is a bound state with a large (negative) binding energy for the electrons. This bound is expected to be strong if the electrons are concentrated in between the two nuclei. How does it work? Imagine two electrons with the same spins (up-up). Then the dependence of the wavefunction on the spins is symmetric, and the dependence of the wavefunction on the position must be antisymmetric. The electrons tend to be further apart, and they don’t want to be concentrated in the middle simultaneously: that causes ”anti-binding”. We’ve shown that it’s the case when the total spin is  S  = 1. But what about the total spin of the electrons equal to S   = 0? In that case, the dependence of the wavefunction on the spins is described by the singlet state (“singlet” means that there exists a single polarization only; the projection of the spin can only be zero because the spin is zero): χ =

√ 12 [ |↑ |↓ −|↓ |↑ ] a

b

9

a

b

The Fermi statistics requires the total wavefunction – that can be usefully written in a factorized form as long as the Hamiltonian does not directly depend on the spins (or if it at least splits to two Hamiltonians that depend on the positions and spins separately) Ψ(x1 , x2 ; s1 , x2 ) =  ψ(x1 , x2 )χ(s1 , s2) to be antisymmetric Ψ(x1 , x2; s1, x2 ) =

 −Ψ(x , x ; s , x ). 2

1

2

1

But because the spin wavefunction is already antisymmetric in our singlet state χ(s1 , s2) = χ(s2 , s1 ),

 −

the spatial wavefunction for the electrons must actually be symmetric ψ(x1 , x2 ) = +ψ(x2 , x1 ). This effectively causes the electrons to behave as bosons. Both of them occupy the space in between the nuclei, and the bound of  H 2  becomes strong. If at least one of the atoms is helium, it contains electrons both with spin up and spin down (the helium is a singlet itself). Therefore the other atom  – either H  or He  – must have at least one electron that shares the same spin state with an electron of the first helium atom. That’s enough of an antibinding to destroy chances for a H e2 or H eH   bound state. We need to perform a more careful analysis for the more complex atoms; see Gasiorowicz.

Nuclei and isospin Spin 1/2 particles can be spinning “up” or “down”. Later we will see that they are spinors – a representation of the SU(2) group. The same mathematics may be applied in other contexts with different interpretations than the actual spinning in space. Consider protons and neutrons. They have a different charge. But the electric charge is not so important in the nuclei where the strong nuclear force is much more significant. From this perspective, a proton and a neutron look like twins; a doublet. They may be described as “nucleons” that only differ by the isospin (isotopical spin): the proton is a nucleon with spin “up” while 10

the neutron’s isospin points “down” (these directions have nothing to do with the real space). Why is there a deuteron - a bound state of one proton and one neutron np (heavy Hydrogen nucleus) but no pp  state (neutron-free helium) or nn (double-neutron)? And why does the deuteron have spin one? The total wavefunction of two nucleons may be written as Ψ = ψspace ψspinψisospin and it must be antisymmetric for nucleons that are fermions because of their spin 1/2. The ground state of the deuteron tries to minimize the angular momentum, much like the ground state of the Hydrogen atom. This implies that ψspace   is symmetric. This means that ψspin ψisospin   must be antisymmetric. We’re left with two possibilities: 1. ψspin   antisymmetric, ψisospin   symmetric: J  = 0 : ψspin  =

| ↑| ↓√  − | ↓| ↑, 2

I  = 1 : ψiso

 | p| p  √   =  (| p|n + |n| p)/ 2  |n|n

2. ψspin   symmetric, ψisospin   antisymmetric J  = 1 : ψspin

 | ↑| ↑  √   =  (| ↑| ↓ + | ↓| ↑)/ 2  | ↓| ↓

,

I  = 0 : ψiso  =

 | p|n√  − |n| p 2

Because of more quantitative and difficult calculations that we can’t do here, nuclear physics prefers the second alternative as the only allowed bound state. This implies that the particle must have one proton and one neutron (deuteron) and its spin must be 1 (a triplet).

Atoms and Hund’s rules You may want to read Griffiths, ch. 5.2. We will only sketch the Hund’s rules here. The atoms are described by the Hamiltonian for  N  electrons (kinetic plus potential energy for the nucleus) plus their pairwise interactions. 11

We can solve a simpler system in which the electrons’ interactions with each others are neglected. In that case, one obtains many copies of the electrons in the Hydrogen eigenstates, described by  n, l, m, sz   where n = 1, 2, 3, . . . is the principal quantum number, l  = 0, 1, 2, . . . n 1 is the orbital quantum number, m = l, l + 1, . . . , l is its z -projection, and s z = 1/2 is the spin. The Pauli principle says that we can only put one electron to each state. The electron’s energy would only depend on their principal quantum numbers. In reality, the self-interactions rearrange the shells a bit, so that the most bound ones are (in this order)

−

− −

±

1s, 2s, 2 p, 3s, 3 p, 4s, 3d , . . . where the letters  s, p, d, f , g , h, i . . . stand for sharp, principal, diffuse, fundamental, g-shaped, h-shaped, i-shaped, and so forth, and denote l = 0, 1, 2, 3, 4, 5, 6 . . . Moreover, the lowest energy eigenstates of the atoms may usually be determined by phenomenological Hund’s rules.

•   couple the valence electrons (those in the highest partially occupied

shell) to give maximum total spin; for chemists: make as many spins as possible parallel, compute the total s z = ms , and call it s (that’s the highest component of a multiplet)

•  given this choice, construct the maximum l compatible with the Pauli

principle; for chemists: try to obey the previous rule but fill in the shells to obtain the maximum lz = ml  and call this maximum value l (again the highest component of the multiplet)

•   the total angular momentum j is j = l + s   if the shell is more than half-full, and j  = l − s otherwise

The explanation why two of these rules “work” is the following: the first rule tries to make the spin wavefunction as symmetric as possible, which means that the spatial wavefunction is antisymmetric as possible. Such an antisymmetry reduces the probability that the electrons are too close to each other, and it reduces the positive energy contribution from their self-interactions. The third rules is implied by a spin-orbital interaction proportional to l s which is a linear function of  j 2 for fixed l, s. The rule tries to minimize this term if the shell is at most half-occupied; if it is more than half-full, it is better to think in terms of holes for which the conclusion reverses.

·

12