Beyond one particle in one dimension

Treating many particles

Particles moving in two, three- and more (!) dimensions

Role of exchange in case of identical particles

Generalizing to the multiparticle case

We had for the one-particle and one-dimensional case before that

The one particle wavefunction

In co-ordinate basis, we can think of it as (x)But, this was without thinking about the activity of other particles, which we presume are independent of this particle. That is fine, but clearly we are interested in systems with many particles and especially in systems in which there is an interaction between the particles. How do we start to talk about those cases? If the particles really are independent and in a sense imagine them far apart spatially, then we want to imagine simply that we can write down a single particle wave function for each one, say

and

But this is not adequate for several reasons...we need to understand this as a description of a single vector (a joint wave function) in a single hilbert space.

Two particles....Imagine the case of just two particles. They could be different particles, it doesn't matter. Let

represent the 1-particle wave function of the first particle. It is a vector in Hilbert space 1 be a wave function associated with the other particle ...it is a

vector in a different Hilbert space, say 2

Together the most simple way to understand a

joint Hilbert space is one in which we form 2-ples of vectors; in symbols,

Where the state would be . Note two important consequences of this choice;

1.) 1 2=

2.) The operators on the separate Hilbert spaces can be thought of as acting on the full in that they only act on their factor, and are 1 on the other factor.

i.e. It is important to realize that this means we are not just adding vector directions to one of the spaces....technically this is called a tensor product and is not a direct sum of vector spaces

.

Two particles....

Stop and think: The joint Hilbert space is not the direct sum

BUT the product

Ex: Suppose both =<a,b,c> and 1 =<x,y,z> are three dimensional

2

Then the full Hilbert space is spanned by 9 elements: |ax>, |ay> , |az>, |bx> ,|by>, bz> ,|cx> , |cy>, |cz>.

SO it cannot be the direct sum, which would only have a 6-dimensional basis!

Two particles....

This is reasonable and sensible in light of the fact that as operators on the full Then we have implemented the commutator relations called the canonical commutation relations (CCR)

Where the 'i' and 'j' label the part of the space the operator is acting on. In what follows we will be dropping the hats almost completely except where it might cause confusion, mostly to simply the notation and emphasize only the index describing which part of the full hilbert space the operator is acting on.

SPECIAL NOTE: there is no guarantee that all the operators on the big hilbert space can be written in terms of operators on each factor as we are discussing! But we focus on this anyways for reasons of clear exposition only...

So, for example, let |x1,x2> be a position basis of states in the full Hilbert space the action of the x-operators are thus,

And the inner product in the co-ordinate basis in the full Hilbert space can be written this way

Note, oddly, that this choice does depend on the fact that we have chosen cartesian co-ordinates.

Special Note In the full Hilbert space there are states that are not simple products of states in the separate factors. For example,

|3,4> + |1,2> Where '1,2,3,4' are separate positions, 1 and 3 of the first particle and 2 and 4 of the second particle.

The fact that the xi all commute with each other is consistent with our ability to write the full state this way (simultaneous diagonalization).

Such a state is called “Entangled”

Two particles....

For a two particle hilbert space (all in 1-d) we often come across two types of problems;

1.) Separable case:

Of course for the multiparticle system we still need

Where the Hamiltonian must make sense for each particle independent of the other. This means that essentially the most complicated Hamiltonian we could have would be

As would be the case for which

2.) Inseperable case: when the H doesn't split as above. These cases can be quite complicated, generally intractable analytically.

to solve:

Two particles....

Separable case- almost trivial one:

through

Separable Case – for simple interaction

For doesn't look separable until we go to

if

center of mass & relative co-ordinates. Define;

SO: solve each system '1' and '2' separately and combine the Hilbert spaces. Done !

Two particles....

Separable Case – for simple interaction (con't)

To keep the CCR, such as

Then the Hamiltonian can be written as

And, lo, it is separable! So the wave function is, in co-ordinates, |xcm, x> = A(xcm)B(x) where in the center of mass co-ordinate the solution A(xcm) is a plane wave (like the example of the translating gaussian minimum uncertainty packet of the last lecture. The B(x) is a solution of the 1-d schroedequation in the potential V(x) but as a massive particle of mass

We take

Three particle caseFor simplicity, take all the masses the same, m (but this can be generalized to different m's

In addition to the usual center of mass and total momentum,

We take One simple relative co-ordinate...causes this to be the other one

and..

So that the CCR are still true,

etc...in

Three particle case

Note further that

So that the kinetic energy part of the Hamiltonian remains separable in these co-ordinates. Generally that will not be true of the inter-particle potential, thoughthere is a special case! Once again in the (isotropic) harmonic oscillator case the potential energy in these co-ordinates is still separable;

N-particle caseConsider the case of N particles that we assume individually interact weakly with an external field but do interact more strongly with one another. In that case it is useful to separate out the center of mass motion and total momemtum from the other 'internal' degrees of freedom.

Note that from the underlying CCR on the 'fundamental' DOF xi and pj

We will have that

N-particle caseSaid another way, the commutator has not scaled away! Going to a large number of particles has not made the commutator vanish. The xcm and pcm of the ensemble, macroscopic quantities though they may be, are constrained by the same quantum mechanical relations and limits as the microscopic degrees of freedom.

Simply, we cannot think of quantum as only applying to the small ! Quantum properties persist at all scales and the separation of the center of mass motion is, for example, an example of why we can write down a quantum mechanical description of a proton although we know that it is made up of many particles.

Multi-dimensional case: single particle

Here, the CCR reduce to the same CCR for the multiparticle case in 1-d, a case we already know how to handle. The upshot here is that since x,y,z co-ordinates are assumed to commute, the vectors in the Hilbert space for a single particle in 3-d can be thought (loosely) as functions of three variables. Rather than have a theoretical discussion here about this, we remind about the development of the Hamiltonian in 3-d in the “Postulates” lecture several weeks ago and build up intuition in this case in the section on the Hydrogen atom that will be coming up soon!!

What does it mean in QM for two particles to be identical?

Return to the two particle case. If we have two identical particles, then quantum mechanics indicates that there is no test or measurement we can perform to distinguish them. We cannot even distinguish them by following their trajectories.

Let the one-particle wave functions for each particle be written |a(x1)> and |b(x2)> if they were independent, distinguishable particles so that we could even talk about the one-particle wave functions. In the full Hilbert space then we could write the state |a(x1)b(x2)> were they distinguishable.

What if they are they are identical particles. Then, no matter where they are or what they are (individually) doing, there is no way to distinguish the state |a(x1)b(x2)> from the state |b(x1)a(x2)>. NOTATION: |ab> = |a(x1)b(x2)> , reading the factors in the multiparticle state from left to right in their position dependence.

What does it mean in QM for two particles to be identical?

Being identical particles must mean that the state |ab> and the state |ba> are physically the same; as vectors in Hilbert space then they must be proportional

|ab> = |ba> where is a phaseOf course, interchanging two particles can be thought of actively, as an operator on the full Hilbert space. It is a linear operator. So, we can think of the above statement as a definition of the exchange operator in the Hilbert space. We can thus interchange the particles twice, finding

|ab> = |ab>

And so must be +1 or -1

Bosons and Fermions

=1 Bosons. So the WF can be written (|ab>+|ba>)/

= -1 Fermions. So the WF can be written (|ab> - |ba>)/

In nature each body is either a Boson or Fermion Identical bosons can have the same wavefunction (and thus the same quantum numbers)Fermions never have the same (one-particle) wave function

(state is identically zero if a=b)

Bosons and Fermions

=1 Bosons. So the WF can be written (|ab>+|ba>)/

= -1 Fermions. So the WF can be written (|ab> - |ba>)/

In nature each body is either a Boson or Fermion Identical bosons can have the same wavefunction (and thus the same quantum numbers)Fermions never have the same (one-particle) wave function

(state is identically zero if a=b)

Photon, higgs, gauge bosons (W+/-, Z0), 87Rb, 85Rb, 23Na, Buckeyball, Phonon, gluons, graviton (?), meson,

Proton, neutron, electron, 86Rb, quark, neutrino, muon

Bosons and Fermions

=1 Bosons. So the WF can be written (|ab>+|ba>)/

= -1 Fermions. So the WF can be written (|ab> - |ba>)/

In nature each body is either a Boson or Fermion Identical bosons can have the same wavefunction (and thus the same quantum numbers)Fermions never have the same (one-particle) wave function

(state is identically zero if a=b)

Photon, higgs, gauge bosons (W+/-, Z0), 87Rb, 85Rb, 23Na, Buckeyball, Phonon, gluons, graviton (?), meson,

Proton, neutron, electron, 86Rb, quark, neutrino, muon

Bose-Einstein Condensates

Fermi Pressure

Bosons and FermionsEx 1: from Modern Class or your chemistry class:

The angular momentum possible in homonuclear molecules...Either all even (bosons) or All odd (fermions) multiples of

Ex 2: Note that < fermion | boson> =0 always.

Bosons and FermionsLet us consider constructing the states of three particles in a 1-d

harmonic oscillator.

The Hilbert space for distinguishable particles is spanned by |n1, n2, n3> where the ni are non-negative integers, and any combination of these values is a state. See the problem 10.2.3 which shares some similarities with this case

Distinguishable Particles

Identical BosonsThe hilbert space of three identical bosons in the 1-d harmonic oscilator is spanned by combinations of quantum numbers that are symmetric under the interchange of any two. As such, these states are a subspace of the Hilbert space of the case in which all the particles are distinguishable. Here are the low lying states in that basis: |0,0,0> the ground state, (|1,0,0> + |0,1,0> + |0,0,1>)/

is the first excited state.

Bosons and FermionsLet us consider constructing the states of three particles in a 1-d

harmonic oscillator.

The Hilbert space for distinguishable particles is spanned by |n1, n2, n3> where the ni are non-negative integers, and any combination of these values is a state. See the problem 10.2.3 which shares some similarities with this case

Distinguishable Particles

Identical BosonsThe hilbert space of three identical bosons in the 1-d harmonic oscilator is spanned by combinations of quantum numbers that are symmetric under the interchange of any two. As such, these states are a subspace of the Hilbert space of the case in which all the particles are distinguishable. Here are the low lying states in that basis: |0,0,0> the ground state, (|1,0,0> + |0,1,0> + |0,0,1>)/

is the first excited state.

Bosons and Fermions

Identical Bosons (con't) Here are some other states in the identical boson case:

(|1,1,0>+ |0,1,1>+|1,0,1>)/

(|2,0,0>+|0,2,0>+|0,0,2>)/

(|0,1,2>+|1,0,2>+|0,2,1>+|1,2,0>+|2,1,0>+|2,0,1>)/ And so on....Identical Fermions

Here we must require that the WF be antisymmetric under the exchange of any two (single particle) states....still a subspace of the distinguishable particle Hilbert space … some examples;(|0,1,2>-|1,0,2>-|0,2,1>+|1,2,0>-|2,1,0>+|2,0,1>)/ This would be the

ground state ! (|0,1,3>-|1,0,3>-|0,3,1>+|1,3,0>-|3,1,0>+ 3,0,1>)/

first excited state(|2,1,3>-|1,2,3>-|2,3,1>+|1,3,2>-|3,1,2>+|3,2,1>)/ another state

Bosons and Fermions

Identical Bosons (con't) Here are some other states in the identical boson case:

(|1,1,0>+ |0,1,1>+|1,0,1>)/

(|2,0,0>+|0,2,0>+|0,0,2>)/

(|0,1,2>+|1,0,2>+|0,2,1>+|1,2,0>+|2,1,0>+|2,0,1>)/ And so on....Identical Fermions

Here we must require that the WF be antisymmetric under the exchange of any two (single particle) states....still a subspace of the distinguishable particle Hilbert space … some examples;(|0,1,2>-|1,0,2>-|0,2,1>+|1,2,0>-|2,1,0>+|2,0,1>)/

(|0,1,3>-|1,0,3>-|0,3,1>+|1,3,0>-|3,1,0>+ 3,0,1>)/first excited state

(|2,1,3>-|1,2,3>-|2,3,1>+|1,3,2>-|3,1,2>+|3,2,1>)/ another state

Bosons and Fermions

Identical Bosons (con't) Here are some other states in the identical boson case:

(|1,1,0>+ |0,1,1>+|1,0,1>)/

(|2,0,0>+|0,2,0>+|0,0,2>)/

(|0,1,2>+|1,0,2>+|0,2,1>+|1,2,0>+|2,1,0>+|2,0,1>)/ And so on....Identical Fermions

Here we must require that the WF be antisymmetric under the exchange of any two (single particle) states....still a subspace of the distinguishable particle Hilbert space … some examples;(|0,1,2>-|1,0,2>-|0,2,1>+|1,2,0>-|2,1,0>+|2,0,1>)/

(|0,1,3>-|1,0,3>-|0,3,1>+|1,3,0>-|3,1,0>+ 3,0,1>)/

(|2,1,3>-|1,2,3>-|2,3,1>+|1,3,2>-|3,1,2>+|3,2,1>)/ another state

FermionsThat concludes our discussion of the case with three particle in the 1-d HO potential. We can however synthesize the fermionic example which in terms of the one-particle states will always be made from exactly 3! = 6 products of single particle states into a general statement. Let represent the 1-particle states in a N-identical fermion system, interacting or not. The hilbert space ofthe fermionic system is then spanned by the “Slater Wavefunctions”

=

This is identical to what we constructed by hand in the preceeding N=2 and N=3 cases. Note that the exchange antisymmetry is guaranteed by the properties of the determinant.

Entanglement and the Mystery of the Quantum Cakes

Read: P. Kwait and L. Hardy, Am. J. of Physics, 68 Pg. 33 (2000)

Read and discuss together in class....This is a very clear exposition of the novelty of entanglement in quantum mechanics, which indicatesthat there are quantum systems with correlations in them that cannotbe thought of in terms of any type of joint classical probability function.

Enjoy!