# Quantum Mechanics Review

• Debroglie - wave particle duality
 Particle Wave
• Heisenberg uncertainty relationship
where characterizes the location of in the x-y plane
• Energy conservation
Classical:
Quantum (Schrodinger Equation):
• Quantum interpretations
E = energy eigenvalues
= eigenvectors
= probability of finding the particle (wave packet) between and
= complex conjugate
= average (expectation) value of observable after many measurements of
example:
• Constraints on Quantum solutions
1. is continuous accross a boundary : and ( if is infinite this second condition can be violated)
2. the solution is normalized:
• Current conservation: the particle current density associated with the wave function is given by

## Schrodinger Equation

### 1-D problems

#### Free particle

If there is no potential field (V(x) =0) then the particle/wave packet is free. The wave function is calculated using the time-dependent Schrodinger equation:

Using separation of variables Let:

Substituting we have

reorganizing you can move all functions of on one side and on the other suggesting that both sides equal some constant which we will call

Solving the temporal (t) part:

: just integrate this first order diff eq.

Solving the spatial (x) part:

Such second order differential equations have general solutions of

where

Now put everything together

Notice
the wave function amplitude does not change with time
also, if the operator for an observable A does not change in time, then
even though particles are not stationary they are in a quantum state which does not change with time (unlike decays).
the term of amplitude represents a wave traveling in the +x direction while the second term represents a wave traveling in the -x direction.
Example
consider a free particle traveling in the +x direction
Then
wave moving to right
if the particles are coming from a source at a rate of particles/sec then

#### Step Potential

Consider a 1-D quantum problem with the Step potential V(x) defined below where

Break these types of problems into regions according to how the potential is defined. In this case there will be 2 regions

##### x<0

When x<0 then V =0 and we have a free particle system which has the solution given above.

where and
##### x>0

separation of variables:

Constant

The time dependent part of the problem is the same as the free particle solution. Only the spatial part changes because the Potential is not time dependent.

Constant

If then we have a wave that traverses the step potential partly reflected and partly transmitted, otherwise it will be reflected back and the part that is transmitted will tunnel through the barrier attenuated exponentially for x>0.

Here is how it works out mathematically

For the case where :

SHM solutions

The above Diff. Eq. is the same form as the free particle but with a different constant

Let

Now apply Boundary conditions:

and

We now have a system of 2 equations and 4 unknowns which we can't solve.

Notice
The coefficient "D" in the above system represent the component of represent a wave moving from the right towards x=0. If we assume the free particle encountered this step potential by originating from the left side, then there is no way we can have a component of moving to the left. Therefore we set .
The coefficient A represent the incident plane wave on the barrier. The remaining coefficients B and C represent the reflected and transmitted components of the traveling wave, respectively.
Know our system of equations is
If I assume that the coefficient A is known (I know what the amplitude of the incoming wave is) then I can solve the above system such that

similarly

###### Reflection (R) and Transmission (T) Coefficients
Let

Then

exponential decay
Assume solution
Recall the solution for x<0
where
Apply Boundary conditions

If

Then

Continuous conditions at x=0

Assuming A is known we have 2 equations and 2 unknowns again

###### Reflection (R) and Transmission (T) Coefficients=
Evanescent waves
Waves like which carry no current. There is a finite probability of penetrating the barrier (tunneling) but no net current is transmitted. A feature which separates Quantum mechanics from classical.

#### Rectangular Barrier Potential

Barrier potentials are 1-D step potentials of height which have a finite step width:

We now have 3 regions in space to solve the schrodinger equation

We know from the free particle solutions that on the left and right side of the barrier we should have

where

But in the region we have the save type of problem as the step in which the solution depends on the Energy of the system with respect to the potential. One solution for the (oscilatory) system and one for the (exponetial decay) system.

where

For the case where :

Before we set because there wasn't a wave moving to the left towards the interface. The rectangular barrier though could have a wave reflect back form the interface.

Apply Boundary conditions

and

and

and

We now have a system of 4 equations and 6 unknowns (A,B,C, D, F and G).

But:

: no source for wave moving to left when x>a

If we treat as being known (you know the incident wave amplitude) then we have 4 unknowns (B,C,D, and F) and the 4 equations:

###### Transmission
= the transmission coefficient

To find the ration of F to A

1. solve the last 2 equations for C & D in terms of F
2. solve the first 2 equations for A in terms C and D
3. 3.)substitute your values for C and D from the last 2 equations so you have the ratio of B/A in terms of F/A

1.)solve the last 2 equations

for C and D

2.) solve the first 2 equations for B in terms of C & D

for A in terms of C and D

3.) Find Reflection Coeff in terms of Transmission Coeff

or

since

Then

### 3-D problems

#### Infinite Spherical Well

What is the solution to Schrodinger's equation for a potential V which only depends on the radial distance (r) from the origin of a coordinate system?

Such a potential lends itself to the use of a Spherical coordinate system in which the schrodinger equation has the form

In spherical coordinates

Note

so

Using separation of variables:

which we can also write as

where

Substitute

##### V=0

We have a constant on the right hand side so the left hand side must also be constant

a "centrifugal" barrier which keeps particles away from r=0

substituting

In the region where V=0

Let

Then we have the "spherical Bessel"differential equation with the solutions:

where

##### and Table
 0 0 1 0 1 2 0 1 2

The general solution for the 3-D spherical infinite potential well problem is

= eigen function(s)

where

are quantization number and quantum energy level = eigen state(s)
##### Energy Levels

To find the Energy eignevalues we need to know the value for "k". We apply the boundary condition

at

to determine the "nodes" of ; ie value of so if you tell me the size of the well then I can tell you the value of k which will satisfy the boundary conditions. This means that "k" is not a "real" quantum number in the sense that it takes on integral values.

We simple label states with an integer representing the zero crossing via:

For example:

In the case
when
You arbitrarily label these state as
In the case
Notice
The angular momentum is degenerate for each level making the degeneracy for each energy

#### Simple Harmonic Oscillator

The potential for a Simple Harmonic Oscillator (SHM) is:

This potential is does not depend on any angles. It's a central potential. Our solutions for Y_{l,m} from the 3-D infinite well potential will work for the SHM potential as well! All we need to do is solve the radial differential equation:

or

When solving the 1-D harmonic oscillator solutions were found which are of the form

where

If you construct the solution

Assume R(r) may be written as

substituting this into the differential equation gives

The above differential equation can be solved using a power series solution

After performing the power series solution; ie find a recurrance relation for the coefficents a_i after substituting into the differential equation and require the coefficent of each power of r to vanish.

You arrive at a soultion of the form

where

polynomial in of degree in which the lowest term in is

these polynomials are solutions to the differential equation

if you do the variable substitution

you get

the above differential equation is called the "associated" Laguerre differential equation with the Laguerre polynomials as its solutions.

The following table gives you the Radial wave functions for a few SHO states:

 0 0 1 1 2 0 2 2 3 1
Note
Again there is a degeneracy of for each
Again E is independent of (central or constant potentials)
if is odd is odd and if is even is even
multiple values of occur for a give such that
The degeneracy is because of the above points

#### The Coulomb Potential for the Hydrogen like atom

The Coulomb potential is defined as :

where

atomic number
charge of an electron
permittivity of free space =

This potential does not depend on any angles. It's a central potential. Our solutions for from the 3-D infinite well potential will work for the Coulomb potential as well! All we need to do is solve the radial differential equation:

or

Use the change of variable to alter the differential equiation

Let

Then the differential equation becomes:

Consider the case where (Bound states)

Bound state also imply that the eigen energies are negative

Let

Rydberg's constant

###### Boundary conditions
• if is large then the diff equation looks like

To keep finite at large you need to have B=0

• if is very small ( particle close to the origin) then the diff equation looks like

The general solution for this type of Diff Eq is

where A =0 so is finite

A general solution is formed using a linear combination of these asymptotic solutions

where

substitute this power series solution into the differential equation gives

which is again the associated Laguerre differential equation with a general series solution containing functions of the form

with the recurrance relation

notice that

now diverges for large .

To keep the solution from diverging as well we need to truncate the coefficients at some by setting the coefficient to zero when

This value of for the truncations identifies a quantum state according to the integer which truncates the solution and gives us our energy eigenvalues

or since \lambda is just a dummy variable

###### Coulomb Eigenfunctions and Eigenvalues
 Spec Not. 1 0 1S 2 1 2S 2 1 2P 3 0 3S 3 1 3P 3 2 3D

The SHO and Coulomb schrodinger equations have Laguerre polynomial solutions for the radial part with the SHO solution polynomials of and the Coulomb solution polynomials linear in . The number of degenerate quantum states differs though, the SHO has 10 degenerate states while the Coulomb potential has 9 states.

## Angular Momentum

As you may have noticed in the quantum solution to the coulomb potential (Hydrogen Atom) problem above, the quantum number plays a big role in the identification of quantum states. In atomic physics the states S,P,D,F,... are labeled according to the value of . Perhaps the best part is that as long as there is no angular dependence to the potential, you can reused the spherical harmonics as the angular component to the wave function for your problem. Furthermore, the angular momentum is a constant of motion because the potential is without angular dependence (central potential), just like the classical case.

The mean angular momentum for a given quantum state is given as

since has its origin in

and the uncertainty principle has

we expect that the uncertainty principle will also impact such that

where characterizes the location of in the x-y plane.

or in other words, once we determine one component of (ie: ) we are unable to determine the remaining components ( and ).

As a result, the convention used is to define quantum states in terms of such that

This means that represents the projection of along the axis of quantization (z-axis).

Notice
: if then we would know and .

### Intrinsic angular Momentum (Spin)

The Stern Gerlach experiment showed us that electrons have an intrinsic angular momentum or spin which affects their trajectory through an inhomogeneous magnetic field. This prperty of a particle has no classical analog. Spin is treated in the same way as angular momentum, namely

Note
Nucleons like electrons are also spin objects.

### Total angular momentum

The total angular momentum of a quantum mechanical system is defined as

such that \vec{j} behaves quantum mechanically jusst like its constituents such that

where

In spectroscopic notation where is labeled by s,p,d,f,g,... the value of j is added as a subsript

for example
state with
with
with

In Atomic systems, the electrons in light element atoms interact strongly according to their angular momentum with their spin playing a small role (you can use separation of variables to have . In heavy atoms, the spin-orbit () interactions are as strong as the individual and interactions. In his case the total angluar momentum () of each constituent is coupled to some , you construct . When there is a very strong external magnetic field, and are even more decoupled.

Note
Nuclei (composed of many spin 1/2 nucleons) have a total angular momentum as well which is usually has the symbol

## Parity

Parity is a principle in physics which when conserved means that the results of an experiment don't change if you perform the experiment "in a mirror". Or in other words, if you alter the experiment such that

the system is unchanged.

If

Then the potential (V(r)) is believed to conserve parity.

and

Positive (Even) parity
Negative (Odd) parity
Note
Thus if is even then is Positive parity, if is odd then is negative parity.

### 3-D SHO

The Radial wave functions of the 3-D SHO oscillator problem can be either positive or negative parity.

Thus
Thus

The conclusion is that the total wave function is positive under parity.

### Parity Violation

In 1957, Chien-Shiung Wu announced her experimental result that beta emission from Co-60 had a preferred direction. In that experiment an external B-field was used to align the total angular momentum of the Co-60 source either towards or away from a scintillator used to detect particles. She reported seeing that only 30% of the particles came out along the direction of the B-field (Co-60 spin direction). In a mirror, the total angular momentum of the Co-60 source would point in the same direction as before ( while the momentum vector of the emitted particles would change sign and hence direction.

Consequence of the experimental observation
The Weak interaction does not conserve parity
Parity Violation for the Strong or E&M force has not been observed

## Transitions

### Stable Particles

For a stable particle its wave function will be in a stationary state that is static in time. This means that if I measure the average energy of this state I will see no fluctuation because the state is stationary.

In other words

The implication of this using the uncertainty principle is that

Or in other words quantum states with live forever.

### Particle Decay

If a quantum state has then it is possible for the quantum state to change (particle decay) within a finite time.

The first excited state of a nucleon (the \Delta particle) is an example of a quantum state with uncertain energy of 118 MeV

by the uncertainty principle we would expect that the particles mean lifetime would be about seconds.

The energy uncertainty is often referred to as the width of the resonance. Below is a plot representing the missing mass (W) of a particle created during the scattering of an electron from a proton. At least two peaks are clearly visible. The highest and most left peak has a mass of about 1 GeV representing elastic scattering and the peak following that as you move to the right represents the first excited state of the proton.

The gaussian shape of the "bump" shows how this state does not have a well defined energy but rather can be created over a range of energies. The "Width" () of an exclusive distribution determined a fit to a Breit Wigner function like

where

Center of Mass energy
Mass of the resonance
resonance's Width