2nd order derivation Schrodinger eqn.docx

7/30/2019 2nd order derivation Schrodinger eqn.docx

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Complex Roots

In this section we will be looking at solutions to the differential equation

in which roots of the characteristic equation,

are complex roots in the form

Now, recall that we arrived at the characteristic equation by assuming that all solutions to the

differential equation will be of the form

Plugging our two roots into the general form of the solution gives the following solutions to the

differential equation.

Now, these two functions are nice enough (theres those words again well get around to

defining themeventually) to form the general solution. We do have a problem however. Since

we started with only real numbers in our differential equation we would like our solution to onlyinvolve real numbers. The two solutions above are complex and so we would like to get our

hands on a couple of solutions (nice enough of course) that are real.

To do this well need Eulers Formula.
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A nice variant of Eulers Formula that well need is.

Now, split up our two solutions into exponentials that only have real exponents and exponentialsthat only have imaginary exponents. Then use Eulers formula, or its variant, to rewrite the

second exponential.

This doesnt eliminate the complex nature of the solutions, but it does put the two solutions into

a form that we can eliminate the complex parts.

Recall from thebasics sectionthat if two solutions are nice enough then any solution can bewritten as a combination of the two solutions. In other words,

will also be a solution.

Using this lets notice that if we add the two solutions together we will arrive at.

This is a real solution and just to eliminate the extraneous 2 lets divide everything by a 2. This

gives the first real solution that were after.
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Note that this is just equivalent to taking

Now, we can arrive at a second solution in a similar manner. This time lets subtract the two

original solutions to arrive at.

On the surface this doesnt appear to fix the problem as the solution is still complex. However,upon learning that the two constants, c1 and c2 can be complex numbers we can arrive at a real

solution by dividing this by 2i. This is equivalent to taking

Our second solution will then be

We now have two solutions (well leave it to you to check that they are in fact solutions) to thedifferential equation.


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It also turns out that these two solutions are nice enough to form a general solution.

So, if the roots of the characteristic equation happen to be the

general solution to the differential equation is.

Lets take a look at a couple of examples now.

Example 1Solve the following IVP.

Solution

The characteristic equation for this differential equation is.

The roots of this equation are . The general solution to the

differential equation is then.

Now, youll note that we didnt differentiate this right away as we did in the last section. The

reason for this is simple. While the differentiation is not terribly difficult, it can get a little

messy. So, first looking at the initial conditions we can see from the first one that if we just

applied it we would get the following.


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In other words, the first term will drop out in order to meet the first condition. This makes thesolution, along with its derivative

A much nicer derivative than if wed done the original solution. Now, apply the second initial

condition to the derivative to get.

The actual solution is then.



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Solution

The characteristic equation this time is.

The roots of this are . The general solution as well as its derivative is

Notice that this time we will need the derivative from the start as we wont be having one of the

terms drop out. Applying the initial conditions gives the following system.

Solving this system gives and . The actual solution to the IVP

is then.



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Solution

The characteristic equation this time is.

The roots of this are . The general solution as well as its

derivative is

Applying the initial conditions gives the following system.

Do not forget to plug the t = into the exponential! This is one of the more common mistakes


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that students make on these problems. Also, make sure that you evaluate the trig functions as

much as possible in these cases. It will only make your life simpler. Solving this system gives

The actual solution to the IVP is then.

Lets do one final example before moving on to the next topic.


Solution

The characteristic equation for this differential equation and its roots are.


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The general solution to this differential equation and its derivative is.

Plugging in the initial conditions gives the following system.

So, the constants drop right out with this system and the actual solution is.

Shrdinger's Equation


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The current fundamental physical model of the atom was created by the Austrian physicist Erwin

Schdinger while he was a young professor at the University of Zurich. Schrdinger's colleague

Victor Henri gave him a copy of de Broglie's doctoral thesis on the wave properies of anelectron. Schdinger was not impressed with the thesis and began looking in other directions. It

was several months later while reading a section of the thesis dealing with Bohr's quantization

rules that he recognized the connection between Bohr's stationary states of the hydrogen atomand deBroglie's wave properties of an electron. Schdinger realized that waves are macroscopicsystems that behave like atoms. A standing wave like and atom can only absorb or release energy

in quantitized amounts. This bold decision was the beginning of quantum mechanics. The results

of Schrdinger's doctoral thesis was a single general equation first published in 1926 (Ann.Physik, 79,361). The invention of quantum mechanics, which Schdinger shares with Heisenberg

and Dirac, has been one the most important advancements in science - to be compared with the

contributions of Galileo, Newton and Einstein.

Like the equations comprising Newton's laws of motion, Schrdinger's equation can not be

derived. His equation is a generalizations of the world as we observe it and is validated by how

well it describes experimental observations. What follows is not a derivation but merely aprocedure by which Schrdinger's equation can be constructed. Schdinger developed his

equation using analogies to the behavior of light. He reasoned that the classical equations used todescribe light waves could be used to describe matter waves if the equations were modified toinclude newly discovered quantum properties of photons. The energy of a photon of light is

related to its frequency by Planck's constant, E light = h and the momentum of a photon of light is

related to its wavelength by Planck's constant, p light = h/Rather than using a metaphor based onlight, as Schdinger did, I am going to construct his equation by the analogy to a standing wavesuch as the vibration of a guitar string.

There are many similarities between the motion of a guitar string and the motions of an electron

trapped in an atom. Both are described by wavefunctions that oscillate in time and space. The

waves are characterized by stationary points called nodes where the wavefunction goes throughzero. The guitar string is fixed at the bridge and neck of the guitar and hence must have a node at

these positions. Attachment of the guitar string at the bridge and neck place a restraint or

boundary condition on the guitar string's movement. The boundary conditions limits the motionof the string to certain special vibrations with fixed energies. The first three special vibrations orovertones of a guitar string are depicted above with nodes represented as black dots. These are

the principle overtones your hear when you pluck a string and the reason that the notes C and G

harmonize with each other. These special vibrations in a quantum mechanical system would becalled eigen states of the system. The sound coming from a single guitar string produces a line

spectrum much like the hydrogen atom. The composite tone that we hear is a superposition of the

eigen states of the system.


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We will begin our construction of Shrdinger's equation with the mathematical equation for a

standing wave:

The result of differentiating twice in respects to x is the second order differential equation for a

wave:

We can begin the transformation of this classical equation to a quantum mechanical wave

equation by using the deBroglie relation p=h/for momentum.

Momentum also plays a central role in classical equations of motion. A standing wave is aconservative system in which the potential energy does not depend on momentum. In such a

system the total energy (kinetic plus potential) is a constant of motion:

The total energy must also be a constant of our bound quantum mechanical system. Applyingthis classical relationship to our wave equation gives:

This is Schrdinger's time independent wave equation. The same equation that is presented at the

top of this page. Schrdinger's equation is a second-order differential equation whose solution is

the wavefunction for the system. The energy of the system will depend on how fast the

wavefunction bends (second derivative of the wavefunction) and the potential of the system (V).

It is instructive to write this equation as an operator equation.


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Classically .H, the Hamiltonian of the system is defined simply as the sum of the kinetic energy

T and potential energy V of the system. The same Hamiltonian applies to quantum mechanical

systems but any terms that are associated with kinetic energy and momentum must be replaced

by their equivalent quantum mechanical operator:

These equations can easily be generalized to three dimensions by using the dell operator:

This ends our construction of Schrdinger's time independent equation. His equation is a simple

operator eigen value equation. The Hamiltonian operating on the wavefunction gives thewavefunction back again times the constant energy of system. Such an equation is called a eigen

equation after the German word for "self". The wavefunction is called the eigen function and the

energy is called the eigen value. For a given Hamiltonian there are many possible eigenfunctions, but not all of these functions will have physical significance. It is a central postulate of

wave mechanics that all of the measurable information about a system is contained in its

wavefunction. To correspond to physical observations we must put some constraints on thewavefunctions for a system.

1. The wavefunction should be single valued2. The wavefunction should be continuous so we can take its derivatives3. The wavefunction should be finite so that we can take its integral4. The wavefunction for a bound electron should vanish at the boundaries of the system

These constraints like the attachments on a guitar limit the energies of the system to certain

quantitized values that we can count with a quantum number n.


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Schdinger guessed that the line spectrum of the hydrogen atom indicated that the equations of

motion must be wave equations with boundary conditions that fix the possible energy levels. The

quantum numbers that describe an atom are a natural consequence of Schrdinger's waveequation that describes the atom.

In summary we can write down the steps we need take in order to apply Schrdinger's equation:

1. Determine the appropriate potential for the system2. Write down the classical Hamiltonian for the system3. Form the quantum mechanical Hamiltonian by replacing the momentum and kinetic

energy in the classical Hamiltonian with their quantum operators

4. Establish the boundary conditions for the wavefunction5. Solve Schrdinger's eigen value equation to determine the eigen functions and eigen

values of the system.

Here is the wave function .

Let the electron cloud form a probability standing

wave since wave cannot be progressive

So

= A sin(kx + z) 1)

Now we have to eliminate A so

/ x = Akcos (kx + z ) .2)

Now again

2 /

2x = -Ak

2sin(kx +z ).3)

Now we get

2/

2x = -k

2 ..4)

Now k (angular wave number ) is k = 2 /

And by de-Broglies hypothesis

= h / mv 5)so

k = 2 (mvx) /h


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so

k2

= 4 2

(mvx)2

/ h2

6)

so

2

/

2

x = - 4

2

(mvx)

2

/ h

2

.7)

Same for y and z axis give us

2 /

2y = - 4

2(mvy)

2 / h

2..8)

2 /

2z = - 4

2(mvz)

2 / h

2.9)

Now summation of 7 , 8, 9 give us

2 /

2x +

2 /

2y +

2 /

2z = -4?

2m

2( vx

2+ vy

2+vz

2) /h

2.10)

The left part can be replace by Peirre Simon Laplaces operator

2 = -4

2m

2v

2 /h

211)

{ v2 = vx2

+ vy2

+vz2

}

Now mv2

= K = E-V (energy is conserved)

So

Mv2

= 2( E-V) ..12)

So we get

2 = -4

2m(2(E-V) ) /h

2.13)

So

2 + 8

2m(E-V) /h

2= 0 ..14)

Equation 14 represents the Erwin Schrodinger Wave equation

- bladeX ( le brave des braves)

2nd order derivation Schrodinger eqn.docx

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