Quantum Mechanics Project I, Spring 2012

FY521 ProjectBy

Daniel Bøgh Drasbæk (090190)&

Rasmus Rexbye Hansen (160889)

February 27, 2012

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Contents

1 Introduction 3

2 The Schrodinger equation 42.1 The Stationary Schrodinger Equation . . . . . . . . . . . . . . . 42.2 The dimensionless Schrodinger Equation . . . . . . . . . . . . . . 72.3 Restrictions and geometric behavior of wave functions . . . . . . 92.4 Specific examples . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.5 The WKB approximation . . . . . . . . . . . . . . . . . . . . . . 15

3 Numerical investigations by the shooting method 173.1 Procedure of the shooting method . . . . . . . . . . . . . . . . . 17

4 Properties of stationary bound state wave functions and thevariational method 194.1 Properties of the Hamilton Operator . . . . . . . . . . . . . . . . 194.2 The variation method . . . . . . . . . . . . . . . . . . . . . . . . 214.3 Analytic example of the variation method . . . . . . . . . . . . . 21

5 Numeric results from the variation principle 24

6 Summary 24

7 Appendix 25.1 dimensionless SE . . . . . . . . . . . . . . . . . . . . . . . . . . . 25.2 The shooting method . . . . . . . . . . . . . . . . . . . . . . . . . 25.3 Test functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

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1 Introduction

In this project we try analyze the different properties of the Schrodinger equa-tion. In doing so, we try to obtain a good idea of some of the basic propertiesof quantum mechanics. We start out by seeing how quantum mechanical sys-tems differ from classical mechanics, and work our way through the essentialsof Schrodingers equation of motion. This is illustrated by a few examples. Fur-thermore, we use different procedures to work our way around the numericdifficulties which arise, when attempting to investigate energies of a quantummechanical system.

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2 The Schrodinger equation

2.1 The Stationary Schrodinger Equation

Let us firstly consider the reduced Planck’s constant; ~. It is a natural constantgoverning the relationship between the energy of a photon and it’s frequency.The Planck relation is written as:

E = hν

E being the energy, h = 2π~ and ν the frequency. The units for ~ is [Js],which corresponds to the units of angular momentum or the action integralover the Lagrangian. From Einstein’s Special Theory of Relativity, we knowthe relativistic energy of an elastic collision, for particles with zero mass, can bewritten as below, and from the classic relation between wavelength, frequencyand velocity, it is possible to obtain the De Broglie relation:

E =√

(pc)2 +m2c4 = pc (1)

λν = c

⇒ E = hν = pc

⇒ hν = λνp

⇒ λ =h

p

This relationship shows how the particle-wave duality of elementary particles isrelated by the linear momentum(p) and the wavelength (λ). One can see, thatif the momentum (p = mv) is high, the wavelength becomes arbitrary small,which is in consistence with the classical approach.

Schrodingers wave equation consists of two parts; a time dependent part anda time independent part. Here we break up the equation of motion into twoparts, each governing their respective variable.

i~∂Ψ(r, t)

∂t= − ~2

2m

∂2Ψ(r, t)

∂r2+ V (r)Ψ(r, t)

⇒ i~∂

∂tT (t)Ψ(r) = − ~2

2m

∂2

∂r2T (t)Ψ(r) + V (r)T (t)Ψ(r)

⇒ i~∂T (t)

∂t

1

T (t)= − ~2

2m

∂2Ψ(r)

∂r2

1

Ψ(r)+ V (r)

Here we have separated the variables. When varying the different componentsone sees that in order for this to still hold, each side must be constant (which

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we will denote E), so:

i~∂T (t)

∂t

1

T (t)= − ~2

2m

∂2Ψ(r)

∂r2

1

Ψ(r)+ V (r) = E

⇒ E = i~∂T (t)

∂t

1

T (t)

⇒ E = − ~2

2m

∂2Ψ(r)

∂r2

1

Ψ(r)+ V (r)

⇒ ET (t) = i~∂T (t)

∂t

⇒ EΨ(r) = − ~2

2m

∂2Ψ(r)

∂r2+ V (r)Ψ(r)

Thus, we can obtain two ordinary differential equations, which both can besolved the trivial way. First we will consider the solution for the stationarySchrodiner Equation (SE) with zero potential, then the solution of the timedependent SE. The stationary SE will have no time dependence, since one isonly looking at how the wave packet behaves. The general solution will be:

EΨ = − ~2

2m

∂2Ψ(r)

∂r2

⇒ ∂2Ψ(r)

∂r2= −2mE

~2Ψ(r) ≡ −κ2Ψ(r)

⇒ Ψ(r) = Aeiκr +Be−iκr

Setting B = 0 is equivalent to a wave moving in the positive direction of r.One must remember that the exponential solution really consists of cosine andsine terms, specified by the restrictions on the specific wave equation. Thus, ageneral solution to the stationary SE could be;

Ψ(r) = Aeiκr

Where the amplitude (A) is determined by normalization. Normalization en-sures that the inner product of ψ amounts to 1. Now, considering the solutionsto the time dependent part of the SE:

ET (t) = i~∂T (t)

∂t(2)

⇒ ∂T (t)

∂t=E

i~T (t)

⇒ T (t) = e−iEt~

We can combine the two solutions obtained for each part of the SE, to oneunique solution;

Ψ(r, t) = Ψ(r)T (t) = Aeiκre−iEt~ = Aei(κr−

Et~ )

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Now, let us look at the probability density in order to prove time independence,but first let us ensure that our wave equation is correct. We start by ensuringnormality; ∫

Ψ(r)∗Ψ(r)dr =

∫Ae−iκrAeiκrdr

= r|A|2 = 1

⇒ |A| = 1√r

⇒ Ψ(r, t) =1√rei(κr−

Et~ )

Now that we have normalized the wave function, we want to prove the timedependency of the function, from the probability density:

|Ψ(r, t)|2 = |Ψ(r)T (t)|2 = Ψ(r)∗T (t)∗Ψ(r)T (t)

= Ψ(r)∗[iE

~

]Ψ(r)

[− iE

~

]= Ψ(r)∗Ψ(r) = |Ψ(r)|2

This means that the probabilities of states are all time independent.When comparing two wave equations of different states, one finds that they willbe mutually orthogonal. We have left out this calculation here, but one wouldfind that the amplitude of one state has to be zero, which means that no wavefunction can exist in the same state. The property of orthonormality may bewritten in terms of the Dirac delta function:

〈ψm|ψn〉 = δmn =

{0 if m 6= n1 if m = n

This completeness of the possible states, and a clear definition of state, are dis-cussed in detail later on, here it is only noted as a property of the wave function.So, since the states of the wave function are orthonormal, the only thing gov-erning the probability of a particle being in the different states, is the expansioncoefficients. These coefficients ensure both linearity and probability. The ex-pansion coefficients can be thought of as a ’probability amplitude of a state’.Together with its complex conjugate it gives the probability of a particle beingat a state, as shown above.Another approach to determine linear independence is by the Wronskian Matrix.We know that there exists two types of states. The bound state (E < V (−∞),E < V (+∞)) restricting the wave equation to a finite spatial region, such asthe stationary SE, and the scatter states (E > V (−∞) and/or E > V (+∞)),which are not restricted.For the scatter state the are two solutions, one for r < 0 and one for r > 0. Forthe bound state there is only one solution, since the function at ±∞ are equal.

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We write the Wronskian for the scatter state as:

W (r) =

∣∣∣∣∣∣ψ1 ψ2∂ψ1

∂r∂ψ2

∂r

∣∣∣∣∣∣ = ψ1∂ψ2

∂r− ψ2

∂ψ1

∂r6= 0

⇒ dW

dr=

(∂ψ1

∂r

∂ψ2

∂r+ ψ1

∂2ψ2

∂r2

)−(∂ψ2

∂r

∂ψ1

∂r+ ψ2

∂2ψ1

∂r2

)= ψ1

∂2ψ2

∂r2− ψ2

∂2ψ1

∂r2

= ψ1

(−2mE

~2

)ψ2 − ψ2

(−2mE

~2

)ψ1 = 0

The same can be done for a bound state, and again we find dWdr = 0. Therefore

we can proclaim that it is linearly independent.

2.2 The dimensionless Schrodinger Equation

For us to make numerical calculations we need to have a dimensionless SE.This is because a program like MatLab cannot calculate with the necessaryunit/dimensions. The standard time independent SE is given by:

− ~2

2m

d2Ψ(r)

dr2+ V (r)Ψ(r) = EΨ(r)

This form of the SE is convenient to do analytically calculations. For us to findan dimensionless SE, we have to rewrite the variables to dimensionless variables,which is done by defining the following:

x =r

L, v(x) =

V (xL)

E∗, ε =

E

E∗, ψ(x) = Ψ(xL)

√L

We now substitute this into the SE:

− ~2

2m

d2Ψ(xL)

dxL2+ V (xL)Ψ(xL) = EΨ(xL)

And since we know that if you differentiate with respect to a variable multipliedby a constant, is the same as differentiating with respect to the variable and

then afterwards multipling with the constant. Thus we can rewrite d2Ψ(xL)d(xL)2 =

1L ·

1Ld2Ψ(xL)dx2 we therefore now have:

− ~2

2mL2

d2Ψ(xL)

dx2+ V (xL)Ψ(xL) = EΨ(xL)

⇒ − ~2

2mL2

d2ψ(x)/√L

dx2+ v(x) · E∗ψ(x)/

√L = Eψ(x)/

√L

⇒ − 1

E∗~2

2mL2

d2ψ(x)

dx2+ v(x)ψ(x) =

E

E∗ψ(x) = εψ(x)

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We now want the term 1E∗

~2

2mL2 to be equal to 1. If it does, we would have thedimensionless SE we have been trying to find. For this to happen we can easily

see that E∗ = ~2

2mL2 . Thus we now have the relation between E∗ and L:

E∗ =~2

2mL2(-4)

And example of how the dimensionless SE could be used, could be a finite squarewell potential, where the wells width is given by 2a. Here we have r = 2a andwe can not determine what L should be equal to. The easiest way is to setL = a, and we therefore have x = r

L = 2aa = 2. If we write the dimensionless

SE we know everything except the potential function. But from what we knowwe can determine its value at certain points:

v(x) =

v0 if −∞ < x < −10 if − 1 ≤ x ≤ 1v0 if 1 < x <∞

A sketch for the finite square well can be seen in the appendix:We could have done the exact same sketch for the infinite square well potential,the only difference would be that v0 would be substituted by ±∞.

We know from theory that a harmonic potential can be described by V (r) =12mω

2r2 If we would like to calculate the harmonic potential, we need to findv(x) in the dimensionless SE. We can do this in the following way:

V (r) =1

2mω2r2 ⇒ V (xL) =

1

2mω2L2x2

We know from what we postulated earlier, that we can describe v(x) = V (r)/E∗.We will further try to use ~ω as the unit for E∗, since we know that it has thesame dimension as energy. We therefore have:

v(x) =V (r)

E∗=mω2L2x2

2~ω=x2L2ωm

2~We now substitute L2 from our earlier definition of E∗ and afterwards (again)put E∗ = ~ω:

v(x) =~2

2mE∗x2ωm

2~=

~2x2ωm

2m~2ω=

1

4x2

This means that instead of calculating a new potential in the dimensionless SEevery time, we can just use this value, as long as we know that the potential isharmonic. The basic unit here is [E∗] = ~ωAnother example on where we can use the same value for the potential everytime, is for the hydrogen atom. If we use the Bohr radius for the length unit.The non-dimensionless SE would have the Coulomb potential, known from elec-

tromagnetism: Vc(r) = − e2

4πε01r . This means that we now have:

L = a =4πε0e2

~2

m, Vc(xL) = − e2

4πε0

1

xL

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We can now find the dimensionless potential for the hydrogenatom, using thedefinition above as a defintion of E∗. We thereby get that v(x) = − 2

x . Here we

use the basic unit for the length as [L] = 4πε0e2

~2

mBoth E∗ and L are independent, but not of each other. This means that youcan chose E∗ > 0 or L > 0 but not both since they are related by;

E∗ =~2

2mL2

You can go from the dimensionless SE back to the normal stationary SE byusing the following:

x =r

L, v(x) =

V (r)

E∗, ε =

E

E∗, ψ(x) = Ψ(r) ·

√L

By substituting this into the dimensionless SE you will come back to the ordi-nary time independent SE.

2.3 Restrictions and geometric behavior of wave functions

In this section, we would like to examine the geometrical behavior of the wavefunctions. We choose to use the dimensionless SE, which makes it easier to dothe analasys. We will first examine whether ψ′ is continuous or discontinuousin a point where v(x) is respectively continuous, and has a finite discontinuity.This means that we should integrate the SE from x− ε to x+ ε:

−d2ψ(x)

dx2+v(x)ψ(x) = εψ(x)⇒ −

∫ x+ε

x−ε

d2ψ(x)

dx2+v(x)ψ(x)dx = ε

∫ x+ε

x−εψ(x)dx

⇒ dψ(x+ ε)

dx− dψ(x− ε)

dx=

∫ x+ε

x−ε(v(x)− ε)ψ(x)dx

We know that this is continuous if limε→0

on the lefthand side is equal to zero,

which means that the two points are the same. It would not be continuous ifthis does not hold, since the points will not meet.We now want to examine three forms of potential:

• v(x) is continuos

• v(x) is discontinuos

• v(x) described by v0δ(x)

In the first case when ε→ 0 it is easy to see that the areal underneath the graphgoes toward 0. It is therefore trivial that ψ′ is continuous. In the second case,we can see that for ε→ 0 the areal does not tend toward zero.If we split up the integral to the following;∫ x

x−ε(v(x)− ε)ψ(x)dx+

∫ x+ε

x

(v(x)− ε)ψ(x)dx

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It becomes clear, that if we let ε goes to 0, we see that both the left and theright expression equals 0 and therefore ψ′ is continuous.In the third case we have the potential described by Dirac’s delta function. Weknow the following about the Dirac delta function:

δ(x) =

{0 if x 6= 0

∞ if x = 0

Further we know that∫∞−∞ δ(x)dx = 1 and that

f(x)δ(x− a) = f(a)δ(x− a) and

∫ ε

−εf(x)δ(x)dx = f(0)

We know that x takes on a value, at the point where there will be discontinuityand therefore setting x = 0, the integration of the SE becomes:∫ ε

−ε(v0δ(x)− ε)ψ(x)dx =

∫ ε

−εv0δ(x)ψ(x)dx+

∫ ε

−ε−εψ(x)dx

= v0

∫ ε

−εδ(x)ψ(x)dx− ε

∫ ε

−εψ(x)dx = v0ψ(0)− ε

∫ ε

−εψ(x)dx = v0ψ(0)

The integral in the second last term is 0, due to the same argumentation as inthe first two cases. We can therefore proclaim that ψ′, is discontinuous as longas ψ(0) 6= 0.We now want to examine ψ for a potential being symmetric around x = 0, Thismeans that v(x) = v(−x), so the SE(x) must equal SE(-x):

d2ψ(−x)

dx2= (v(−x)− ε)ψ(−x) = (v(x)− ε)ψ(−x)

We can therefore deduce that ψ(x) = ψ(−x)α. We now use that we know that:∫ ∞−∞

σdx = 1⇔∫ ∞−∞| ψ(x) |2 d(−x) = 1⇔

∫ ∞−∞| ψ(−x)α |2 d(−x) = 1

⇔∫ −∞∞

| ψ(x)α |2 d(−x) = 1⇔∫ ∞−∞| ψ(x)α |2 dx = 1

⇔ α2

∫ ∞−∞| ψ(x) |2 dx = 1⇔ α2

∫ ∞−∞

σdx = 1

So, α must by all means equal ±1, if this has to hold. Which leads to ψ(x) =±ψ(−x); thus it either an even or odd function.

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Rewriting the stationary SE in the form below, makes the geometric approxi-mations easier too study;

d2ψ(x)

dx2= (v(x)− ε)ψ(x)

The second order derivative can be thought of as an infinitesimal change withinthe slope of the wave function (How ψ′(x) curves, that is). This is of coursedependent of the states energy relation between v and ε. Below is a sketch ofhow the function could look, for various values of v compared with ε.

There can be no stationary state with energy less than the minimal potentialenergy. If this is the case, the energy state will tend to ∞. This is illustrated inthe figure above, where the potential is set too zero and ε to be less than zero;this results in a positive value for the wave function no matter the values of x,since ε < 0 for a bound state.It should be noted here, that v > ε and v < ε on the figure, are only a pictureof what happens, when the position dependent potential varies. One has toremember that the potential is a function of x and the energy of the specificstate is a constant. Should ε > v = 0, it will correspond to a particle movingfreely without influence.A useful approximation, for understanding how the energy interaction works is,that the more potential a function has, the harder it is to move it away from itsequilibrium point, or better yet; if the state of the wave equation has a ’high’potential, the harder it is to excite it from its ground state.Another restriction on the wave function comes from Heisenbergs uncertaintyprinciple. It states that the more you know about a particles position, the lessyou know about its momentum. There are two ways this can be illustrated:

σxσp ≥~2

and σEσt ≥~2

The standard deviation of position and momentum are intertwined by this for-mula, so if one knows the almost exact position, the standard deviation ofmomentum will become proportionally large. A clever way around this uncer-tainty is to, for example, trap the particle in a box of some specific size, then

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assuming the box contains the wave, then it will be easier to avoid violating thisprinciple.

2.4 Specific examples

One way to illustrate the energy interaction, is by ’trapping’ the particle in apotential well, by setting restrictions on the potential. This is done by statingthat the potential is infinity, except in a small region where it will be zero. Butsince this is a rather deep well, we would like to obtain a finite well and see howthe energies of the two differ. so we have;

v0(x) =

{0 if 0 < x < av0 if elsewhere

v∞(x) =

{0 if 0 < x < a∞ if elsewhere

Notice, that the wave function on the boundaries is zero. If this was not true, wewould not have trapped the entire wave function and the outcome possibilitiesmight be wrong. If one looks at the equation for a stationary state, it is easyto see what happens if the potential is infinite of finite. The oscillations of thisstate is depend on the forms of energy at a specific point. Should the potentialbe infinite, we have v > ε, and so on. The graph above shows how part of thewave functions behaves for the different values of ε and v.For instance, one could dig a hole in the bottom of the quantum well, whichwill decrease the potential of the wave function, and again we can consult thegraph for how the function would act, when comparing the potential energy ata specific point, with the energy’s state. If we dug a hole in the quantum well,it would lead to an increase in the ψ′, where the hole was dug. This is becauseif v(x)→ 0 we have:

d2ψ

dx2= −εψ

But remembering that ε < 0 for this bound state, we have d2ψdx2 > 0.

If we for instance, have an infinite potential well with walls positioned at; x =0 and x = 2, we can obtain the wave function by some of the restrictionsmentioned above. We know that v(x) = 0 in this interval, and that the wavefunction must be zero at the end points.

d2ψ

dx2= −εψ

√ε ≡ κ

⇒ ψ(x) = Asin(κx) +Bcos(κx)

⇒ ψ(0) = ψ(2) = 0→ B = 0

⇒ ψ(x) = Asin(√εx)

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We know that in the interval, L = {0 < x < 2}, and ψ will be a π-periodicfunction, so we can break it into positive integers(n) in this interval:

ψ(x) = Asin(κL) = Asin(nπ)

⇒ κL = nπ

⇒ ε =n2π2

L2

L=2=

n2π2

4

⇒ ψ(x) = Asin(nπLx)

Since the potential is zero here, the only energy can come from the momentum,with the relation ε = p2. So for the product λp, we kan obtain the dimensionalunits;

λp = λ√ε = λ

√2mE = 2π

λ =2π

κand E =

~2κ2

2m

⇒ pλ =2π

κ

√2m

~2κ2

2m= ~2π = h

Which is the De Broglie relation, so the units of the product will be the sameas Planck’s constant; [Js].

Now, assume we have a potential, that is zero everywhere, except in a sin-gle point. This can be thought of as Dirac-delta function, with some initialpotential (v(x) = −v0δ(x)). We know that for a bound state, we have ε < 0.Substituting this into the stationary wave function above, one obtains:

d2ψ(x)

dx2= − (v0δ(x) + ε)ψ(x)

l2 ≡ (v0δ(x) + ε)

ψ(x) =

Ael(x)x for x < 0 B = 0Be−l(x)x for x > 0 A = 0

A for x = 0 A = B

Choosing the potential to be at x = 0, which we have done, is to be ableto eliminate every where else, since ψ(x 6= 0) → 0. Now we need to ensureorthonormal amplitudes of the bound state. A lot of terms can be neglected,since ψ(x) is discontinuous at x = 0. We assume that ψ(x) is symmetric aroundzero, so below zero the wave function increases and above it decreases. This givesus an opportunity to simplify the problem. Knowing that ψ is discontinuous atx = 0, we include a small real number (β) and let it tend to zero to be able to

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evaluate the integral.

d2ψ

dx2= [v(x)− ε]ψ

⇒∫ x+β

x−β

d2ψ

dx2dx =

∫ x+β

x−β−[v0δ(x) + ε]ψdx

⇒ dψ

dx(x+ β)− dψ

dx(x− β) =

∫ x+β

x−β−v0δ(x)ψ −

∫ x+β

x−β−εψ

Setting x = 0

⇒ dψ(0)

dx+β −

dψ(0)

dx−β = lim

β→ 0

[−∫ +β

−βv0ψ(0) +

∫ +β

−βεψ(0)

]⇒ −lψ(0)− lψ(0) = −v0ψ(0)

⇒ 2l = v0

⇒ ε = l2 =(v0

2

)2

The first order differentials, on the lefthandside, corresponds to the differentialof the two solutions if x 6= 0. It is crucial to keep in mind that the potentialonly exists at x = 0. Elsewhere the inclination and decent only depend on ε.

Now, consider the potential square well, with width 2a, the potential may bewritten as;

V (x) =

{0 for |x| > a−v0 for a ≤ x ≤ a

We derived the solutions to the wave equation, to be the following:

ψeven(x) =

Fe−κx for x > aDcos(lx) for a > x > 0ψ(−x) for x < 0

ψodd(x) =

Fe−κx for x > aCsin(lx) for a > x > 0−ψ(x) for x < 0

Here F , D and C are arbitrary constants, with κ ≡√−2mE

~2 and l ≡√

2m(E+V0)~2 .

From conditions of continuity and some algebra, it is possible to find the con-nection of allowed energies, between κ and l to be:

κ = ltan(la)

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Now defining k ≡ la and k0 ≡ a~√

2mV0, we can obtain an easier expression forhow the energy acts if ψ is even or odd, which is represented below;

ψeven → tan(k) =

√k2

0

k2− 1

ψodd → tan(k) =1√

1− k20k2

The two points which have been marked by a red circle, are the two points inwhich the equation for the even bound state energies are fulfilled. And thereforewe have boundstates in these points. The graph of the odd wave function havebeen excluded in this report.

2.5 The WKB approximation

This type of approximation is useful when attempting to calculate solutions topartial derivatives dependent on a small parameter. When used in quantummechanics the WKB technique can be applied to the stationary SE to obtainapproximate solutions to it. The general formula for this approximation is;

ψ(x) ≈ C√p(x)

e±i~∫p(x)dx

p(x) ≡√

2m[E − V (x)]

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Where V (x) is approximated to be constant, or slowly varying in comparisonto the wavelength. The constant C includes all constants that may arise fromthe calculations. The momentum, is assumed to be real, as in the classic region,so E > V (x). However ψ might not be real. The approximation comes from

assuming that the amplitude (A) is slowly varying, so: A′′

A << p2

~2 .Underneath we derive the WKB for the harmonic potential analytically. We

know that the potential is given by v(x) = 14x

2 for the dimensionless SE. Wefurther know from theory that:

p · L = nπ~ = nhπ

2⇒∫ L

0

pdx = nhπ

2

So we may now make a semi classical approximation:

E =p2

2m+ V (r)⇒ p =

√2m(e− V (r))

We now put this into our integral and evaluate it to obtain:∫ L

0

pdx =

∫ L

0

√2m(E − V (r))dx = nh

π

2

This is only valid for the non-dimensionless SE, with a potential equivalent tothat of a square well with an infinite potential. For a dimensionless SE it statesthe following:

p(x) =√ε− v(x)

In classical physic we can find the turning points x+ and x− so in our case withthe harmonic potential we have that:

ε =1

4x2 ⇒ x± = ±2

√ε

Thus we have the integral:∫ x+

x−

√ε− x2

4dx =

∫ √x2

+

4−x2−4

=1

2

∫ x+

−x+

√x2

+ − x2

In order to do this calculation, we must substitute,

x = sinφx+ ⇒ dx = x+cosφdφ

and plug it back in, to obtain:∫ π2

0

x2+

√1− sin2φcosφdφ =

∫ π2

0

x2+cos

2φdφ =x2

+

2

∫ π2

0

dφ

x2+

1

2

π

2=x2

+π

4

16

And since we know that ε = x2

4 , we can rewrite the expression as:

x2+

4π = επ

So, by using the integral we determined in the start, we can find the same valuethat the WKB program does. That is;

πε =

(n− 1

2

)⇒ ε = n− 1

2

3 Numerical investigations by the shooting method

In this section we describe how to use the dimensionless SE to make numericalcalculations of the stationary SE. A Maple file was created for us by J. BoidenPedersen, in which the basic functions was already described and a library hadbeen made. So the general setup for our calculations was pre-made. We shalldescribed the numerical calculations for the following four potentials: ”FiniteSquare Well”, ”Harmonic Oscillator”, ”Triangle Potential” and ”Coulomb Po-tential”. Further more, we shall use the WKB approximation on some of thesepotentials.

3.1 Procedure of the shooting method

First of we define the potential as being V = 50 ∗ (1 − Box(x,−1, 1)). Where50 declares the depth of our well, while the Box command describe how wideour well is. We start out by doing the shooting method to find the interval inwhich the ground state lies. The shooting method needs to know the potential,the energy in which it should investigate, the maximum value of x, how manycalculations it must do and if it should look for an odd or even wave function.If we set the energy to high we see that ψ(x)→ −∞ and if we set it to low thenψ(x)→∞ and since this is not a possible solution we can discard these valuesfor our bound-state energy, but we now know in which interval to search.Then using the AutoShooting method in maple; it does the exact same thing asthe shooting method, except it searches for the solution in a specified interval.Its output is a graph with the mean energy of the wave function. If you write’true’ in the end of the input, one will also obtain the probability density of thefunction. We now repeat these steps until we have the first four excited statesand the ground state. We know that the ground state is an even function, thuswe give AutoShoot the command +1, the first exited state is an odd and there-fore we give the AutoShoot the command -1 etc. We can do the exact sameprocedure for the harmonic oscillator, only we now have the potential function:

V = x2

4 .The triangle potential is a little different; here we want to examine the asym-metric infinite triangle, this is a potential which is infinity for x < 0 and x

10 for

17

x > 0. The tricky part of this is, that we cannot calculate exactly with infinities;instead we can use some symmetry arguments to describe how the energy couldbehave, from the symmetric infinite triangle.We know that at x = 0 the wave function should equal 0, for x > 0 it is exactlythe same as the symmetric infinity triangle. Therefore, if we use infinite triangleenergies as a reference, we know that we can only use those whose solution is0 at x = 0, since the wave function should be continuous. It is therefore onlythe first and third excited state of the symmetric infinite triangle, which aresolutions to the asymmetric infinite triangle.We can create a finite asymmetric triangle potential, in which the potential forx < 0 is equal to 1 and x

10 for x > 0, but we need to put a limit on this lastterm, so it will not be able to exceed 1.The fourth potential is the Coulomb potential. This potential also needs to beanalyzed in the right way, since for x=0 the potential is minus infinity. There-fore we need to tell Maple that it should not calculate any further than to minus100, so the potential can not be less than -100. As well as in the asymmetrictriangle potential, we should only use the first and third excited state, as actualvalues for the Coulomb potential, for our ground state and our first excitedstate, respectively. This is because the Coulomb potential will be −∞ at x = 0.Furthermore, it is worth noticing that the energies will all be negative. Thesame procedure as the other potentials is then executed.Here we have shown the different energies found and add’ed some comments onthem. Further we have done a comparison between the results found from theshooting method with the ones from the WKB:

When we did the calculations for the deviation of the harmonic potential, wenoticed that the deviation rose with the rising quantum number, but still the

18

deviation only increased a little. Therefore we can conclude that there is acoherence between the WKB values and the shooting method values. The reasonfor this can be seen in the WKB section where we make the calculations for thisWKB.For the symmetric infinite triangle potential, the deviation is again very low,since it is approximated zero for three of the four values. So, again, the WKBis a good approximation. It is the same with the asymmetric infinite trianglepotential. But the WKB is not a good approximation for the Coulomb potentialsince it has a large deviation. But still we saw that the approximation becamebetter with the rising quantum number. The Coulomb potential can be seenas an analogy to the hydrogen atom; the the higher the quantum number, themore excited the atom is and therefore the less it is bound to the nucleus.We cannot do the same comparison with the WKB for the final square well, sincewe havent done any exact calculations for the infinite square well potential, onlythe finite square well. But again we can see that the deviation would have risenwith the rising quantum number, if we had done the calculations for the exactenergy. The values for the energy can be seen underneath.

The different graphs can be located in the appendix, with comments.

4 Properties of stationary bound state wave func-tions and the variational method

4.1 Properties of the Hamilton Operator

The bound SE consists of smaller, mutually orthonormal, waves states and theentire solution is the superposition of these states. This assumption of com-pleteness of all states, yield different properties, which we will explore in thissection.A state can be thought of as one possible outcome of a system. This can berepresented as a point in a phase space; here the phase space consists of allpossible values of momentum and position. These states are governed by a setof dynamic variables, needed to describe the system fully. The representationfor a state can be written as a ket-vector in Hilbert space, containing all infor-mation about the specific state. The points in the phase space are determinedby the observables of the system, usually represented as a self-adjoint operator,known as Hermitian. These operators are what we can observe (such as positionor momentum), represented by a matrix acting on each vector in the Hilbertspace. So, knowing all possible states of a system, and the states’ probability,

19

one has the necessary means to describing a quantum system.

Suppose we have an Hermitian operator acting on the wave function, then,corresponding to the operator, one would obtain different eigenvalues for a spe-cific system. The eigenvectors/eigenfunctions will then be the solutions to thedifferent states of the system. Consider for instance the Hamiltonian operatorworking on the bound SE;

ψ(x) =∑n

cnψn(x)

Hψn(x) = εnψn(x)

Here the eigenfunctions are ψn(x) with their corresponding eigenvalues εn. Thisis a crucial and fundamental property of the Hamiltonian; when applied, itsingles out the different eigen-energies of each state.Knowing that the set of eigenfunctions to the matrix H is complete, we can writethe solution in terms of expansion coefficients and stationary states. However, ifeach state is represented with an expansion coefficient, governing the probabilityof the specific state, will this also be so for the specific energy of the state? Letus apply the Hamiltonian to the stationary SE (done here in Dirac notation),to see how it works:

〈H〉 ≡ 〈ψ|H |ψ〉 = 〈∑n

cnψn|H |∑n

cnψn〉 =∑n

c∗ncn 〈ψn|H |ψn〉

=∑n

c∗ncnεn 〈ψn|ψn〉 =∑n

|cn|2εn = 〈E〉

There is a lot of information hidden in this calculation. Most importantly; theexpectation value of the Hamiltonian is equivalent to the mean expectation valueof the systems energy. 〈E〉 is really nothing more than the superposition of allindividual eigenstates energies (εn)and their probabilities (c∗ncn), for a systemdescribed entirely by the wave function |ψ(x)〉.Another important property of this calculation is, that the wave function canbe expressed as a quantum mechanical superposition of all states. For example,if one has a die and throws it, then (without looking at it) one can expressthe possible outcomes as a sum over the six outcomes (states) with a certainpossibility of being in each state (c1 = ... = c6 = 1

6 ).The expansion coefficients are ensured to relate discrete probability to the sys-tem. Just to be safe, let us look at what happens if we take the inner productof two different states;

1 = 〈ψ(x)|ψ(x)〉 = 〈∑k

ckψk(x)|∑n

cnψn(x)〉

=∑k

∑n

c∗kcn 〈ψk|ψn〉 =∑k

∑n

c∗kcnδk,n

k=n=∑n

c∗ncn =∑n

|cn|2

20

The squared expansion coefficients sum to 1, as supposed to in statistics. Westarted by looking at two different states (k, n), and saw that the inner productof two linearly independent system states is a delta function, and that this deltafunction may exclude one of the states.

4.2 The variation method

Knowing all this, we can formulate and upper bound for the ground state (ε0);∑n

|cn|2εn ≥∑n

|cn|2ε0

⇒ 〈ψ|H|ψ〉 ≥ ε0⇒ 〈H〉 ≥ ε0

This is known as the variation principle. It states, that if one has any normalizedwave function, it is possible to find a good approximation for the ground state,since it will be the lowest possible expectation value of H. All other values willcorrespond to the particle being in an excited state.Then choosing different test functions for the wave equation, one starts to ap-proximate the correct wave function. Then, seeing how they behave, on canpick the most likely equation and find the unknown parameter on which it isbased. Then by minimizing 〈H〉, one obtains the minimal expectation value.

4.3 Analytic example of the variation method

Here we will observe a particle in a one dimensional well with the potential:

V (x) =

{∞ for x < 0

αx for x > 0 α > 0

We have the following three test functions:

ψ1 = e−βx2

for x > 0

ψ2 = x · e−βx2

for x > 0

ψ3 = x2 · e−βx2

for x > 0

For all three three test functions, we have that; ψi = 0 for x < 0 ∀i.We can resonate, that we must have an asymmetric infinite triangle potential.Since the wave function must be continuous (ψ = 0 at x = 0), we can discardψ1 since it is not zero for x = 0. The best test-function must be ψ2, sinceit has the lowest energy. This is because the curve of ψ3 needs to be higherthan that of ψ2, and since the curve and energy is proportional, we know that〈E〉ψ2

< 〈E〉ψ3. ψ2 must therefor be the best test function, since you always hit

a higher energy with a test function than that of the wave function.We will now show how to find an expression for the mean kinetic energy andthe mean potential energy, as functions of β and α. We start by finding the

21

normalization factor, this is done in the following way:We define the wave function to be described as follows, where N is the normal-ization factor: ψ = N · e−βx2

. We then state that the integral of ψ multipliedby its complexed conjugated, from minus infinity to infinity, should equal one.But since we know that ψ = 0 for x < 0, so we only have to evalute the integralfrom zero to infinity:∫ ∞

0

ψ∗ · ψdx = 1⇒∫ ∞

0

ψ2dx = 1⇒ N ·∫ ∞

0

(x · e−βx

2)2

dx = 1⇒

N ·∫ ∞

0

x2 · e−2βx2

dx = 1⇒ N ·√π

(2 · 1)!

1!

(1

2√

2β

)2·1+1

= 1⇒

N√π · 2 ·

(1

2√

2β

)3

= 1⇒ N2√π

8

(1√2β

)3

= 1⇒ N =8β3/2

√2√

π

The integration we did here was done by using the Gauss integral for evenexponents.We are now ready to start the calculations for the kinetic energy, but insteadof multiplying this normalization factor from the beginning, we have chosen todo it in the end instead. Therefore we have that:

〈K〉N

=−~2

2m

∫ ∞0

ψ∗[d2

dx2

]ψdx =

−~2

2m

∫ ∞0

ψ ·(−6βxe−βx

2

+ 4x3β2e−βx2)dx

=−~2

2m

∫ ∞0

−6βx2e−2βx2

+ 4x4β2e−2βx2

dx

−~2

2m

(∫ ∞0

−6βx2e−2βx2

dx+

∫ ∞0

4x4β2e−2βx2

dx

)Gauss integral even

=−~2

2m

(−6β√π · 2

(1

2√

2β

)3

+ 4 · 12√πβ2

(1

2√

2β

)5)

−~2

2m

(−3

2

√πβ

β3/2

(1√2

)3

+3

2

√πβ2

β5/2

(1√2

)5)

−~2

2m

3

2

√π

β

(1√2

)3(1

2− 1

)=

3

8

~2

m

√π

β

(1√2

)3

We shall now only multiply this with our normalization factor to obtain themean value for the kinetic energy:

〈K〉 =〈K〉N·N =

3

8

~2

m

√π

β

(1√2

)3

· 8β3/2√

2√π

3

2

~2

m

√π√ββ3/2

√π

=3

2m~2β

22

And now the calculations of the mean value of the potential energy:

〈V 〉N

=

∫ ∞0

ψ∗ [αx]ψdx = α

∫ ∞0

x3e−2βx2

dx

Gauss integral odd= α

1!

2

(1√2β

)4

=α

2

1

4β2=

α

8β2

⇒ 〈V 〉 =〈V 〉N·N =

α

8β2∗ 8β3/2

√2√

π=α√

2√πβ

We can see here, that if we make β large the kinetic energy will rise, whilethe potential will fall. This makes sense since β describes how ’fast’ the wavefunction converges to zero. So, with large β follows a larger curve of the wavefunction and therefor also a higher kinetic energy, since these are proportional.This also means that the wave functions position is more well defined and there-fore the momentum must be proportionally undefined; this relates to Heisenberguncertainty principle. If β is small then our wave function will be more ’flat’ andtherefore the probability of finding the particle further out is therefore larger,hence the position will be undefined, which will yield a larger potential energy.We now want to determine the best estimate of the ground state, using ψ2 asa test function. This is done by using of the Hamilton operator, since we knowthat the energy of the ground state must be smaller or equal to minimum valueof the hamiltonian.

〈H〉 = 〈K〉+ 〈V 〉 =3

2

~2β

m+α√

2√βπ

Now to determine 〈H〉min; this is done by taking the derivative of the Hamil-tonian with respect to β and setting this equal to zero:

d〈H〉dβ

=3

2

~2

m− 1

2

α√

2π

(βπ)3/2= 0⇒ 3

2

~2

m=

1

2

α√

2π

(βπ)3/2⇒ β =

(α√

2m

3~2√π

)2/3

Now we only have to substitute this into our definition of the Hamiltonian:

〈H〉min =3

2

~2

m

(α√

2m

3~2√π

)2/3

+α√

2√π

(α√2m

3~2√π

)2/3−1/2

Which can be reduced to:

〈H〉min =(m

2+ 1)·(

6α2~2

πm

)1/3

23

5 Numeric results from the variation principle

Below we show a table over the different test-functions on the different poten-tials, further we compare them with the energy values found by the shootingmethod.

A graph of the best estimates can be seen in the appendix.We can see here that the trigonometric test function (cosine) is the best for thefinite square well, the gaussian test function is not bad at all. Below we havechosen to show the trigonometric test function together with the exact wavefunction.Furthermore, we can see that for the harmonic potential the gaussian test func-tion is the best. In fact, it is the exact solution, which has to do with the shapeof the gaussian test function. This can also be seen on the following graphs.At last we can see that all three of the test functions are good approximations,but the gaussian test function is still the best, which can be seen here:

24

6 Summary

In this project we learned the basic properties of quantum mechanics. Startingby analyzing the stationary SE and seeing how the propagation of it behaves,then how to bring it into a dimensionless equation, to make it easier to workwith. After analyzing the geometric properties of the SE, the restrictions onit, and working through some chosen examples, we saw how some useful ap-proximation can come in handy. One of these approximations is the WKBapproximation, which we tried a hands on approach to, by analyzing some nu-merical results. Lastly, we worked with the Hamiltonian operator, to get anidea of how operators work and how these observables could be used. This wasalso followed by some numerical calculations. All of this has created a solidfoundation on how energy interactions occur on a quantum scale.

25

7 Appendix

.1 dimensionless SE

Illustration of the potential well:

.2 The shooting method

In all the graphs, the blue line inidcates the potential, green is the probabilitydensity and the black is the wave function.

The 4th excited state of a finite square well:

Here we can see that this is the fourth excited state, since we have four pointsof interaction with the mean energy.

26

The third excited state of the harmonic potential:

Here we can see the third excited state of the harmonic potential. We can alsosee that there is a larger possibility of finding the particle in the end-points.

A Gaussian wave packet in an infinite symmetric triangle potential:

On this graph, we see the ground state for the symmetric infinite triangle po-tential. We can here see that the wave actually moves outside the potential, thisis not possible in classical mechanics, but in quantum mechanics it can. This isone of the hard things to visualize.

27

The potential for a finite assymetric triangle:

Here we see the first excited state in the asymmetric finite well. We cannot seethat at x=10 the potential goes no further then to V=1

The ground state of the asymmetric infinite triangle:

Here we have the first excited state of the symmetric infinite triangle but aswe described in the shooting method theory it is also the ground state of theasymmetric infinite well if we only look at x > 0

28

The first excited state of the Coulomb potential:

Here we see the second excited state of the coulomb potential. As well as in theexample with the asymmetric infinite well we here only look at x > 0.

29

.3 Test functions

Square well potential with cosinus as testfunction

We have chosen only to show the cosine test function because it was the bestfor the square well potential. Here the black line is the exact, while the red isour test function.

The harmonic potential, with a Gaussian testfunction :

Here we have the Gaussian test function on the harmonic potential, we canonly see one line since the Gaussian test function is the exact solution for theharmonic potential.

30

The symmetric triangle potential with a Gaussian test-function:

Here we have the gaussian test function in the symmetric triangle potential, wesee that it fits really well, but we are still able to distinguish the test functionfrom the exact function.

31