In Class Exercises 7

In Class Exercises 7, Physics 113 B 1. Use the wavefunction of the form Ψ HxL =

A CosA

Πx a

E, if

0,

-a 2

< x <

a 2

otherwise

to obtain a bound on the ground state of the one-dimensional harmonic oscillator. Determine the best value of a and compare with the exact result. Let us first determine the normalization constant A. In[30]:=

SolveBA2 IntegrateBCosB

2

Πx

-a

F , :x, a

2 Out[30]=

::A ® -

a ,

2

>, Assumptions ® a > 0F Š 1, AF 2

2 >, :A ®

a

>> a

2 In[31]:=

A =

; a

To find the expectation value for the energy we find the the expectation values of the kinetic and potential energies and sum them. The Kinetic Energy Ñ2 =

T

<Ψ

Ψ> =

<Ψ

∆2

-

Ψ> 2 m ∆ x2

< T > Ñ2 In[35]:=

Texp =

Π

2

A 2m

2

Πx IntegrateBCosB

a

-a

F , :x, a

a ,

2

>, Assumptions ® a > 0F 2

Π2 Ñ2 Out[35]=

2 a2 m

This is our expectation value of kinetic energy. Notice how it is the same as the ground state energy of the infinite square well. The Potential Energy < V > = < Ψ | V | Ψ > = < Ψ |

1 2

m Ω2

Ψ >

< V > 1 In[36]:=

Vexp =

m Ω2 A2 IntegrateBx2 CosB

2

Πx

2

-a

F , :x, a

a ,

2

>, Assumptions ® a > 0F 2

a2 m I- 6 + Π2 M Ω2 Out[36]=

24 Π2

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2

In Class Exercises 7.nb

In[49]:=

Eexp = Texp + Vexp a2 m I- 6 + Π2 M Ω2

Π2 Ñ2 +

Out[49]=

24 Π2

2 a2 m

Here is our energy for our wavefunction in the one - dimensional harmonic osciallator. We now need to minimize this energy to find the ground state energy with this particular guessed wavefunction. In[51]:=

Solve@D@Eexp, aD Š 0, aD 2 ΠJ

Out[51]=

3 -6+Π2

14

N

Ñ

::a ®

2 314 Π

Ñ

>, :a ® m ä

>, 2 14

m I- 6 + Π M

Ω

2 314 Π

Ñ

ä

:a ® -

2 314 Π

Ω Ñ

>, :a ® 2 14

m I- 6 + Π M

>> 2 14

m I- 6 + Π M

Ω

Ω

We shall take the postive real solution. 2 ΠJ In[52]:=

Eexp . a ®

3 -6+Π 2

m 1

1

2

3

1

1

2

3

Out[52]=

In[54]:=

Out[54]=

14

N

Ñ  FullSimplify

Ω

I- 6 + Π2 M Ω Ñ

I- 6 + Π2 M Ω Ñ  N

0.567862 Ω Ñ

Here is the ground state energy of the cosine wavefunction in the one 1 dimensional harmonic oscillator. Notice that it is pretty close to Ñ Ω. 2 Let us calculate the error percentage. In[55]:=

H0.567862 - 0.5L  0.5 * 100

Out[55]=

13.5724

2 . Use Ψ HxL = Πx B sin on the interval H- a, aL to obtain a bound on the first excited state of the one a dimensional simple harmonic oscillator. Compare with the exact answer. We carry out the same process as before. The key is that we know that sine as a guess wavefunction is an odd function. Odd functions in this case are the excited states and such. We then find our energy same way as before and minimize with respect to a. Then it like we have found the ground excited energy, otherwise known as the first excited state. We start by solving for B using the normalization condition.

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In Class Exercises 7.nb

In[56]:=

SolveBB2 IntegrateBSinB

Πx

2

F , 8x, - a, a<, Assumptions ® a > 0F Š 1, BF a

1 Out[56]=

1

::B ® -

>, :B ®

>>

a

a

1 In[57]:=

B =

; a

The Kinetic Energy < T > Ñ2 In[58]:=

2

Π

Texp2 =

B 2m

2

Πx IntegrateBSinB

F , 8x, - a, a<, Assumptions ® a > 0F

a

a

Π2 Ñ2 Out[58]=

2 a2 m

Once again this is the same as the ground state energy of the infinite square well. The Potential Energy < V > 1 In[60]:=

m Ω2 IntegrateBx2 SinB

Vexp2 = 2

Πx

2

F , 8x, - a, a<, Assumptions ® a > 0F a

3

2

2

a m I- 3 + 2 Π M Ω Out[60]=

In[77]:=

12 Π2 Eexp2 = Texp2 + Vexp2 a3 m I- 3 + 2 Π2 M Ω2

Π2 Ñ2 +

Out[77]=

12 Π2

2 a2 m

Now let us minimize with respect to a and solve for a In[78]:=

D@Eexp2, aD  FullSimplify a2 m I- 3 + 2 Π2 M Ω2

Π2 Ñ2 -

Out[78]=

4 Π2

a3 m

a m I- 3 + 2 Π2 M Ω2 In[66]:=

SolveB 6 Π2 ΠJ

Out[66]=

Π2 Ñ2 Š 0, aF

6 -3+2 Π2

a3 m 14

N

Ñ

::a ® -

14

6 -3+2 Π2

N

Ñ

>, :a ® m

äΠJ

äΠJ

6 -3+2 Π2

>, m

Ω 14

N

:a ®

Ñ

ΠJ

6 -3+2 Π2

Ω 14

N

Ñ

>, :a ® m

Ω

>> m

Ω

We shall take the postive real solution to a.

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3

4

In Class Exercises 7.nb

m Ω2 a2 I2 Π2 - 3M

Ñ2 Π2 In[74]:=

Eexp2 = 2 m a2 1

Out[74]=

6

14

Ñ

. a ® Π

+ 12 Π2

2 Π2 - 3

mΩ

I- 3 + 2 Π2 M Ω Ñ

6 In[75]:=

N@Eexp2D

Out[75]=

1.67029 Ω Ñ

Notice how the energy is fairly close to the energy of the first excited state of the one - dimensional simple harmonic oscillator. We should calculate the percentage error. In[76]:=

H1.67029 - 1.5L  1.5 * 100

Out[76]=

11.3527

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