In Class Exercises 7

8/14/2019 In Class Exercises 7

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In Class Exercises 7, Physics 113 B

1. Use the wavefunction of the form

x A Cos

x

a , if

a

2

x a

2

0, 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]:= SolveA2 IntegrateCos x

a

2

, x, a

2,

a

2, Assumptions a 0 1, A

Out[30]= A 2

a

, A 2

a

In[31]:= A 2

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

T T 2

2 m

2

x2

< T >

In[35]:= Texp

2

2 mA

a

2

IntegrateCos x

a

2

, x, a

2,

a

2, Assumptions a 0

Out[35]=

2 2

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 > = < | V | > = < |1

2m 2

< V >

In[36]:= Vexp 1

2m 2 A2 Integratex2 Cos

x

a

2

, x, a

2,

a

2, Assumptions a 0

Out[36]=

a2 m 6 2 2

24 2

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In[49]:= Eexp Texp Vexp

Out[49]=

a2 m 6 2 2

24 2

2 2

2 a2 m

Here is our energy for our wavefunction in the one - dimensional harmonic oscialla-

tor. We now need to minimize this energy to find the ground state energy with thisparticular guessed wavefunction.

In[51]:= SolveDEexp, a 0, a

Out[51]= a 2 3

6214

m

, a 2 314

m 6 214 ,

a 2 314

m 6 214 , a

2 314

m 6 214

We shall take the postive real solution.

In[52]:= Eexp . a 2 3

6214

m

FullSimplify

Out[52]=

1

2

1

36 2

In[54]:=

1

2

1

3 6 2 N

Out[54]= 0.567862

Here is the ground state energy of the cosine wavefunction in the one

dimensional harmonic oscillator. Notice that it is pretty close to1

2 .

Let us calculate the error percentage.

In[55]:= 0.567862 0.5 0.5 100

Out[55]= 13.5724

2 . Use x B sin

x

aon the interval a, a to obtain a bound on the first excited state of the one

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.

2 In Class Exercises 7.nb



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In[56]:= SolveB2 IntegrateSin x

a

2

, x, a, a, Assumptions a 0 1, B

Out[56]= B 1

a

, B 1

a

In[57]:= B 1

a

;

The Kinetic Energy < T >

In[58]:= Texp2

2

2 mB

a

2

IntegrateSin x

a

2

, x, a, a, Assumptions a 0

Out[58]=

2 2

2 a2 m

Once again this is the same as the ground state energy of the infinite square well.

The Potential Energy < V >

In[60]:= Vexp2 1

2m 2 Integratex2 Sin

x

a

2

, x, a, a, Assumptions a 0

Out[60]=

a3 m 3 2 2 2

12 2

In[77]:= Eexp2 Texp2 Vexp2

Out[77]=

a3 m 3 2 2 2

12 2

2 2

2 a2 m

Now let us minimize with respect to a and solve for a

In[78]:= DEexp2, a FullSimplify

Out[78]=

a2 m 3 2 2 2

4 2

2 2

a3 m

In[66]:= Solvea m 3 2 2 2

6 2

2

2

a3 m 0, a

Out[66]= a 6

32 214

m

, a 6

32 214

m

,

a 6

32 214

m , a

632 2

14

m

We shall take the postive real solution to a.

In Class Exercises 7.nb 3



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In[74]:= Eexp2

2

2

2 m a2

m 2 a2 2 2 312 2

. a 6

2 2 3

14

m

Out[74]=

1

63 2 2

In[75]:= NEexp2

Out[75]= 1.67029

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]:= 1.67029 1.5 1.5 100

Out[76]= 11.3527

4 In Class Exercises 7.nb


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In Class Exercises 7