3)
Find the expectation value for the momentum of a particle with a wave function:

U(x)
= A exp [i(a -

**a**^{2 }**ħ**t/ 2m]
Solution:

By definition:
[p] =

(N.B. Square brackets [] used for expectation values because blogger code rejects the standard ones)

And since we know the oscillator (on account of
tunneling) can penetrate beyond the:

**ò**^{¥}_{-}_{¥}_{ }U(x) p^_{x}U(x)* dx
Where: p^

_{x }= -i**ħ****¶****/****¶****x**
Then:

[p]=

**ò**^{¥}_{-}_{¥}_{ }A exp [-i(a -**a**^{2 }**ħ**t/ 2m] (-i**ħ****¶****/****¶****x**) A exp [i(a -**a**^{2 }**ħ**t/ 2m] dx
[p]=

-i

-i

**ħ****ò**^{¥}_{-}_{¥}_{ }A exp (-ia x)**¶****/****¶****x**A exp (ia x) dx
But:

**¶****/****¶****x [**A exp (ia x)] = ia A exp (ia x)
Therefore:

[p] = (-i

=

**ħ ) (**ia )**ò**^{¥}_{-}_{¥}_{ }AA* dx=

**ħ**a**ò**^{¥}_{-}_{¥}_{ }AA* dx
But,

**ò**^{¥}_{-}_{¥}_{ }AA* dx = 1 is just the normalization condition so:
[p]=

**ħ**a(N.B. Square brackets [] used for expectation values because blogger code rejects the standard ones)

4)Using
the operators,

**p^r**and**ℓ**^^{2 }=**-****ħ**^{2}**/ sin**^{2}**q****[ sin****q****¶****/****¶****q****(sin****q****¶****/****¶****q****) +****¶**^{2}**/****¶****j**^{2 }**]**
Write
out the full form of the Schrodinger equation for the hydrogen atom in
spherical coordinates. Thence, obtain
the final form such that

**H**=_{op}
- [E – V] y (

**r,****q****,****j**)
Hint:
Replace m by the reduced mass,

m
= mM/ m+M

__Solution__:

Write out:

**H**=

_{op}**p^r**/2m +

^{2}**ℓ**^

^{2 }/2mr

^{2}+ V(r)

And y = y (

**r,****q****,****j**)
Then:

**H**y = E y =_{op}
[

**p^r**/2m +^{2}**ℓ**^^{2 }/2mr^{2}+ V(r)] y (**r,****q****,****j**) = Ey (**r,****q****,****j**)
Where:

**p^r**= (-iħ 1/r (¶ (r)/ ¶r )

^{2}^{2 }= - ħ

**(1/r**

^{2 }**¶ /¶r (r**

^{2}**¶ /¶r )**

^{2}**But ℓ**^

^{2 }=

**-**

**ħ**

^{2}**/ sin**

^{2}**q**

**[ sin**

**q**

**¶**

**/**

**¶**

**q**

**(sin**

**q**

**¶**

**/**

**¶**

**q**

**) +**

**¶**

^{2}**/**

**¶**

**j**

^{2 }**]**

Then we may write out the Schrodinger equation
in spherical coordinates:

**-**

**ħ**

^{2}**/ 2m (**¶ /¶r (r

**¶ /¶r ) -**

^{2}**ħ**

^{2}**/ 2mr**

^{2}**sin**

^{2}**q**

**[ sin**

**q**

**¶**

**/**

**¶**

**q**

**(sin**

**q**

**¶**

**/**

**¶**

**q**

**)**

**+**

**¶**

^{2}**/**

**¶**

**j**

^{2 }**]**y (

**r,**

**q**

**,**

**j**) + V(r) y (

**r,**

**q**

**,**

**j**)

Replace
m by the reduced mass, m
= mM/ m+M, then:

- ħ

^{2}/ 2m [ ¶ /¶r (r^{2}¶ /¶r ) - 1/ r^{2}sin^{2}q sin q ¶ / ¶ q (sin q ¶ / ¶ q)**+**

**¶**

^{2}**/**

**¶**

**j**

^{2 }**]**y (

**r,**

**q**

**,**

**j**) + V(r) y (

**r,**

**q**

**,**

**j**) = Ey (

**r,**

**q**

**,**

**j**)

Which is the appropriate Schrodinger equation in
spherical coordinates.

6)
In the case of the first solution (5(i))
we demand that the function be single-valued and continuous so that:

F(

**j****+ 2****p****) =**F(**j****)**

So
that: exp
[ i m

_{ℓ }**(**j**+ 2****p****)] =**exp (i m_{ℓ}_{ }j**)**
where m

_{ℓ}is the magnetic quantum number. Dividing by
exp
(i m

_{ℓ }j**) we obtain:**
exp
[ i m

_{ℓ }**(2****p****)] = 1**

Indicate
the condition on m

_{ℓ}for which this is satisfied.__Solution__:

exp [ i m

_{ℓ }**(**j**+ 2****p)] =**exp (i m_{ℓ }j**) = 1**

**Then by de Moivre’s theorem:**

cos

**2****p****m**_{ℓ}**+ i**sin**2****p****m**_{ℓ}**= 1**
But: sin 2p

**m**_{ℓ}**= 0 so:**cos 2p**m**_{ℓ}**= 1**
Which requirement is satisfied provided the
absolute value of m

_{ℓ}**has one of the following values:****ç**m

_{ℓ}

**ç**

**=**0, 1, 2, 3, 4……

(10) The Hamiltonian operator for the quantum
harmonic oscillator is:

**H^**=**p^**^{2}/ 2m + mw^{2}x^{2}/ 2. Also:**p^ =**i h (¶ /¶x). Use these to write out the applicable Schrodinger equation then indicate the form of the expected solutions – with a diagram for the potential.__Solution__:

We have for the Hamiltonian:

**H^**=

**p^**

^{2}/ 2m + mw

^{2}x

^{2}/ 2

And the kinetic energy operator:

**p^ =**i h (¶ /¶x).
It is useful to sketch the quantum potential at
this point and also get the normalization condition. We have:

–A

__<__x__<__A limits of the potential, then we need to use:
y

_{o}(y) = A exp [-y^{2}/ 2] where y = (mw/ h)^{1/2}x
To get the value of A the normalizing factor, we
have:

**ò**

^{¥}

_{- }

_{¥}_{ }‖y(y) ‖

^{2 }dy = 1

Then:

‖y(y) ‖

^{2 }= [A exp (-y^{2}/ 2)]^{ 2}= A^{2}exp (-y^{2})
So that:
A

^{2}**ò**^{¥}_{- }_{¥}**exp (-y**_{ }^{2}) dy = 1
And we may use the well known integral:

**ò**

^{¥}

_{- }

_{¥}**exp (-y**

_{ }^{2}) dy = Öp

Simplifying:

A

^{2}Öp = 1 and A = (1/Öp )^{1/2}
We now modify the energy operator

**p^ =**i h (¶ /¶x).
Instead we use:

**p**_{y}^ = p_{y}^^{2}/ 2m= (-i h Öa ¶ /¶y)^{ 2}/ 2m
= a ħ

^{2 }/ 2m [¶^{2}/¶y^{ 2 }]
(Note: Öa = mw/ ħ )

^{1/2}
The appropriate operator equation is then:

**H**= (-i h Öa ¶ /¶y)

_{ y}^^{ 2}/ 2m + y (mw/ h)

^{-}

^{1/2}

And Schrodinger’s equation becomes – in terms of
y (y) and a:

d

^{2}y/ dy^{2}+ (a^{2}– y^{2}) y = 0
The acceptable solutions for the above are
constrained by the condition:

**ò**

^{¥}

_{- }

_{¥}_{ }‖y(y) ‖

^{2 }dy = 1

Which shows that as y ® 0, y ® ¥

The properties of the given Schrodinger’s
equation are such that the normalization condition can only be fulfilled if:

a = 2n + 1 and n = 0, 1,
2, 3 ….

Since Öa = (mw/ ħ )

^{1/2 }and w = 2pn
But E = ħ n then: a
= 2E/ ħ n

The
energy levels of the harmonic oscillator are:

E

_{n}= (n + ½) ħ n n = 0, 1, 2, 3
The
energy is thereby quantized in steps of
ħ n with the zero point energy:

E

_{o}= ħ n/2
Because now : a

_{n}= 2 E_{n}/ ħ n
We see that each choice of an a

_{n}leads to a different wave function. Each function then introduces a polynomial called a Hermite polynomial of the form: H_{n}(y) yielding either odd or even powers of y. Thus, the Hermite polynomial must be introduced as part of any solutions for the wave function.
The form for any nth wave function then is:

**y**

_{n}**= (2m**

**n**

**/ ħ )**

^{¼}(2^{n}n!)^{-1/2}H_{n}(y) exp (-y^{2}/2)
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