# #6A Solutions

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### Q1

The energy of a single electron around a proton is

$E_n = -\dfrac {m_e e^4}{8\epsilon_0^2 h^2 n^2}\tag{1}$

with $$n=1,2,3 ...\infty$$. The corresponding eigenstates can be expressed as the product of a angular and radial functions

$|\psi(r,\theta,\phi) \rangle = | R(r) \rangle |Y_{\ell}^{m_l}(\theta,\phi) \rangle$

which have three quantum numbers to describe an unique wavefunction, $$n$$, $$l$$ and $$m_l$$ each with unique range permissible. The fact that energy depends only on $$n$$ indicates that this system has degenerate wavefunction. List all the eigenstates (explicitly pointing out the appropriate quantum numbers) that have the following energies:

1. $$E_n = -\dfrac {m_e e^4}{32\epsilon_0^2 h^2}$$
2. $$E_n = -\dfrac {m_e e^4}{72\epsilon_0^2 h^2}$$
3. $$E_n = -\dfrac {m_e e^4}{128\epsilon_0^2 h^2}$$

### Solution:

To find $$n$$ corresponding to the energies in points a,b and c one needs to compare $$(1)$$ to the expressions in each point. The numeric coefficient in denominator is a product of 8 and $$n^{2}$$:

1. $$32 = 8\times n^{2} \rightarrow n^{2}=4 \rightarrow n=2$$
2. $$72 = 8\times n^{2} \rightarrow n^{2}=9 \rightarrow n=3$$
3. $$128 = 8\times n^{2} \rightarrow n^{2}=16\rightarrow n=4$$

In order to list all the eigenstates corresponding to the $$n$$'s found in points a, b and c, one needs to remember the values the other two quantum numbers $$\ell$$ and $$m_{\ell}$$ could get:

• $$n = 1, 2, 3, ...\infty$$
• $$\ell = 0, 1, 2, ... (n-1)$$
• $$m_{\ell} = 0, \pm 1, \pm 2, ... \pm \ell$$

Thus, the lists for eigenstates are:

a. $$| \psi_{200} \rangle$$

$$| \psi_{21-1} \rangle$$, $$| \psi_{210} \rangle$$, $$| \psi_{211} \rangle$$

b. $$| \psi_{300} \rangle$$

$$| \psi_{31-1} \rangle$$, $$| \psi_{310} \rangle$$, $$| \psi_{311} \rangle$$

$$| \psi_{32-2} \rangle$$, $$| \psi_{32-1} \rangle$$, $$| \psi_{320} \rangle$$, $$| \psi_{321} \rangle$$,$$| \psi_{322} \rangle$$

c. $$| \psi_{400} \rangle$$

$$| \psi_{41-1} \rangle$$, $$| \psi_{410} \rangle$$, $$| \psi_{411} \rangle$$

$$| \psi_{42-2} \rangle$$, $$| \psi_{42-1} \rangle$$, $$| \psi_{420} \rangle$$, $$| \psi_{421} \rangle$$,$$| \psi_{422} \rangle$$

$$| \psi_{43-3} \rangle$$, $$| \psi_{43-2} \rangle$$, $$| \psi_{43-1} \rangle$$, $$| \psi_{430} \rangle$$, $$| \psi_{431} \rangle$$,$$| \psi_{432} \rangle$$, $$| \psi_{433} \rangle$$

### Q2

How many radial nodes (i.e., nodes in $$R(r)$$) are there in the following hydrogen orbitals?

1. $$| \psi_{1s} \rangle$$
2. $$| \psi_{2s} \rangle$$
3. $$| \psi_{3s} \rangle$$
4. $$| \psi_{2p_x} \rangle$$
5. $$| \psi_{3d_{z^2}} \rangle$$
6. $$| \psi_{100} \rangle$$ (the subscript refers to the $$n$$,$$l$$,$$m_l$$ triad of quantum numbers)

### Solution:

The number of radial nodes is given by the formula $$n-l-1$$, where $$n$$,$$l$$ are principal and angular quantum number correspondingly.

Thus,

1. 0
2. 1
3. 2
4. 0
5. 0
6. 0

Alternative: To find the number of nodes of radial function in each state you need to know the representation of the radial function in each state (see Table Hydrogen-like atomic wavefunctions for n values 1,2,3 http://chemwiki.ucdavis.edu/Wikitext..._Hydrogen_Atom). To use the Table Hydrogen-like atomic wavefunctions for n values 1,2,3 properly you need to remember that the $$\psi_{n,l,m_l}$$ is a wavefunction of a particular state, described by the $$n$$,$$l$$,$$m_l$$ triad of quantum numbers. You need to analyze how many zeros with respect to $$r$$ the radial functions in Table Hydrogen-like atomic wavefunctions for n values 1,2,3 have in each particular state given by the set of subscripts $$n$$,$$l$$,$$m_l$$. Then you eliminate the zeros which correspond to $$r=0$$. As a result, you get the number of radial nodes.

1. 0
2. 1
3. 2
4. 0
5. 0
6. 0

### Q3

Explain why the radial distribution function $$r^2R(r)^*R(r)$$ should be used to discuss the probability of finding an electron from a nucleus rather than the square of the radial component of the wavefunction $$R^2$$.

### Solution:

In order to find the probability one needs to compute integral

$\iiint \limits_{space}^{ }\psi^{*}(x,y,z) \psi(x,y,z)\, dx\,dy\,dz$

where $$dxdydz$$ is a volume element.

In order to compute this integral in a different coordinate system, e.g. $$(r, \theta,\phi)$$, one needs to compute:

$\iiint \limits_{space}^{ }\psi^{*}(r,\theta,\phi) \psi(r,\theta,\phi) r^2 \sin{\theta}\, dr\,d\theta\,d\phi$

Here, $$r^2 \sin{\theta}$$ is the Jacobian or a function that represents the transformation element of unit volume in one coordinate system to another. In the case described in Q3, for the radial distribution function only the part dependent on $$r$$ could serve as Jacobian, i.e. the $$r^2$$ part. That's why $$r^2R(r)^*R(r)$$ should be used to discuss the probability of finding an electron from a nucleus rather than the square of the radial component of the wavefunction $$R^2(r)$$.

### Q4

Show that the Hamiltonian operator $$\hat{H}=(p^2/2m)$$ of a rigid rotor commutes with all three components of $$\vec{L}$$. Thus $$H$$, $$L^2$$, and $$L_z$$ are mutually compatible observables.

$H = \frac{p^2}{2I} = \frac{L^2}{2I}\tag{1.1}$

We want to prove that $$[L, H] = 0$$.

$[L, H] = [L_x, L^2] + [L_y, L^2] + [L_z, L^2]\tag{1.2}$

Where we have ignored the factor of $$\frac{1}{2I}$$ as it is a constant that multiplies the entire equation and is thus irrelevant as our goal is to show that 1.2 is equal to zero.

$[L_x, L^2] = [L_x, L_x^2] + [L_x, L_y^2] + [L_x, L_z^2]\tag{1.3}$

Commutator of $$L_x$$ with itself is zero, leaving us the other two terms which we expand.

$[L_x, L^2] = L_y[L_x, L_y] - [L_x, L_y]L_y + L_z[L_x, L_z] - [L_x, L_z]L_z\tag{1.4}$

$[L_x, L^2] = L_yL_z + L_zL_y - L_zL_y - L_yL_z = 0\tag{1.5}$

Where the $$i\hbar$$'s are ignored.

$[L_y, L^2] = [L_y, L_x^2] + [L_y, L_y^2] + [L_y, L_z^2]\tag{1.6}$

Commutator of $$L_y$$ with itself is zero, leaving us the other two terms which we expand.

$[L_y, L^2] = L_z[L_y, L_z] - [L_y, L_z]L_z + L_x[L_y, L_x] - [L_y, L_x]L_x\tag{1.7}$

$[L_y, L^2] = L_zL_x + L_xL_z - L_xL_z - L_zL_x = 0\tag{1.8}$

Where again the $$i\hbar$$'s are ignored.

$[L_z, L^2] = [L_z, L_x^2] + [L_z, L_y^2] + [L_z, L_z^2]\tag{1.9}$

Commutator of $$L_z$$with itself is zero, leaving us the other two terms which we expand.

$[L_z, L^2] = L_x[L_z, L_x] - [L_z, L_x]L_x + L_y[L_z, L_y] - [L_z, L_y]L_y\tag{1.10}$

$[L_z, L^2] = L_xL_y + L_yL_x - L_yL_x - L_xL_y = 0\tag{1.11}$

Where again the $$i\hbar$$'s are ignored.

### Q5

Derive the following commutation relationship (one of three in 3-D space) using the representation as differential operators

$[\hat{L}_z,\hat{L}_x] =i \hbar \hat{L}_y$

What does this equation and the other two commutation relationships mean (in your words)?

$[L_z, L_x] = L_zL_x - L_xL_z = (yp_z - zp_y)(xp_y - yp_x) + (xp_y - yp_x)(yp_z - zp_y)$

$= yp_zxp_y - y^2p_zp_x - zxp_yp_y + 2p_yyp_x + xp_yyp_z - xp_yzp_y - yp_xyp_z + yzp_xp_y = z[y, p_y]px + xp_z[p_y, y]$

$= i\hbar(zp_x - xp_z) = i\hbar L_y$

From this we can that only one component of the angular momentum can be defined at any one time since they do not commute with each other.

### Q6

Evaluate the z component of the angular momentum and kinetic energy for a particle on a ring that is described by the following wavefunctions:

1. $$| \psi(\phi) \rangle = e^{i\phi}$$
2. $$| \psi(\phi) \rangle = e^{-2\pi i\phi}$$
3. $$| \psi(\phi) \rangle = \cos(\phi)$$

$L_z = -i\hbar \frac{\partial}{\partial \phi}$

$H = -\frac{\hbar^2}{2I}(\frac{1}{sin\theta}\frac{\partial}{\partial \theta}sin\theta\frac{\partial}{\partial \theta} + \frac{1}{sin\theta^2}\frac{\partial^2}{\partial \phi^2})$

A)

$H| \psi(\phi) \rangle = \frac{\hbar^2}{2I}\frac{1}{sin^2\theta}e^{i\phi}$

$L_z| \psi(\phi) \rangle = \hbar e^{i\phi}$

B)

$H| \psi(\phi) \rangle = \frac{2\pi^2\hbar^2}{I}\frac{1}{sin^2\theta}e^{-2i\pi\phi}$

$L_z| \psi(\phi) \rangle = -2\pi \hbar e^{-2i\pi\phi}$

C)

$H| \psi(\phi) \rangle = \frac{\hbar^2}{2I}\frac{1}{sin^2\theta}cos\phi$

$L_z| \psi(\phi) \rangle = i\hbar sin\phi$

### Q8

To find the average of any observable in quantum mechanics we need to take an expectation value. In Dirac notation:

$\langle \psi | r | \psi \rangle\tag {1}$

Given that

$| \psi(r,\theta,\phi) \rangle = |R_n(r) \rangle |Y_{\ell}^{m}(\theta,\phi) \rangle \tag{2}$

$\langle R_n(r) \langle |Y_{\ell}^{m}(\theta,\phi) | r | R_n(r) \rangle |Y_{\ell}^{m}(\theta,\phi) \rangle \tag{3}$

Which we can rewrite by combining terms in the same coordinates

$\langle Y_{\ell}^{m}(\theta,\phi) | Y_{\ell}^{m}(\theta,\phi) \rangle \langle R_n(r) |r| R_n(r) \rangle \tag{4}$

Because the Spherical Harmonics are already normalized in the wavefunctions

$1 \times \langle R_n(r) |r| R_n(r) \rangle \tag{5}$

Now we need to evaluate the expression, with the right functions for 1s and 2s

1s

$\int^\infty_0 2 \left (\dfrac {Z}{a_0} \right )^{3/2} e^{\dfrac{-Zr}{a_0}}\times r \times 2 \left (\dfrac {Z}{a_0} \right )^{3/2} e^{\dfrac{-Zr}{a_0}} \times r^2dr \tag{6}$
The final $$r^2 dr$$ is the Jacobian in Spherical Coordinates.
$4 \left (\dfrac {Z}{a_0} \right )^3 \int^\infty_0 e^{\dfrac{-2Zr}{a_0}}\times r^3dr \tag{7}$
Using the great integral:
$\int^\infty_0 x^ne^{-bx} dx = \dfrac{n!}{b^{n+1}} \tag{8}$
$4 \left (\dfrac {Z}{a_0} \right )^3 \times 3! \times \dfrac{a_0^4}{16Z^4} \tag{9}$
$\langle r \rangle = \dfrac{3a_0}{2Z} \tag{11}$
2s
$\int^\infty_0 \dfrac {1}{2 \sqrt {2}}\left (\dfrac {Z}{a_0} \right )^{3/2} (2 - \dfrac{Zr}{a_0}) e^{\dfrac{-Zr}{2a_0}} \times r \times \dfrac {1}{2 \sqrt {2}}\left (\dfrac {Z}{a_0} \right )^{3/2} (2 - \dfrac{Zr}{a_0}) e^{\dfrac{-Zr}{2a_0}} \times r^2dr \tag{12}$
$\dfrac {1}{8} \left (\dfrac {Z}{a_0} \right )^{3} \int^\infty_0 (2 - \dfrac{Zr}{a_0})^2 e^{\dfrac{-Zr}{a_0}} \times r^3dr \tag{13}$
After much simplifying you get
$\dfrac {1}{2} \left (\dfrac {Z}{a_0} \right )^{3} \int^\infty_0 r^3 e^{\dfrac{-Zr}{a_0}} dr - \dfrac {1}{2} \left (\dfrac {Z}{a_0} \right )^{4} \int^\infty_0 r^4 e^{\dfrac{-Zr}{a_0}} dr + \dfrac {1}{8} \left (\dfrac {Z}{a_0} \right )^{5} \int^\infty_0 r^5 e^{\dfrac{-Zr}{a_0}} dr \tag{14}$
$\dfrac {3Z}{a_0} - \dfrac {12Z}{a_0} + \dfrac {15Z}{a_0} \tag {15}$
$\langle r \rangle = \dfrac{6a_0}{Z} \tag{16}$

## Q9

The least probable radius is $$r=0$$. The electron will not be found inside the proton.

The most probable radius for 2p will be found by finding the maximum of the radial probability density.

$r^2 R(r)^* R(r)\tag{17}$

$r^2 \times \dfrac {1}{2 \sqrt {6}}\left (\dfrac {Z}{a_0} \right )^{3/2} \dfrac {Zr}{a_0} e^{\dfrac {-Zr}{2a_0}} \times \dfrac {1}{2 \sqrt {6}}\left (\dfrac {Z}{a_0} \right )^{3/2} \dfrac{Zr}{a_0} e^{\dfrac{-Zr}{2a_0}} \tag{18}$

$r^4 \dfrac {1}{24} \left (\dfrac {Z}{a_0} \right )^{5} e^{\dfrac {-Zr}{a_0}} \tag{19}$

Use the product rule to find the derivative

$\dfrac{d}{dr} r^4 \dfrac {1}{24} \left (\dfrac {Z}{a_0} \right )^{5} e^{\dfrac {-Zr}{a_0}} = \dfrac{-Zr^4}{a_0}e^{\dfrac{-Zr}{a_0}} + 4r^3e^{\dfrac{-Zr}{a_0}} \tag{20}$

Simplify

$r^3 \left( 4 -\dfrac{-Zr}{a_0} \right) e^{\dfrac{-Zr}{a_0}} \tag{21}$

So the function will be zero at $$r=0$$ - not helpful - or when:

$4 -\dfrac{-Zr}{a_0} = 0 \tag{22}$

$r = \dfrac{4a_0}{Z} \tag{23}$

## Q10

The selections rules are

Any n values allowed

$$\ell = \pm 1$$

$$m_\ell = \pm 1 , 0$$

1. $$2p_x \rightarrow 1s$$ Allowed
2. $$2s \rightarrow 1s$$ Forbidden
3. $$3s \rightarrow 1s$$ Forbidden
4. $$3d_{z^2} \rightarrow 2s$$ Forbidden
5. $$3d_{x^2-y^2} \rightarrow 1s$$ Forbidden

## Q11

Angular Momentum is

$L = \hbar \left[ \ell(\ell +1) \right]^{\dfrac{1}{2}} \tag{24}$

And for the number of nodes

Total Nodes $$= n -1$$

Angular Nodes $$= \ell$$

Radial Nodes $$= n -1 - \ell$$

1. 1s, $$L = 0$$, Angular Nodes $$= 0$$, Radial Nodes $$= 0$$
2. 2s, $$L =0$$, Angular Nodes $$= 0$$, Radial Nodes $$= 1$$
3. 2p, $$L =\hbar \sqrt{2}$$, Angular Nodes $$= 1$$, Radial Nodes $$= 0$$
4. 3d, $$L =\hbar \sqrt{6}$$, Angular Nodes $$= 2$$, Radial Nodes $$= 0$$
5. 5f, $$L =\hbar \sqrt{12}$$, Angular Nodes $$= 3$$, Radial Nodes $$= 1$$

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