# Compute normalization constant A if n=1?

## First of all, those are energy eigenvalues same as eigenstates ,Energy is the operator but we do not know the eigenfunction. Are hydrogen energy eigenstates orthonormal?Why?Does this have any effect when we separate ψ(x, t) into spatial wavefunction and time function I suppose (Q1) is to be solved by the properties of orthonormality and i get an answer of $\frac{1}{\sqrt{18}}$ but can it be wrong? Then i get an answer of ${\alpha}^{2} \mu {c}^{2} \frac{4.22}{18}$ for $< E >$ I think the wavefunction remains same at time t because the time part ${e}^{i \omega t}$ becomes 1.

Feb 22, 2018

$\psi \left(r\right) = A \left[3 i {\psi}_{100} \left(r\right) - 4 {\psi}_{211} \left(r\right) - i {\psi}_{210} \left(r\right) + \sqrt{10} {\psi}_{21 - 1} \left(r\right)\right]$

This wavefunction, or state function, is constructed as a linear combination of eigenstates that are themselves orthonormal ... if they are not, then nobody would want to do this problem because it would take them days.

CLEARLY, $n \ne 1$, as we have $n = 2$ eigenstates mixed in... we are given ${\psi}_{n l m}$, so... what is $n$ for the latter three?

Regardless, we set it up the easy way as

??? = |c_1|^2 + |c_2|^2 + ...

$= {\left(3 i\right)}^{\text{*"(3i) + (-4)^2 + (-i)^"*}} \left(i\right) + {\left(\sqrt{10}\right)}^{2}$
$36 = 9 + 16 + 1 + 10$

So the magnitude of this wave function is $36$ times unity. Therefore, to normalize ${\psi}^{\text{*}} \psi$, ${A}^{2} = \frac{1}{36}$, or...

$\implies A = \frac{1}{6}$

The hard way is to do this...

$1 = {A}^{2} {\int}_{0}^{2 \pi} {\int}_{0}^{\pi} {\int}_{0}^{\infty} {\left[3 i {\psi}_{100} \left(r\right) - 4 {\psi}_{211} \left(r\right) - i {\psi}_{210} \left(r\right) + \sqrt{10} {\psi}_{21 - 1} \left(r\right)\right]}^{\text{*}} \left[3 i {\psi}_{100} \left(r\right) - 4 {\psi}_{211} \left(r\right) - i {\psi}_{210} \left(r\right) + \sqrt{10} {\psi}_{21 - 1} \left(r\right)\right] {r}^{2} \mathrm{dr} \sin \theta d \theta d \phi$

Here this is made easier because the wave functions apparently, somehow, are invariant to angular coordinates...

Next, we ignore cross-terms, which are orthogonal, and only obtain the diagonal terms, which are normalized.

$1 = {A}^{2} \cdot 4 \pi {\int}_{0}^{\infty} \left[9 {\psi}_{100}^{\text{*"(r)psi_(100)(r) +16psi_(211)^"*"psi_(211)(r) + psi_(210)^"*"(r)psi_(210)(r) + 10psi_(21-1)^"*}} \left(r\right) {\psi}_{21 - 1} \left(r\right)\right] {r}^{2} \mathrm{dr}$

Separate integrals to get:

$= 9 {A}^{2} \cdot 4 \pi {\int}_{0}^{\infty} {\psi}_{100}^{\text{*"(r)psi_(100)(r) r^2dr + 16A^2 cdot 4pi int_(0)^(oo)psi_(211)^"*"psi_(211)(r)r^2dr + A^2 cdot 4pi int_(0)^(oo)psi_(210)^"*"(r)psi_(210)(r)r^2dr + 10A^2 cdot 4pi int_(0)^(oo)psi_(21-1)^"*}} \left(r\right) {\psi}_{21 - 1} \left(r\right) {r}^{2} \mathrm{dr}$

Each of these are normalized after including the $4 \pi$ in the angular coordinates by construction, so:

$1 = 9 {A}^{2} \cdot 1 + 16 {A}^{2} \cdot 1 + {A}^{2} \cdot 1 + 10 {A}^{2} \cdot 1$

$= {A}^{2} \left(9 + 16 + 1 + 10\right)$

$= 36 {A}^{2}$

Therefore, $\textcolor{b l u e}{A = \frac{1}{6}}$.

Feb 23, 2018

This is to address the rest of the questions, except $f$. I don't want to do $f$ because most of this question is not well-inspected...

b)

Well, this should be trivial. Keep it simple. You tack on Gamma(t) = e^(-iE_nt//ℏ), i.e. ignore the rest of the quantum numbers.

color(blue)(Psi(r,t)) = c_1psi_(nlm)(r)e^(-iE_nt//ℏ) + c_2psi_(nlm)(r)e^(-iE_nt//ℏ) + . . .

= color(blue)(1/6 [3ipsi_(100)(r)e^(-iE_1t//ℏ) - 4psi_(211)(r)e^(-iE_2t//ℏ) - ipsi_(210)(4)e^(-iE_2t//ℏ)+ sqrt(10)psi_(21-1)(r)e^(-iE_2t//ℏ)])

We can't say any more because these eigenstates aren't degenerate in Hydrogen atom.

c)

The expectation value is well-known as a form of the Rydberg equation for hydrogen atom:

${E}_{n} = - \text{13.61 eV} \cdot {Z}^{2} / {n}^{2}$

Wave functions are normalized to an arbitrary 100% probability density, so no matter what constant is in front, the eigenvalues are not modified.

(Otherwise, it would be a huge issue that surely would not have gone unnoticed amongst physical chemists everywhere!)

The Hamiltonian is linear, so

$\hat{H} \frac{1}{6} \left[3 i {\psi}_{100} \left(r\right) - 4 {\psi}_{211} \left(r\right) - i {\psi}_{210} \left(r\right) + \sqrt{10} {\psi}_{21 - 1} \left(r\right)\right]$

$= \frac{1}{6} \left[3 i \hat{H} {\psi}_{100} \left(r\right) - 4 \hat{H} {\psi}_{211} \left(r\right) - i \hat{H} {\psi}_{210} \left(r\right) + \sqrt{10} \hat{H} {\psi}_{21 - 1} \left(r\right)\right]$

$= \frac{1}{6} \left[3 i {E}_{1} {\psi}_{100} \left(r\right) - 4 {E}_{2} {\psi}_{211} \left(r\right) - i {E}_{2} {\psi}_{210} \left(r\right) + \sqrt{10} {E}_{2} {\psi}_{21 - 1} \left(r\right)\right]$

And since for $Z = 1$,

${E}_{1} = - \text{13.61 eV}$

${E}_{2} = - \frac{\text{13.61 eV}}{4} = {E}_{1} / 4$

it follows that

$\hat{H} \psi \left(r\right) = \frac{1}{6} \left[3 i \cdot 4 {E}_{1} / 4 {\psi}_{100} \left(r\right) - {E}_{1} / 4 \cdot 4 {\psi}_{211} \left(r\right) - i {E}_{1} / 4 {\psi}_{210} \left(r\right) + \sqrt{10} {E}_{1} / 4 {\psi}_{21 - 1} \left(r\right)\right]$

$= \textcolor{red}{{E}_{1} / 4} \cdot \frac{1}{6} \left[\textcolor{red}{4} \cdot 3 i {\psi}_{100} \left(r\right) - 4 {\psi}_{211} \left(r\right) - i {\psi}_{210} \left(r\right) + \sqrt{10} {\psi}_{21 - 1} \left(r\right)\right]$

$\textcolor{red}{\ne {E}_{1} / 4 \cdot \psi \left(r\right)}$

This is a hugely problematic question, because you cannot get ONE expectation value. $\psi \left(r\right)$ is not an eigenstate of the Hamiltonian, because you have three triply degenerate $n = 2$ states and one separate $1 s$ state.

The eigenvalue of a linear combination is only the same as the individual eigenvalues of each wave function if all of them are degenerate eigenstates.

d)

The eigenvalue of ${\hat{L}}^{2}$ corresponding to a properly normalized eigenstate is:

hatL^2Y_l^(m_l)(theta,phi) = ℏ^2l(l+1)Y_l^(m_l)(theta,phi)

Therefore,

${\hat{L}}^{2} \psi \left(r\right)$

$= {\hat{L}}^{2} \left\{\frac{1}{6} \left[3 i {\psi}_{100} \left(r\right) - 4 {\psi}_{211} \left(r\right) - i {\psi}_{210} \left(r\right) + \sqrt{10} {\psi}_{21 - 1} \left(r\right)\right]\right\}$

But all of these functions have no angular dependence apparently, so the eigenvalue is just zero.

This is again problematic and rather unphysical, because states with $l > 0$ should have angular components...

e)

Hydrogen atom is spherically symmetric, so angular momentum is conserved, i.e. it is a constant of the motion. Energy is presumably conserved; the question stated upfront that each $\psi$ within $\psi \left(r\right)$ is a normalized energy eigenstate.

Only position is in flux over time.

Mar 5, 2018

I will answer (c), and (d) - the others have already been answered correctly (and (f) will take too much time to type out).
(c) $< E > = \frac{7}{16} {E}_{1}$
(d) < hat{L}^2 > = 3/2ℏ^2

#### Explanation:

${\psi}_{n l m} \left(\vec{r}\right)$ are simultaneous eigenstates of $\hat{H}$, ${\hat{L}}^{2}$ and ${\hat{L}}_{z}$ :

$\hat{H} {\psi}_{n l m} = {E}_{n} {\psi}_{n l m}$,
hat{L}^2 psi_{nlm} = l(l+1)ℏ^2 psi_{nlm},
hat{L}_z psi_{nlm} = mℏ psi_{nlm}

(c) We can calculate the expectation value $< E >$ by evaluating the triple integral

$\int \psi {\left(\vec{r} , 0\right)}^{\ast} \hat{H} \psi \left(\vec{r} , 0\right) {d}^{3} r$

which is not really a very difficult task because the ${\psi}_{n l m}$s are orthonormal. We can, however, get to the answer more quickly by using the measurement theory in quantum mechanics:

If a wavefunction $\psi$ is expanded in eigenfunction ${\phi}_{i}$ of a hermitian operator $\hat{A}$ (such that $\hat{A} {\phi}_{i} = {a}_{i} {\phi}_{i}$) in the form
$\psi = {\sum}_{n} {c}_{n} {\phi}_{n}$
then the measurement of $\hat{A}$ will lead to a result of ${a}_{i}$ with probability $| {c}_{i} {|}^{2}$.

Thus, for the given function, the probability that an energy measurement will yield ${E}_{1}$ is given by $| \frac{3 i}{6} {|}^{2} = \frac{1}{4}$. The other three pieces in the given wavefunctions are each eigenfunctions of $\hat{H}$ with the eigenvalue ${E}_{2}$. So the probability of getting ${E}_{2}$ on measuring the energy is given by $1 - \frac{1}{4} = \frac{3}{4}$ (you can also get this result by adding together the $| - \frac{4}{6} {|}^{2} + | - \frac{i}{6} {|}^{2} + | \frac{\sqrt{10}}{6} {|}^{2}$).
Note: the normalization constant is $\frac{1}{6}$

So, the expectation value of the energy is

$< E > = \frac{1}{4} \times {E}_{1} + \frac{3}{4} \times {E}_{2}$

Since ${E}_{2} = {E}_{1} / 4$, it follows that

$< E > = \left(\frac{1}{4} + \frac{3}{16}\right) {E}_{1} = \frac{7}{16} {E}_{1}$

(d) It is easy to see that the probabilities of getting the results 0 = 0(0+1)ℏ^2, and 2ℏ^2 = 1(1+1)ℏ^2 are again $\frac{1}{4}$ and $\frac{3}{4}$ respectively.

So < hat{L}^2 > = 1/4 times 0 + 3/4 times 2ℏ^2 = 3/2ℏ^2