The energy is proportional to the mass so the percent error is given by:
$$
\frac{\mu}{m}-1 = \frac{mM/(m+M)}{m} = \frac{M}{m+M} = \frac{1}{1+\frac{m}{M}}-1 \approx 1 - \frac{m}{M}-1 = - \frac{.511\text{MeV}}{938\text{MeV}} \approx .05\% \text{ error}
$$
where $m$ and $M$ are the masses of the electron and proton respectively
b)
Define $E_o,\lambda_o$ to be the transition energy and wavelength in the limit $\mu \to m$.
$$
\lambda_H - \lambda_D = \frac{hc}{E_o}(\frac{E_o}{E_H}-\frac{E_o}{E_D}) = \lambda_o(\frac{1}{1-.00054}-\frac{1}{1-.00027}) \approx = \lambda_o(.00054-.00027) = .00027 \lambda_o
$$
since the balmer alpha line is 656nm, the separation between the hydrogen and deuterium balmer alpha lines is .18nm.
c)
$$
\mu = \frac{mm}{m+m} = \frac{m}{2}
$$
so the binding energy is half that of hydrogen, i.e. $\frac{13.6\text{eV}}{2}$ or 7.8eV.
d)
ignoring the reduced mass correction
$$
E_n \to \frac{m_{\mu}}{m_e} E_n
$$
Since wavelength is inversely proportional to energy, we divide the hydrogen lyman wavelength by the ratio $\frac{m_{\mu}}{m_e}$.
We get $\lambda_{\mu} \approx \frac{122 \text{nm}}{207} \approx .59 \text{nm}$. This would be an x-ray.
Q2 - G5.5
(note: I have a detailed walk through of this problem in my discussion notes for week 1)
a)
In general, the hamiltonian for $N$ non-interacting particles, in one dimension and neglecting spin, is an operator that acts on functions $\Psi(x_1,\ldots,x_N)$ is the sum $\sum_n^N h_n$ where $h_n$ is the hamiltonian for the $n^{\text{th}}$ individual particle, i.e. $-\frac{\hbar^2}{2m_i}\frac{\partial^2}{\partial x_i^2} + V(x_i)$.
For our particular case we have $N=2$ particles that have the same mass m and live in the same box so the hamiltonian is
$$
-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x_1^2}+-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x_2^2}
$$
(we can ignore the potential by requiring the wave function vanish whenever $x_1$ or $x_2$ lie outside $(0,a)$.
Instead of operating this directly on the fermion wave function in example 5.1, I would instead notice that the wave function is a linear combination of products $\psi_1(x_1) \psi_2(x_2)$ and vice versa, where $\psi_n(x)$ is the $n^{\text{th}}$ eigenfunction for the single particle in a box.
These you should check are eigenfunctions of $H$, with the same energy $E_1+E_2=K(1^2+2^2)=5K$. Since a linear combination of two eigenvectors with the same eigenvalue is also an eigenvector with the same eigenvalue, our work is done.
b)
The next energy level of $H$ is $8K$, where $n_1 = 2, n_2 = 2$ i.e. $\psi_2(x_1)\psi_2(x_2)$. If we swap the arguments we get the same state back, so this state has even permutation symmetry and is this a valid state for bosons and distinguishable particles (who have no restriction on permutation symmetry). The next level is $n_1=1,n_2=3$ and $n_1=3,n_2=1$, i.e. two-fold degenerate with an energy $10K$. This level contains one bosonic and one fermionic state, both of which are valid distinguishable particle states. The next level would be $n_1=2,n_2=3$ and vice versa, again two fold degenerate and which energy $13K$. It also contains one odd and one even state.
Any linear combination $cos\theta \psi_0(x_1) \psi_1(x_2) + e^{i \phi} sin\theta \psi_1(x_1) \psi_0(x_2)$ is a valid distinguishable state, where $\psi_n(x)$ are the eigenfunctions of the single particle harmonic oscillator.
probability density plot for $\Psi(x_1,x_2) = \psi_0(x_1)\psi_1(x_2)$.
ii)
$ \frac{1}{\sqrt{2}}( \psi_0(x_1) \psi_1(x_2) + \psi_1(x_1) \psi_0(x_2) ) $
probability density plot for bosonic state.
iii)
$ \frac{1}{\sqrt{2}}( \psi_0(x_1) \psi_1(x_2) - \psi_1(x_1) \psi_0(x_2) ) $
probability density plot for fermionic state.
b)
We make the following observations:
$\langle x \rangle = 0$, by symmetry of the potential, for either particle
From Example 2.5 in Griffiths
$$
\langle V \rangle = \frac{1}{2} E_n
$$
But $V = \frac{1}{2}m \omega^2 x^2$ so
$$
\langle x^2 \rangle = \frac{\hbar \omega (n + \frac{1}{2})}{m \omega^2} = (n + \frac{1}{2})x_o^2
$$
where $x_o = \sqrt{\frac{\hbar}{m \omega}}$ is the "length scale" of the harmonic oscillator.
We can immediately solve for the distinguishable case, which should be $x_o^2( 0 + \frac{1}{2} + 1 + \frac{1}{2} ) = 2 x_o^2$.
Using the ladder operators we can easily solve $\langle x \rangle_{ab}$
\begin{align}
& \langle x \rangle_{ab} \\
= & \int dx \psi_a^*(x) x \psi_b(x) \\
= & \int dx \psi_a^*(x) \frac{1}{\sqrt{2}} x_o ( \hat{a}_+ + \hat{a}_- ) \psi_b(x) \\
= & \frac{1}{\sqrt{2}} x_o \int dx \psi_a^*(x) ( \sqrt{b+1} \psi_{b+1}(x) + \sqrt{b} \psi_{b-1}(x) ) \\
= & \frac{1}{\sqrt{2}} x_o (\sqrt{b+1} \delta_{a,b+1} + \sqrt{b} \delta_{a,b-1})
\end{align}
Taking $a=0$ and $b=1$ we get $\langle x \rangle_{01} = \frac{1}{\sqrt{2}} x_o$ or $\langle x \rangle^2_{01} = \frac{1}{2} x_o^2$. So for bosons and fermions we get answer of $\frac{5}{2}x_o^2$ and $\frac{3}{2}x_o^2$, respectively.
b)
Substituting in $R$ and $r$ for $x_1$ and $x_2$:
$$
|\psi_0(x_1)\psi_1(x_2)|^2 = \frac{2}{\pi} e^{-2R^2-\frac{r^2}{2}} ( R - \frac{r}{2} )^2
$$
Integrating this over $R$:
$$
\frac{2}{\pi} e^{-\frac{r^2}{2}} \int dR e^{-2R^2} ( R - \frac{r}{2} )^2 =
\frac{2}{\pi} e^{-\frac{r^2}{2}} \int dR e^{-2R^2} ( R^2 - 2rR + \frac{r^2}{4} )
$$
The $rR$ term in the parentheses is odd in $R$ and so its contribution to the integral is zero (since $e^{-R^2}$ is even).
We substitute $R \to R' = \sqrt{2} R$ to get
$$
\frac{\sqrt{2}}{\pi} e^{-\frac{r^2}{2}} \int dR' e^{-R'^2} ( R'^2 + \frac{r^2}{2} )
$$
Looking up the relevant integrals on wikipedia we find:
$\int_{-\infty}^{\infty} dR e^{-R^2} = \sqrt{\pi}$
$\int_{-\infty}^{\infty} dR R^2 e^{-R^2} = \frac{\sqrt{\pi}}{2}$
So that we find for the distinguishable case we get a probability $P_d(r)$:
$$
P_d(r) = \frac{1}{2} \frac{1}{\sqrt{2 \pi}} e^{-\frac{r^2}{2}} ( 1 + r^2 )
$$
For the bosonic / fermionic cases, we note that P(r) is the same for $\psi_1(x_1)\psi_0(x_2)$, so we only need to look at the cross term:
$$
\frac{2}{\pi} e^{-\frac{r^2}{2}} \int dR e^{-2R^2} ( R - \frac{r}{2} )( R + \frac{r}{2} ) =
\frac{2}{\pi} e^{-\frac{r^2}{2}} \int dR e^{-2R^2} ( R^2 - \frac{r^2}{4} )
$$
Which contains the same integral we just computed except the second term is negated.
For bosons the second term from the cross term will cancel and for fermions the first term will cancel, i.e.
$$
P_{\pm}(r) = \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{\sqrt{2\pi}} e^{-\frac{r^2}{2}} \left( 2 ( 1 + r^2 ) \pm 2 ( 1 - r^2 ) \right)
= \frac{e^{-\frac{r^2}{2}}}{\sqrt{2\pi}} \begin{cases} 1 & \text{bosons} \\ r^2 & \text{fermions} \end{cases}
$$
Multiply all these by 2 if you want the distribution for $|r|$ instead of $r$.
Probability distribution of particle separations
Q4 - G5.8
a)
We are restricted to states, that, upon performing a simultaneous measurement of the state of the three (presumably distinguishable) particles, return the results $a$,$b$, and $c$, in any order. This restricts us to the basis obtained by applying all possible permutations to the arguments of $Psi_o(x_1,x_2,x_3)\equiv \psi_a(x_1) \psi_b(x_2) \psi_c(x_3)$. Since $\Psi_o$ is what is obtained by performing the "identity" permutation ($1 \to 1, 2 \to 2, 3 \to 3$) on $\Psi_o$, it is a valid state for a system of distinguishable particles.
b) & c)
We can project any identical (but perhaps distinguishable) $N$-particle wavefunction $\Psi(x_1,\ldots,x_N)$ onto the symmetric (+) or antisymmetric (-) subspace of wavefunctions using the following operator $A_{\pm}$:
$$
(A_{\pm} \Psi)(x_1, \ldots, x_N) :=
\frac{1}{\sqrt{N!}} \sum_{\pi} (\pm 1)^{\varepsilon(\pi)} \Psi(x_{\pi(1)},\ldots,x_{\pi(N)})
$$
Where the sum is over all permutations $\pi$ of $N$ objects, and $\varepsilon(\pi)$ is +1 for even permutations and -1 for odd.
Let's check that this operator acts as claimed by permuting arguments $i$ and $j$ of the wavefunction $A_{\pm} \Psi$, i.e. $( A_{\pm} \Psi )(x_1,\ldots,x_i,\ldots,x_j,\ldots,x_N) \to ( A_{\pm} \Psi )(x_1,\ldots,x_j,\ldots,x_i,\ldots,x_N)$
\begin{align}
& (A_{\pm} \Psi)(x_1,\ldots,x_j,\ldots,x_i,\ldots,x_N) \\
= & \frac{1}{\sqrt{N!}} \sum_{\pi} (\pm 1)^{\varepsilon(\pi)} \Psi(x_{\pi(1)},\ldots,x_{\pi(j)},\ldots,x_{\pi(i)},\ldots,x_{\pi(N)}) \\
= & \frac{1}{\sqrt{N!}} \sum_{\pi} (\pm 1)^{\varepsilon(\pi)} \Psi(x_{\pi \circ \pi_{ij}(1)},\ldots,x_{\pi \circ \pi_{ij}(i)},\ldots,x_{\pi \circ \pi_{ij}(j)},\ldots,x_{\pi \circ \pi_{ij}(N)}) \\
= & \frac{1}{\sqrt{N!}} \sum_{\pi} (\pm1) (\pm 1)^{\varepsilon(\pi \circ \pi_{ij})} \Psi(x_{\pi \circ \pi_{ij}(1)},\ldots,x_{\pi \circ \pi_{ij}(i)},\ldots,x_{\pi \circ \pi_{ij}(j)},\ldots,x_{\pi \circ \pi_{ij}(N)}) \\
= & \frac{\pm 1}{\sqrt{N!}} \sum_{\pi} (\pm 1)^{\varepsilon(\pi)} \Psi(x_{\pi(1)},\ldots,x_{\pi(i)},\ldots,x_{\pi(j)},\ldots,x_{\pi(N)}) \\
= & \pm 1 (A_{\pm} \Psi)(x_1,\ldots,x_i,\ldots,x_j,\ldots,x_N) \\
\end{align}
This is definitely a tricky demonstration if you've never seen one like it before. I'll walk through it line by line:
Apply the definition of operator $A_{\pm}$ on the wavefunction $\Psi$ who's had its $i$ and $j$ arguments permuted.
This step is the crux.
$\pi_{ij}$ is the permutation which swaps $i$ and $j$ and nothing else.
The $\circ$ symbol means to apply the functions one after the other, i.e.
$\pi \circ \pi_{ij}(n) = \pi(\pi_{ij}(n))$.
We used $\pi_{ij}$ to swap the $i$ and $j$ back to their original slots, but we can got ahead and insert $\pi_{ij}$ in all the other slots as well because $\pi_{ij}$ does not act on these arguments.
Next we insert $\pi_{ij}$ into the argument $\varepsilon$. Since an interchange $i \leftrightarrow j$ is an odd permutation, $\pi \circ \pi_{ij}$ will have a parity (i.e. even-ness or odd-ness) opposite to $\pi$. We insert the factor $(\pm1)$ to compensate for this.
Here we exploit the fact that the set of all permutations $\{\pi\}$ is equal to the set obtained by taking each $\pi$ and acting $\pi_{ij}$ beforehand, i.e. $\pi \to \pi \circ \pi_{ij}$. This maneuver, incidentally, lies at the heart of almost all applications of group theory to physics (the set of permutations of $N$ objects forms a group, mathematically).
After pulling the critical $\pm1$ out of sum, we apply the definition of $A_{\pm}$ in reverse.
So particle interchange $i \leftrightarrow j$ gives the wavefunction back along with the $\pm1$ factor, as promised.
The operation is equivalent to performing the Slater determinant that Griffiths mentions. Carrying this procedure out on $\Psi_o$ we get:
\begin{align}
& (A_{\pm} \Psi_o)(x_1,x_2,x_3) = \frac{1}{\sqrt{6}} ( \\
& \psi_a(x_1)\left( \psi_b(x_2)\psi_c(x_3) \pm \psi_b(x_3)\psi_c(x_2) \right) + \\
& \psi_a(x_2) \left( \psi_b(x_3) \psi_c(x_1) \pm \psi_b(x_1) \psi_c(x_3) \right) + \\
& \psi_a(x_3) \left( \psi_b(x_1) \psi_c(x_2) \pm \psi_b(x_2) \psi_c(x_1) \right) \\
& )
\end{align}
Q5 - G.10
a)
The dimension of the space of antisymmetric $N$ identical particles, where the individual particles live in spaces of dimension $M$, is given by the binomial coefficient $\binom{M}{N}$. Here we have $N=3$ particles each living in a spin $\frac{1}{2}$ space of dimension $M=2$. $\binom{2}{3}=0$, so there are no antisymmetric states.
