Chapter 4: Quantum Mechanics in Three Dimensions

What is the probability that an electron in the ground state of hydrogen will be found inside the nucleus?

1. First calculate the exact answer, assuming the wave function is correct all the way down to$\mathbf{r}\mathbf{=}\mathbf{0}$. Let b be the radius of the nucleus.
2. Expand your result as a power series in the small number$\mathit{a}\mathbf{=}\mathbf{2}\mathit{b}\mathbf{}\mathit{l}\mathbf{}\mathit{a}$, and show that the lowest-order term is the cubic:$\mathbf{P}\mathbf{\approx }\left(4l3\right){\left(bla\right)}^{\mathbf{3}}$. This should be a suitable approximation, provided that$\mathbf{b}\mathbf{\ll }\mathbf{a}$(which it is).
3. Alternatively, we might assume that$\mathbf{\psi }\left(r\right)$is essentially constant over the (tiny) volume of the nucleus, so that$\mathit{P}\mathbf{\approx }\left(4l3\right)\mathit{\pi }{\mathit{b}}^{\mathbf{3}}\mathbf{}\mathit{l}\mathit{\psi }\left(0\right)\mathbf{}{\mathit{l}}^{\mathbf{2}}$.Check that you get the same answer this way.
4. Use$\mathbf{b}\mathbf{\approx }{\mathbf{10}}^{\mathbf{-}\mathbf{15}}\mathbf{}\mathbf{m}$and${\mathbf{a}}^{\mathbf{\approx }\mathbf{0}}\mathbf{\approx }\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{m}$to get a numerical estimate for$\mathbf{P}$. Roughly speaking, this represents the fraction of its time that the electron spends inside the nucleus:"

(a) Use the recursion formula (Equation 4.76) to confirm that when$\mathit{I}\mathbf{=}\mathit{n}\mathbf{-}\mathbf{1}$ the radial wave function takes the form

${\mathit{R}}_{\mathbf{n}\left(n-1\right)}\mathbf{=}{\mathit{N}}_{\mathbf{n}}{\mathit{r}}^{\mathbf{n}\mathbf{-}\mathbf{1}}{\mathit{e}}^{\mathbf{-}\mathbf{r}\mathbf{/}\mathbf{n}\mathbf{a}}$ and determine the normalization constant by direct integration.

(b) Calculate 200a and $<r^2>$ for states of the form ${\mathit{\psi }}_{\mathbf{n}\left(n-1\right)\mathbf{m}}\mathbf{·}$

(c) Show that the "uncertainty" in $\mathit{r}\left({\delta }_r\right)$ is$<r>\mathbf{/}\sqrt{\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{1}}$for such states. Note that the fractional spread in decreases, with increasing (in this sense the system "begins to look classical," with identifiable circular "orbits," for large ). Sketch the radial wave functions for several values of, to illustrate this point.

Coincident spectral lines. ${}^{\mathbf{43}}$According to the Rydberg formula (Equation 4.93) the wavelength of a line in the hydrogen spectrum is determined by the principal quantum numbers of the initial and final states. Find two distinct pairs$\left\{n_i,n_f\right\}$ that yield the same $\mathit{\lambda }$. For example,role="math" localid="1656311200820" $\left\{6851,6409\right\}$ and$\left\{15283,11687\right\}$will do it, but you're not allowed to use those!

Consider the observables$\mathbf{A}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}$and$\mathbf{B}\mathbf{=}{\mathbf{L}}_{\mathbf{z}}$ .

(a) Construct the uncertainty principle for${\mathbf{\sigma }}_{\mathbf{A}}{\mathbf{\sigma }}_{\mathbf{B}}\mathbf{}$

(b) Evaluate${\mathit{\sigma }}_{\mathbf{B}}$ in the hydrogen state${\mathbf{\psi }}_{\mathbf{n}\mathbf{/}\mathbf{m}}\mathbf{}$ .

(c) What can you conclude about$\mathbf{<}\mathbf{xy}\mathbf{>}$in this state?

An electron is in the spin state

$\mathcal{x}\mathbf{=}\mathbf{A}\left(\begin{array}{c}1-2i\\ 2\end{array}\right)$

(a) Determine the constant by normalizing $\mathcal{x}$.

(b) If you measured ${\mathit{S}}_{\mathbf{z}}$on this electron, what values could you get, and what is the probability of each? What is the expectation value of ${\mathbf{S}}_{\mathbf{z}}$?

(c) If you measured ${\mathbf{S}}_{\mathbf{x}}$on this electron, what values could you get, and what is the probability of each? What is the expectation value of ${\mathbf{S}}_{\mathbf{x}}$?

(d) If you measured ${\mathbf{S}}_{\mathbf{y}}$on this electron, what values could you get, and what is the probability of each? What is the expectation value of${\mathbf{S}}_{\mathbf{y}}$?

Show that$\mathit{\Theta }\mathbf{=}\mathit{A}\mathit{I}\mathit{n}\mathbf{\left[}\mathit{t}\mathit{a}\mathit{n}\mathbf{\right(}\frac{\mathbf{\theta }}{\mathbf{2}}\mathbf{\left)}\mathbf{\right]}$satisfies the $\mathit{\theta }$equation (Equation 4.25), for l = m = 0. This is the unacceptable "second solution" -- whats wrong with it?

Suppose two spin -1/2particles are known to be in the singlet configuration (Equation Let ${\mathbf{S}}_{\mathbf{a}}^{\mathbf{\left(}\mathbf{1}\mathbf{\right)}}$be the component of the spin angular momentum of particle number 1 in the direction defined by the unit vector$\hat{\mathbf{a}}$ Similarly, let${\mathbf{S}}_{\mathbf{b}}^{\mathbf{\left(}\mathbf{2}\mathbf{\right)}}$ be the component of 2’s angular momentum in the direction$\hat{\mathbf{b}}$ Show that

$\mathbf{⟨}{\mathbf{S}}_{\mathbf{a}}^{\mathbf{\left(}\mathbf{1}\mathbf{\right)}}{\mathbf{S}}_{\mathbf{b}}^{\mathbf{\left(}\mathbf{2}\mathbf{\right)}}\mathbf{⟩}\mathbf{=}\mathbf{-}\frac{{\mathbf{\hslash }}^{\mathbf{2}}}{\mathbf{4}}\mathbf{cos\theta }$

where $\mathit{\theta }$ is the angle between $\hat{\mathbf{a}}$ and$\hat{\mathbf{b}}$

(a) Work out the Clebsch-Gordan coefficients for the case ${\mathbf{s}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{/}\mathbf{2}\mathbf{,}{\mathbf{s}}_{\mathbf{2}}$=anything. Hint: You're looking for the coefficients A and Bin

$\mathbf{|}\mathbf{sm}\mathbf{⟩}\mathbf{=}\mathbf{A}\mathbf{|}\frac{\mathbf{1}}{\mathbf{2}}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{⟩}\mathbf{|}{\mathbf{s}}_{\mathbf{2}}\mathbf{}\left(m-\frac{1}{2}\right)\mathbf{⟩}\mathbf{+}\mathbf{B}\mathbf{|}\frac{\mathbf{1}}{\mathbf{2}}\left(-\frac{1}{2}\right)\mathbf{⟩}\mathbf{|}{\mathbf{s}}_{\mathbf{2}}\left(m+\frac{1}{2}\right)\mathbf{⟩}$

such that$\mathbf{|}\mathbf{sm}\mathbf{⟩}$ is an eigenstate of . Use the method of Equations 4.179 through 4.182. If you can't figure out what${\mathit{S}}_{\mathbf{x}}^{\left(2\right)}$ (for instance) does to$\mathbf{|}{\mathbf{s}}_{\mathbf{2}}{\mathbf{m}}_{\mathbf{2}}\mathbf{⟩}$ , refer back to Equation 4.136 and the line before Equation 4.147. Answer:

;role="math" localid="1658209512756" $\mathbf{A}\mathbf{=}\sqrt{\frac{{\mathbf{s}}_{\mathbf{2}}\mathbf{+}{\displaystyle \frac{\mathbf{1}}{\mathbf{2}}}\mathbf{\pm }\mathbf{m}}{\mathbf{2}{\mathbf{s}}_{\mathbf{2}}\mathbf{+}\mathbf{1}}}\mathbf{;}\mathbf{}\mathbf{B}\mathbf{=}\mathbf{\pm }\sqrt{\frac{{\mathbf{s}}_{\mathbf{2}}\mathbf{+}{\displaystyle \frac{1}{2}}\mathbf{\pm }\mathbf{m}}{\mathbf{2}{\mathbf{s}}_{\mathbf{2}}\mathbf{+}\mathbf{1}}}$

where, the signs are determined by$\mathbf{s}\mathbf{=}{\mathbf{s}}_{\mathbf{2}}\mathbf{\pm }\mathbf{1}\mathbf{/}\mathbf{2}$ .

(b) Check this general result against three or four entries in Table 4.8.

Find the matrix representing${\mathbf{S}}_{\mathbf{x}}$for a particle of spin3/2 (using, as

always, the basis of eigenstates of${\mathit{S}}_{\mathbf{z}}$). Solve the characteristic equation to

determine the eigenvalues of${\mathbf{S}}_{\mathbf{x}}$.

Work out the spin matrices for arbitrary spin , generalizing spin (Equations 4.145 and 4.147), spin 1 (Problem 4.31), and spin (Problem 4.52). Answer:

$$\begin{gathered} {\mathbf{S}}_{\mathbf{z}}\mathbf{=}\mathbf{\hslash }\mathbf{\left(}\begin{array}{ccccc}\mathbf{s}& \mathbf{0}& \mathbf{0}& \mathbf{\cdots }& \mathbf{0}\\ \mathbf{0}& \mathbf{s}\mathbf{-}\mathbf{1}& \mathbf{0}& \mathbf{\cdots }& \mathbf{0}\\ \mathbf{0}& \mathbf{0}& \mathbf{s}\mathbf{-}\mathbf{2}& \mathbf{\cdots }& \mathbf{0}\\ \mathbf{⋮}& \mathbf{⋮}& \mathbf{⋮}& \mathbf{\cdots }& \mathbf{⋮}\\ \mathbf{0}& \mathbf{0}& \mathbf{0}& \mathbf{0}& \mathbf{-}\mathbf{s}\end{array}\mathbf{\right)} \\ {\mathbf{S}}_{\mathbf{x}}\mathbf{=}\frac{\mathbf{\hslash }}{\mathbf{2}}\mathbf{\left(}\begin{array}{ccccccc}\mathbf{0}& {\mathbf{b}}_{\mathbf{s}}& \mathbf{0}& \mathbf{0}& \mathbf{\cdots }& \mathbf{0}& \mathbf{0}\\ {\mathbf{b}}_{\mathbf{s}}& \mathbf{0}& {\mathbf{b}}_{\mathbf{s}\mathbf{-}\mathbf{1}}& \mathbf{0}& \mathbf{\cdots }& \mathbf{0}& \mathbf{0}\\ \mathbf{0}& {\mathbf{b}}_{\mathbf{s}\mathbf{-}\mathbf{1}}& \mathbf{0}& {\mathbf{b}}_{\mathbf{s}\mathbf{-}\mathbf{2}}& \mathbf{\cdots }& \mathbf{0}& \mathbf{0}\\ \mathbf{0}& \mathbf{0}& {\mathbf{b}}_{\mathbf{s}\mathbf{-}\mathbf{2}}& \mathbf{0}& \mathbf{\cdots }& \mathbf{0}& \mathbf{0}\\ \mathbf{⋮}& \mathbf{⋮}& \mathbf{⋮}& \mathbf{⋮}& \mathbf{\cdots }& \mathbf{⋮}& \mathbf{⋮}\\ \mathbf{0}& \mathbf{0}& \mathbf{0}& \mathbf{0}& \mathbf{\cdots }& \mathbf{0}& {\mathbf{b}}_{\mathbf{-}\mathbf{s}\mathbf{+}\mathbf{1}}\\ \mathbf{0}& \mathbf{0}& \mathbf{0}& \mathbf{0}& \mathbf{\cdots }& {\mathbf{b}}_{\mathbf{-}\mathbf{s}\mathbf{+}\mathbf{1}}& \mathbf{0}\end{array}\mathbf{\right)} \\ {\mathbf{S}}_{\mathbf{y}}\mathbf{=}\frac{\mathbf{\hslash }}{\mathbf{2}}\mathbf{\left(}\begin{array}{ccccccc}\mathbf{0}& \mathbf{-}{\mathbf{ib}}_{\mathbf{s}}& \mathbf{0}& \mathbf{0}& \mathbf{\cdots }& \mathbf{0}& \mathbf{0}\\ {\mathbf{ib}}_{\mathbf{s}}& \mathbf{0}& \mathbf{-}{\mathbf{ib}}_{\mathbf{s}\mathbf{-}\mathbf{1}}& \mathbf{0}& \mathbf{\cdots }& \mathbf{0}& \mathbf{0}\\ \mathbf{0}& \mathbf{-}{\mathbf{ib}}_{\mathbf{s}\mathbf{-}\mathbf{1}}& \mathbf{0}& \mathbf{-}{\mathbf{ib}}_{\mathbf{s}\mathbf{-}\mathbf{2}}& \mathbf{\cdots }& \mathbf{0}& \mathbf{0}\\ \mathbf{0}& \mathbf{0}& \mathbf{-}{\mathbf{ib}}_{\mathbf{s}\mathbf{-}\mathbf{2}}& \mathbf{0}& \mathbf{\cdots }& \mathbf{0}& \mathbf{0}\\ \mathbf{⋮}& \mathbf{⋮}& \mathbf{⋮}& \mathbf{⋮}& \mathbf{\cdots }& \mathbf{⋮}& \mathbf{⋮}\\ \mathbf{0}& \mathbf{0}& \mathbf{0}& \mathbf{0}& \mathbf{\cdots }& \mathbf{0}& \mathbf{-}{\mathbf{ib}}_{\mathbf{s}\mathbf{+}\mathbf{1}}\\ \mathbf{0}& \mathbf{0}& \mathbf{0}& \mathbf{0}& \mathbf{\cdots }& \mathbf{-}{\mathbf{ib}}_{\mathbf{s}\mathbf{+}\mathbf{1}}& \mathbf{0}\end{array}\mathbf{\right)}\mathbf{} \end{gathered}$$