Chapter 8: Spin and Atomic Physics

Verify that the normalization constant given in Example 8.2is correct for both symmetric and antisymmetric states and is independent of$\mathbf{n}$and ${\mathbf{n}}^{\mathbf{\text{'}}}$?

The general form for symmetric and antisymmetric wave functions is${\mathbf{\psi }}_{\mathbf{n}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{2}}\mathbf{\right)}{\mathbf{\psi }}_{\mathbf{n}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{2}}\mathbf{\right)}\mathbf{\pm }{\mathbf{\psi }}_{\mathbf{n}}\mathbf{\cdot }\mathbf{\left(}{\mathbf{x}}_{\mathbf{1}}\mathbf{\right)}{\mathbf{\psi }}_{\mathbf{n}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{2}}\mathbf{\right)}$ but it is not normalized.

(a) In applying quantum mechanics, we usually deal with quantum states that are "orthonormal." That is, if we integrate over all space the square of any individual-particle function, such as${\mathbf{\psi }}_{\mathbf{n}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{2}}\mathbf{\right)}{\mathbf{\psi }}_{\mathbf{n}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{2}}\mathbf{\right)}\mathbf{\pm }{\mathbf{\psi }}_{\mathbf{n}}\mathbf{\cdot }\mathbf{\left(}{\mathbf{x}}_{\mathbf{1}}\mathbf{\right)}{\mathbf{\psi }}_{\mathbf{n}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{2}}\mathbf{\right)}$, we get 1, but for the product of different individual-particle functions, such as${\mathbf{\psi }}_{\mathbf{n}}^{\mathbf{\phi }}\mathbf{\left(}\mathbf{x}\mathbf{\right)}{\mathbf{\psi }}_{\mathbf{n}}\mathbf{,}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$, we get 0. This happens to be true for all the systems in which we have obtained or tabulated sets of wave functions (e.g., the particle in a box, the harmonic oscillator, and the hydrogen atom). Assuming that this holds, what multiplicative constant would normalize the symmetric and antisymmetric functions?

(b) What value$\mathit{A}$gives the vector$\mathbf{V}\mathbf{=}\mathbf{A}\mathbf{(}\widehat{\mathbf{x}}\mathbf{\pm }\widehat{\mathbf{y}}\mathbf{)}$unit length?

(c) Discuss the relationship between your answers in (a) and (b)?

Two particles in a box have a total energy $\mathbf{5}{\mathbf{\pi }}^{\mathbf{2}}{\mathbf{\hslash }}^{\mathbf{2}}\mathbf{/}\mathbf{2}{\mathbf{mL}}^{\mathbf{2}}$.

(a) Which states are occupied?

(b) Make a sketch of${\mathbf{P}}_{\mathbf{5}}\mathbf{(}{\mathbf{x}}_{\mathbf{1}}\mathbf{,}{\mathbf{x}}_{\mathbf{2}}\mathbf{)}$versus${\mathbf{x}}_{\mathbf{1}}$for points along the line${\mathbf{x}}_{\mathbf{2}}\mathbf{=}{\mathbf{x}}_{\mathbf{1}}$.

(c) Make a similar sketch of${\mathbf{P}}_{\mathbf{\text{A}}}\mathbf{(}{\mathbf{x}}_{\mathbf{1}}\mathbf{,}{\mathbf{x}}_{\mathbf{2}}\mathbf{)}$.

(d) Repeat parts (b) and (c) but for points on the line${\mathbf{x}}_{\mathbf{2}}\mathbf{=}\mathbf{L}\mathbf{-}{\mathbf{x}}_{\mathbf{1}}$

The wave functions for the ground and first excited states of a simple harmonic oscillator are ${\mathbf{Ae}}^{\mathbf{-}{\mathbf{bx}}^{\mathbf{2}}\mathbf{/}\mathbf{2}}$and${\mathbf{Bxe}}^{\mathbf{-}{\mathbf{bx}}^{\mathbf{2}}\mathbf{/}\mathbf{2}}$. Suppose you have two particles occupying these two states.

(a) If distinguishable, an acceptable wave function would berole="math" localid="1659955524302" ${\mathbf{Ae}}^{\mathbf{-}{\mathbf{bx}}_{\mathbf{1}}^{\mathbf{2}}\mathbf{/}\mathbf{2}}{\mathbf{Bx}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{-}{\mathbf{bx}}_{\mathbf{2}}^{\mathbf{2}}\mathbf{/}\mathbf{2}}$. Calculate the probability that both particles would be on the positive side of the origin and divide by the total probability for both being found over all values of${\mathbf{x}}_{\mathbf{1}}$,${\mathbf{x}}_{\mathbf{2}}$. (This kind of normalizing-as-we-go will streamline things.)

(b) Suppose now that the particles are indistinguishable. Using the$\mathbf{\pm }$symbol to reduce your work. calculate the same probability ratio, but assuming that their multiparticle wave function is either symmetric or antisymmetric. Comment on your results.

Summarize the connection between angular momentum quantization and the stem-Gerlach experiment.

Here we consider adding two electrons to two "atoms," represented as finite wells. and investigate when the exclusion principle must be taken into account. In the accompanying figure, diagram (a) shows the four lowest-energy wave functions for a double finite well that represents atoms close together. To yield the lowest energy. the first electron added to this system must have wave function $\mathbf{A}$and is shared equally between the atoms. The second would al so have function $\mathbf{A}$and be equally shared. but it would have to be opposite spin. A third would have function B. Now consider atoms far a part diagram(b) shows, the bumps do not extend much beyond the atoms - they don't overlap-and functions $\mathbf{A}$and $\mathbf{B}$approach equal energy, as do functions $\mathbf{C}$and $\mathbf{D}$. Wave functions$\mathbf{A}$and$\mathbf{B}$in diagram (b) describe essentially identical shapes in the right well. while being opposite in the left well. Because they are of equal energy. sums or differences ofandare now a valid alternative. An electron in a sum or difference would have the same energy as in either alone, so it would be just as "happy" inrole="math" localid="1659956864834" $\mathbf{A}\mathbf{,}\mathbf{B}\mathbf{,}\mathbf{A}\mathbf{+}\mathbf{B}$, or$\mathbf{A}$- B. Argue that in this spread-out situation, electrons can be put in one atom without violating the exclusion principle. no matter what states electrons occupy in the other atom.

What is the minimum possible energy for five (non-interacting) spin $\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}$particles of massmin a one dimensional box of length L ? What if the particles were spin-1? What if the particles were spin $\mathbf{-}\frac{\mathbf{3}}{\mathbf{2}}$?

Slater Determinant: A convenient and compact way of expressing multi-particle states of anti-symmetric character for many fermions is the Slater determinant:

$\left|\begin{array}{ccccc}{\psi }_{n_1}\left(x_1\right)m_{31}& {\psi }_{n_2}\left(x_1\right)m_{32}& {\psi }_{n_3}\left(x_1\right)m_{33}& ···& {\psi }_{n_N}\left(x_1\right)m_{sN}\\ {\psi }_{n_1}\left(x_2\right)m_{11}& {\psi }_{n_2}\left(x_2\right)m_{32}& {\psi }_{n_3}\left(x_2\right)m_{33}& ···& \psi {\psi }_{n_1}\left(x_2\right)m_{sN}\\ {\psi }_{n_3}\left(x_3\right)m_{31}& {\psi }_{n_2}\left(x_3\right)m_{12}& {\psi }_{n_3}\left(x_3\right)m_{33}& & {\psi }_{n_N}\left(x_3\right)m_{sN}\\ ···& ···& ···& ···& ···\\ {\psi }_{n_1}\left(x_N\right)m_{11}& {\psi }_{n_2}\left(x_N\right)m_{32}& {\psi }_{n_3}\left(x_N\right)m_{33}& ···& {\psi }_{n_N}\left(x_N\right)m_{sN}\end{array}\right|$

It is based on the fact that for N fermions there must be Ndifferent individual-particle states, or sets of quantum numbers. The ith state has spatial quantum numbers (which might be ${\mathit{n}}_{\mathbf{i}}\mathbf{,}{\mathit{\ell }}_{\mathbf{i}}$, and ${\mathit{m}}_{\mathbf{f}\mathbf{i}}$) represented simply by${\mathit{n}}_{\mathbf{i}}$ and spin quantum number ${\mathit{m}}_{\mathbf{s}\mathbf{i}}$. Were it occupied by the ith particle, the slate would be${\mathit{\psi }}_{\mathbf{n}\mathbf{i}}\left(x_j\right){\mathit{m}}_{\mathbf{s}\mathbf{i}}$ a column corresponds to a given state and a row to a given particle. For instance, the first column corresponds to individual particle state ${\mathit{\psi }}_{\mathbf{n}}\left(x_j\right){\mathit{m}}_{\mathbf{s}\mathbf{1}}$. Where jprogresses (through the rows) from particle 1 to particle N. The first row corresponds to particle I. which successively occupies all individual-particle states (progressing through the columns). (a) What property of determinants ensures that the multiparticle state is 0 if any two individual particle states are identical? (b) What property of determinants ensures that switching the labels on any two particles switches the sign of the multiparticle state?

The Slater determinant is introduced in Exercise 42. Show that if states $\mathit{n}$and $\mathbf{n}\mathbf{\text{'}}$of the infinite well are occupied and both spins are up, the Slater determinant yields the antisymmetric multiparticle state:${\mathbf{\psi }}_{\mathbf{n}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{1}}\mathbf{\right)}\mathbf{\uparrow }{\mathbf{\psi }}_{{\mathbf{n}}^{\mathbf{\text{'}}}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{2}}\mathbf{\right)}\mathbf{\uparrow }\mathbf{-}{\mathbf{\psi }}_{{\mathbf{n}}^{\mathbf{\text{'}}}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{1}}\mathbf{\right)}\mathbf{\uparrow }{\mathbf{\psi }}_{\mathbf{n}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{2}}\mathbf{\right)}\mathbf{\uparrow }$

The Slater determinant is introduced in Exercise 42. Show that if states $\mathbf{n}$ and $\mathbf{n}\mathbf{\text{'}}$of the infinite well are occupied. with the particle in state $\mathbf{n}$ being spin up and the one in being spin down. then the Slater determinant yields the antisymmetric multiparticle state: ${\mathbf{\psi }}_{\mathbf{n}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{1}}\mathbf{\right)}\mathbf{\uparrow }{\mathbf{\psi }}_{{\mathbf{n}}^{\mathbf{\text{'}}}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{2}}\mathbf{\right)}\mathbf{\downarrow }\mathbf{-}{\mathbf{\psi }}_{{\mathbf{m}}^{\mathbf{2}}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{1}}\mathbf{\right)}{\mathbf{\psi }}_{\mathbf{n}}\mathbf{\left(}{\mathbf{x}}_{\mathbf{2}}\mathbf{\right)}\mathbf{\uparrow }$.