5.5 Multiple-Particle Systems Including Spin

Spin will turn out to have a major effect on how quantum particles behave. Therefore, quantum mechanics as discussed so far must be gener­alized to include spin. Just like there is a proba­bility that a particle is at some position ${\skew0\vec r}$, there is the additional proba­bility that it has spin angular momentum $S_z$ in an arbitrarily chosen $z$-direction and this must be included in the wave function. This section discusses how.

5.5.1 Wave function for a single particle with spin

The first question is how spin should be included in the wave function of a single particle. If spin is ignored, a single particle has a wave function $\Psi({\skew0\vec r};t)$, depending on position ${\skew0\vec r}$ and on time $t$. Now, the spin $S_z$ is just some other scalar variable that describes the particle, in that respect no different from say the $x$-position of the particle. The “every possible combin­ation” idea of allowing every possible combin­ation of states to have its own proba­bility indicates that $S_z$ needs to be added to the list of variables. So the complete wave function $\Psi$ of the particle can be written out fully as:

$$\fbox{$\displaystyle \Psi \equiv \Psi({\skew0\vec r},S_z;t) $}$$ (5.16)

The value of $\vert\Psi({\skew0\vec r},S_z;t)\vert^2{\,\rm d}^3{\skew0\vec r}$ gives the proba­bility of finding the particle within a vicinity ${\rm d}^3{\skew0\vec r}$ of ${\skew0\vec r}$ and with spin angular momentum in the $z$-direction $S_z$.

But note that there is a big difference between the spin “coordinate” and the position coordinates: while the position variables can take on any value, the values of $S_z$ are highly limited. In particular, for the electron, proton, and neutron, $S_z$ can only be $\frac12\hbar$ or $-\frac12\hbar$, nothing else. You do not really have a full $S_z$ “axis”, just two points.

As a result, there are other meaningful ways of writing the wave function. The full wave function $\Psi({\skew0\vec r},S_z;t)$ can be thought of as consisting of two parts $\Psi_+$ and $\Psi_-$ that only depend on position:

$$\fbox{$\displaystyle \Psi_+({\skew0\vec r};t) \equiv \Ps... ...uiv \Psi({\skew0\vec r},-{\textstyle\frac{1}{2}}\hbar;t) $}$$ (5.17)

These two parts can in turn be thought of as being the components of a two-di­mensional vector that only depends on position:

$$\skew{-1}\vec\Psi({\skew0\vec r};t) \equiv \left( \b... ... r};t) \\ \Psi_-({\skew0\vec r};t) \end{array} \right)$$

Remarkably, Dirac found that the wave function for particles like electrons has to be a vector, if it is assumed that the relativistic equations take a guessed simple and beautiful form, like the Schrödinger and all other basic equations of physics are simple and beautiful. Just like relativity reveals that particles should have build-in energy, it also reveals that particles like electrons have build-in angular momentum. A description of the Dirac equation is in chapter 12.12 if you are curious.

The two-di­mensional vector is called a “spinor” to indicate that its components do not change like those of ordinary physical vectors when the coordinate system is rotated. (How they do change is of no importance here, but will eventually be described in derivation {D.69}.) The spinor can also be written in terms of a magnitude times a unit vector:

$$\skew{-1}\vec\Psi({\skew0\vec r};t) = \Psi_m({\skew0\vec... ...r};t) \\ \chi_2({\skew0\vec r};t) \end{array} \right).$$

This book will just use the scalar wave function $\Psi({\skew0\vec r},S_z;t)$; not a vector one. But it is often convenient to write the scalar wave function in a form equivalent to the vector one:

$$\Psi({\skew0\vec r},S_z;t) = \Psi_+({\skew0\vec r};t) {\uparrow}(S_z) + \Psi_-({\skew0\vec r};t) {\downarrow}(S_z).$$ (5.18)

The square magnitude of function $\Psi_+$ gives the proba­bility of finding the particle near a position with spin-up. That of $\Psi_-$ gives the proba­bility of finding it with spin-down. The “spin-up” function ${\uparrow}(S_z)$ and the “spin-down” function ${\downarrow}(S_z)$ are in some sense the equivalent of the unit vectors ${\hat\imath}$ and ${\hat\jmath}$ in normal vector analysis; they have by definition the following values:

$$\fbox{$\displaystyle {\uparrow}({\textstyle\frac{1}{2}}\... ...=0\quad {\downarrow}(-{\textstyle\frac{1}{2}}\hbar)=1. $}$$

The function arguments will usually be left away for concise­ness, so that

$$\fbox{$\displaystyle \Psi = \Psi_+ {\uparrow}+ \Psi_- {\downarrow} $}$$

is the way the wave function of, say, an electron will normally be written out.

Key Points

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Spin must be included as an independent variable in the wave function of a particle with spin.

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Usually, the wave function $\Psi({\skew0\vec r},S_z;t)$ of a single particle with spin $\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ will be written as

$$\Psi = \Psi_+ {\uparrow}+ \Psi_- {\downarrow}$$

where $\Psi_+({\skew0\vec r};t)$ determines the proba­bility of finding the particle near a given location ${\skew0\vec r}$ with spin up, and $\Psi_-({\skew0\vec r};t)$ the one for finding it spin down.

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The functions ${\uparrow}(S_z)$ and ${\downarrow}(S_z)$ have the values

$${\uparrow}({\textstyle\frac{1}{2}}\hbar)=1\quad {\uparro... ...hbar)=0\quad {\downarrow}(-{\textstyle\frac{1}{2}}\hbar)=1$$

and represent the pure spin-up, respectively spin-down states.

5.5.1 Review Questions

1. What is the normal­ization requirement of the wave function of a spin $\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ particle in terms of $\Psi_+$ and $\Psi_-$?

   Solution complexsa-a

5.5.2 Inner products including spin

Inner products are important: they are needed for finding normal­ization factors, expec­tation values, uncertainty, approximate ground states, etcetera. The additional spin coordinates add a new twist, since there is no way to integrate over the few discrete points on the spin “axis”. Instead, you must sum over these points.

As an example, the inner product of two arbitrary electron wave functions $\Psi_1({\skew0\vec r},S_z;t)$ and $\Psi_2({\skew0\vec r},S_z;t)$ is

$$\langle\Psi_1\vert\Psi_2\rangle = \sum_{S_z = \pm\frac... ..._z;t) \Psi_2({\skew0\vec r},S_z;t) {\,\rm d}^3 {\skew0\vec r}$$

or writing out the two-term sum,

$$\langle\Psi_1\vert\Psi_2\rangle = \int_{{\rm all}\ {\s... ...},-{\textstyle\frac{1}{2}}\hbar;t) {\,\rm d}^3 {\skew0\vec r}$$

The individual factors in the integrals are by definition the spin-up components $\Psi_{1+}$ and $\Psi_{2+}$ and the spin down components $\Psi_{1-}$ and $\Psi_{2-}$ of the wave functions, so:

$$\langle\Psi_1\vert\Psi_2\rangle = \int_{{\rm all}\ {\s... ...};t) \Psi_{2-}({\skew0\vec r};t) {\,\rm d}^3 {\skew0\vec r}$$

In other words, the inner product with spin evaluates as

$$\fbox{$\displaystyle \langle \Psi_{1+}{\uparrow}+ \Psi... ...+}\rangle + \langle\Psi_{1-}\vert\Psi_{2-}\rangle $} %$$ (5.19)

It is spin-up components together and spin-down components together.

Another way of looking at this, or maybe remembering it, is to note that the spin states are an ortho­normal pair,

$$\fbox{$\displaystyle \langle{\uparrow}\vert{\uparrow}\ra... ...uad \langle{\downarrow}\vert{\downarrow}\rangle = 1 $} %$$ (5.20)

as can be verified directly from the definitions of those functions as given in the previous subsection. Then you can think of an inner product with spin as multi­plying out as:

$$\begin{eqnarray*} \lefteqn{ \langle\Psi_{1+}{\uparrow}+ \Psi_{1-}{\downarrow... ...ert\Psi_{2+}\rangle + \langle\Psi_{1-}\vert\Psi_{2-}\rangle \end{eqnarray*}$$

Key Points

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In inner products, you must sum over the spin states.

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For spin $\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ particles:

$$\langle \Psi_{1+}{\uparrow}+ \Psi_{1-}{\downarrow} \ve... ...\Psi_{2+}\rangle + \langle\Psi_{1-}\vert\Psi_{2-}\rangle$$

which is spin-up components together plus spin-down components together.

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The spin-up and spin-down states ${\uparrow}$ and ${\downarrow}$ are an ortho­normal pair.

5.5.2 Review Questions

1. Show that the normal­ization requirement for the wave function of a spin $\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ particle in terms of $\Psi_+$ and $\Psi_-$ requires its norm $\sqrt{\langle\Psi\vert\Psi\rangle}$ to be one.

   Solution complexsai-a

2. Show that if $\psi_{\rm {l}}$ and $\psi_{\rm {r}}$ are normalized spatial wave functions, then a combin­ation like $\left(\psi_{\rm {l}}{\uparrow}+\psi_{\rm {r}}{\downarrow}\right)$$\raisebox{.5pt}{$/$}$$\sqrt 2$ is a normalized wave function with spin.

   Solution complexsai-b

5.5.3 Commutators including spin

There is no known “inter­nal physical mechanism” that gives rise to spin like there is for orbital angular momentum. Fortunately, this lack of detailed infor­mation about spin is to a considerable amount made less of an issue by knowledge about its commutators.

In particular, physicists have concluded that spin components satisfy the same commu­tation relations as the components of orbital angular momentum:

$$\fbox{$\displaystyle [{\widehat S}_x,{\widehat S}_y] = {... ...hat S}_z,{\widehat S}_x] = {\rm i}\hbar{\widehat S}_y $} %$$ (5.21)

These equations are called the “fundamental commu­tation relations.” As will be shown in chapter 12, a large amount of infor­mation about spin can be teased from them.

Further, spin operators commute with all functions of the spatial coordinates and with all spatial operators, including position, linear momentum, and orbital angular momentum. The reason why can be understood from the given description of the wave function with spin. First of all, the square spin operator ${\widehat S}^2$ just multi­plies the entire wave function by the constant $\hbar^2s(s+1)$, and everything commutes with a constant. And the operator ${\widehat S}_z$ of spin in an arbitrary $z$-direction commutes with spatial functions and operators in much the same way that an operator like $\partial$$\raisebox{.5pt}{$/$}$$\partial{x}$ commutes with functions depending on $y$ and with $\partial$$\raisebox{.5pt}{$/$}$$\partial{y}$. The $z$-component of spin corre­sponds to an additional “axis” separate from the $x$, $y$, and $z$ ones, and ${\widehat S}_z$ only affects the variation in this additional direction. For example, for a particle with spin one half, ${\widehat S}_z$ multi­plies the spin-up part of the wave function $\Psi_+$ by the constant $\frac12\hbar$ and $\Psi_-$ by $-\frac12\hbar$. Spatial functions and operators commute with these constants for both $\Psi_+$ and $\Psi_-$ hence commute with ${\widehat S}_z$ for the entire wave function. Since the $z$-direction is arbitrary, this commu­tation applies for any spin component.

Key Points

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While a detailed mechanism of spin is missing, commutators with spin can be evaluated.

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The components of spin satisfy the same mutual commu­tation relations as the components of orbital angular momentum.

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Spin commutes with spatial functions and operators.

5.5.3 Review Questions

1. Are not some commutators missing from the fundamental commu­tation relationship? For example, what is the commutator $[{\widehat S}_y,{\widehat S}_x]$?

   Solution complexsac-a

5.5.4 Wave function for multiple particles with spin

The extension of the ideas of the previous subsections towards multiple particles is straight­forward. For two particles, such as the two electrons of the hydrogen molecule, the full wave function follows from the “every possible combin­ation” idea as

$$\fbox{$\displaystyle \Psi = \Psi({\skew0\vec r}_1,S_{z1},{\skew0\vec r}_2,S_{z2};t) $}$$ (5.22)

The value of $\vert\Psi({\skew0\vec r}_1,S_{z1},{\skew0\vec r}_2,S_{z2};t)\vert^2{\,\rm d}^3{\skew0\vec r}_1{\rm d}^3{\skew0\vec r}_2$ gives the proba­bility of simulta­neously finding particle 1 within a vicinity ${\rm d}^3{\skew0\vec r}_1$ of ${\skew0\vec r}_1$ with spin angular momentum in the $z$-direction $S_{z1}$, and particle 2 within a vicinity ${\rm d}^3{\skew0\vec r}_2$ of ${\skew0\vec r}_2$ with spin angular momentum in the $z$-direction $S_{z2}$.

Restricting the attention again to spin $\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ particles like electrons, protons and neutrons, there are now four possible spin states at any given point, with corre­sponding spatial wave functions

$$\fbox{$\displaystyle \begin{array}{c} \displaystyle ... ...\vec r}_2,-{\textstyle\frac{1}{2}}\hbar;t) \end{array} $}$$ (5.23)

For example, $\vert\Psi_{+-}({\skew0\vec r}_1,{\skew0\vec r}_2;t)\vert^2{\,\rm d}^3{\skew0\vec r}_1{\,\rm d}^3{\skew0\vec r}_2$ gives the proba­bility of finding particle 1 within a vicinity ${\rm d}^3{\skew0\vec r}_1$ of ${\skew0\vec r}_1$ with spin up, and particle 2 within a vicinity ${\rm d}^3{\skew0\vec r}_2$ of ${\skew0\vec r}_2$ with spin down.

The wave function can be written using purely spatial functions and purely spin functions as

$$\begin{eqnarray*} \Psi({\skew0\vec r}_1,S_{z1},{\skew0\vec r}_2,S_{z2};t) & ... ...1,{\skew0\vec r}_2;t) {\downarrow}(S_{z1}) {\downarrow}(S_{z2}) \end{eqnarray*}$$

As you might guess from this multi-line display, usually this will be written more concisely as

$$\fbox{$\displaystyle \Psi = \Psi_{++} {\uparrow}{\upar... ...wnarrow}{\uparrow}+ \Psi_{--} {\downarrow}{\downarrow} $}$$

by leaving out the arguments of the spatial and spin functions. The under­standing is that the first of each pair of arrows refers to particle 1 and the second to particle 2.

The inner product now evaluates as

$$\begin{eqnarray*} \lefteqn{\langle\Psi_1\vert\Psi_2\rangle =} \\ &&\!\! \... ...;t) {\,\rm d}^3 {\skew0\vec r}_1 {\,\rm d}^3 {\skew0\vec r}_2 \end{eqnarray*}$$

This can be written in terms of the purely spatial components as

$$\fbox{$\displaystyle \langle\Psi_1\vert\Psi_2\rangle = ... ...{2-+}\rangle + \langle\Psi_{1--}\vert\Psi_{2--}\rangle $}$$ (5.24)

It reflects the fact that the four spin basis states ${\uparrow}{\uparrow}$, ${\uparrow}{\downarrow}$, ${\downarrow}{\uparrow}$, and ${\downarrow}{\downarrow}$ are an ortho­normal quartet.

Key Points

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The wave function of a single particle with spin gener­alizes in a straight­forward way to multiple particles with spin.

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The wave function of two spin $\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ particles can be written in terms of spatial components multi­plying pure spin states as

$$\Psi = \Psi_{++} {\uparrow}{\uparrow}+ \Psi_{+-} {\upa... ...{\downarrow}{\uparrow}+ \Psi_{--} {\downarrow}{\downarrow}$$

where the first arrow of each pair refers to particle 1 and the second to particle 2.

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In terms of spatial components, the inner product $\langle\Psi_1\vert\Psi_2\rangle$ evaluates as inner products of matching spin components:

$$\langle\Psi_{1++}\vert\Psi_{2++}\rangle + \langle\Psi_{1... ...Psi_{2-+}\rangle + \langle\Psi_{1--}\vert\Psi_{2--}\rangle$$

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The four spin basis states ${\uparrow}{\uparrow}$, ${\uparrow}{\downarrow}$, ${\downarrow}{\uparrow}$, and ${\downarrow}{\downarrow}$ are an ortho­normal quartet.

5.5.4 Review Questions

1. As an example of the ortho­normality of the two-particle spin states, verify that $\langle{\uparrow}{\uparrow}\vert{\downarrow}{\uparrow}\rangle$ is zero, so that ${\uparrow}{\uparrow}$ and ${\downarrow}{\uparrow}$ are indeed orthogonal. Do so by explicitly writing out the sums over $S_{z1}$ and $S_{z2}$.

   Solution complexsb-a

2. A more concise way of under­standing the ortho­normality of the two-particle spin states is to note that an inner product like $\langle{\uparrow}{\uparrow}\vert{\downarrow}{\uparrow}\rangle$ equals $\langle{\uparrow}\vert{\downarrow}\rangle\langle{\uparrow}\vert{\uparrow}\rangle$, where the first inner product refers to the spin states of particle 1 and the second to those of particle 2. The first inner product is zero because of the orthogonality of ${\uparrow}$ and ${\downarrow}$, making $\langle{\uparrow}{\uparrow}\vert{\downarrow}{\uparrow}\rangle$ zero too.

   To check this argument, write out the sums over $S_{z1}$ and $S_{z2}$ for $\langle{\uparrow}\vert{\downarrow}\rangle\langle{\uparrow}\vert{\uparrow}\rangle$ and verify that it is indeed the same as the written out sum for $\langle{\uparrow}{\uparrow}\vert{\downarrow}{\uparrow}\rangle$ given in the answer for the previous question.

   The underlying mathematical principle is that sums of products can be factored into separate sums as in:
   

   $$\sum_{{\rm all}\ S_{z1}} \sum_{{\rm all}\ S_{z2}} f(S_{z1}) ... ...S_{z1})\right] \left[\sum_{{\rm all}\ S_{z2}} g(S_{z2})\right]$$
   
   
   This is similar to the observ­ation in calculus that integrals of products can be factored into separate integrals:
   $$\begin{eqnarray*}\lefteqn{\int_{{\rm all}\ {\skew0\vec r}_1} \int_{{\rm all}\ {\... ...ec r}_2} g({\skew0\vec r}_2) {\,\rm d}^3 {\skew0\vec r}_2\right] \end{eqnarray*}$$
   
   

   Solution complexsb-b

5.5.5 Example: the hydrogen molecule

As an example, this section considers the ground state of the hydrogen molecule. It was found in section 5.2 that the ground state electron wave function must be of the approximate form

$$\psi_{{\rm gs},0} = a \left[ \psi_{\rm {l}}({\skew0\... ...}({\skew0\vec r}_1)\psi_{\rm {l}}({\skew0\vec r}_2) \right]$$

where $\psi_{\rm {l}}$ was the electron ground state of the left hydrogen atom, and $\psi_{\rm {r}}$ the one of the right one; $a$ was just a normal­ization constant. This solution excluded all consid­eration of spin.

Including spin, the ground state wave function must be of the general form

$$\psi_{{\rm gs}} = \psi_{++} {\uparrow}{\uparrow}+ \psi_{... ...} {\downarrow}{\uparrow}+ \psi_{--} {\downarrow}{\downarrow}.$$

As you might guess, in the ground state, each of the four spatial functions $\psi_{++}$, $\psi_{+-}$, $\psi_{-+}$, and $\psi_{--}$ must be propor­tional to the no-spin solution $\psi_{{\rm {gs}},0}$ above. Anything else would have more than the lowest possible energy, {D.24}.

So the approximate ground state including spin must take the form

$$\psi_{{\rm gs}} = a \left[ \psi_{\rm {l}}({\skew0\ve... ...narrow}{\uparrow}+ a_{--}{\downarrow}{\downarrow} \right]$$ (5.25)

where $a_{++}$, $a_{+-}$, $a_{-+}$, and $a_{--}$ are constants.

Key Points

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The electron wave function $\psi_{{\rm {gs}},0}$ for the hydrogen molecule derived previously ignored spin.

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In the full electron wave function, each spatial component must separately be propor­tional to $a(\psi_{\rm {l}}\psi_{\rm {r}}+\psi_{\rm {r}}\psi_{\rm {l}})$.

5.5.5 Review Questions

1. Show that the normal­ization requirement for $\psi_{\rm {gs}}$ means that
   
   $$\vert a_{++}\vert^2 + \vert a_{+-}\vert^2 + \vert a_{-+}\vert^2 + \vert a_{--}\vert^2 = 1$$
   
   

   Solution complexsc-a

5.5.6 Triplet and singlet states

In the case of two particles with spin $\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$, it is often more convenient to use slightly different basis states to describe the spin states than the four arrow combin­ations ${\uparrow}{\uparrow}$, ${\uparrow}{\downarrow}$, ${\downarrow}{\uparrow}$, and ${\downarrow}{\downarrow}$. The more convenient basis states can be written in $\big\vert s\:m\big\rangle$ ket notation, and they are:

$$\fbox{$\displaystyle \underbrace{ \big\vert 1\:1\big\r... ...rrow}{\uparrow}\right) }_{\mbox{the singlet state}} $} %$$ (5.26)

A state $\big\vert s\:m\big\rangle$ has net spin $s$, giving a net square angular momentum $s(s+1)\hbar^2$, and has net angular momentum in the $z$-direction $m\hbar$. For example, if the two particles are in the state $\big\vert 1\:1\big\rangle$, the net square angular momentum is $2\hbar^2$, and their net angular momentum in the $z$-direction is $\hbar$.

The ${\uparrow}{\downarrow}$ and ${\downarrow}{\uparrow}$ states can be written as

$${\uparrow}{\downarrow}= \frac{1}{\sqrt2}\left(\big\vert 1\... ...t(\big\vert 1\:0\big\rangle - \big\vert\:0\big\rangle \right)$$

This shows that while they have zero angular momentum in the $z$-direction; they do not have a value for the net spin: they have a 50/50 proba­bility of net spin 1 and net spin 0. A consequence is that ${\uparrow}{\downarrow}$ and ${\downarrow}{\uparrow}$ cannot be written in $\big\vert s\:m\big\rangle$ ket notation; there is no value for $s$. (Related to that, these states also do not have a definite value for the dot product of the two spins, {A.10}.

Incidentally, note that $z$ components of angular momentum simply add up, as the Newtonian analogy suggests. For example, for ${\uparrow}{\downarrow}$, the $\frac12\hbar$ spin angular momentum of the first electron adds to the $-\frac12\hbar$ of the second electron to produce zero. But Newtonian analysis does not allow square angular momenta to be added together, and neither does quantum mechanics. In fact, it is quite a messy exercise to actually prove that the triplet and singlet states have the net spin values claimed above. (See chapter 12 if you want to see how it is done.)

The spin states ${\uparrow}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\big\vert\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em /\... ...\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\big\rangle$ and ${\downarrow}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\big\vert\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em /\... ...\kern-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\big\rangle$ that apply for a single spin-${\textstyle\frac{1}{2}}$ particle are often referred to as the “doublet” states, since there are two of them.

Key Points

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The set of spin states ${\uparrow}{\uparrow}$, ${\uparrow}{\downarrow}$, ${\downarrow}{\uparrow}$, and ${\downarrow}{\downarrow}$ are often better replaced by the triplet and singlet states $\big\vert 1\:1\big\rangle$, $\big\vert 1\:0\big\rangle$, $\big\vert 1\:\rule[2.5pt]{5pt}{.5pt}1\big\rangle$, and $\big\vert\:0\big\rangle$.

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The triplet and singlet states have definite values for the net square spin.

5.5.6 Review Questions

1. Like the states ${\uparrow}{\uparrow}$, ${\uparrow}{\downarrow}$, ${\downarrow}{\uparrow}$, and ${\downarrow}{\downarrow}$; the triplet and singlet states are an ortho­normal quartet. For example, check that the inner product of $\big\vert 1\:0\big\rangle$ and $\big\vert\:0\big\rangle$ is zero.

   Solution complexse-a