5.3 Two-State Sys­tems

Two-state sys­tems are sys­tems in which only two quan­tum states are of im­por­tance. That makes such sys­tems the sim­plest non­triv­ial quan­tum sys­tems. A lot of qual­i­ta­tive un­der­stand­ing can be ob­tained from them. Among oth­ers, this sec­tion will shed some light on the rea­son why chem­i­cal bonds tend to in­volve pairs of elec­trons.

As seen in chap­ter 4.6, the pro­tons in the H$_2^+$ hy­dro­gen mol­e­c­u­lar ion are held to­gether by a sin­gle shared elec­tron. How­ever, in the H$_2$ neu­tral hy­dro­gen mol­e­cule of the pre­vi­ous sec­tion, they are held to­gether by a shared pair of elec­trons. In both cases a sta­ble bond was formed. So why are chem­i­cal bonds in­volv­ing a sin­gle elec­tron rel­a­tively rare, while bonds in­volv­ing pairs of shared elec­trons are com­mon?

The uni­fy­ing con­cept re­lat­ing the two bonds is that of two-state sys­tems. Such sys­tems in­volve two in­tu­itive ba­sic states $\psi_1$ and $\psi_2$.

For the hy­dro­gen mol­e­c­u­lar ion, one state, $\psi_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\psi_{\rm {l}}$, de­scribed that the elec­tron was in the ground state around the left pro­ton. A phys­i­cally equiv­a­lent state, $\psi_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\psi_{\rm {r}}$, had the elec­tron in the ground state around the right pro­ton. For the hy­dro­gen mol­e­cule, $\psi_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\psi_{\rm {l}}\psi_{\rm {r}}$ had elec­tron 1 around the left pro­ton and elec­tron 2 around the right one. The other state $\psi_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\psi_{\rm {r}}\psi_{\rm {l}}$ was phys­i­cally the same, but it had the elec­trons re­versed.

There are many other phys­i­cal sit­u­a­tions that may be de­scribed as two state sys­tems. Co­va­lent chem­i­cal bonds in­volv­ing atoms other than hy­dro­gen would be an ob­vi­ous ex­am­ple. Just sub­sti­tute a pos­i­tive ion for one or both pro­tons.

As an­other ex­am­ple of a two-state sys­tem, con­sider the C$_6$H$_6$ “ben­zene mol­e­c­u­lar ring.” This mol­e­cule con­sists of a hexa­gon of 6 car­bon atoms that are held to­gether by 9 co­va­lent bonds. The log­i­cal way that 9 bonds can be arranged be­tween the atoms of a 6 atom ring is to make every sec­ond bond a dou­ble one. How­ever, that still leaves two pos­si­bil­i­ties; the lo­ca­tions of the sin­gle and dou­ble bonds can be swapped. So there are once again two dif­fer­ent but equiv­a­lent states $\psi_1$ and $\psi_2$.

The NH$_3$ “am­mo­nia mol­e­cule” con­sists of an ni­tro­gen atom bonded to three hy­dro­gen atoms. By sym­me­try, the log­i­cal place for the ni­tro­gen atom to sit would surely be in the cen­ter of the tri­an­gle formed by the three hy­dro­gen atoms. But it does not sit there. If it was in the cen­ter of the tri­an­gle, the an­gles be­tween the hy­dro­gen atoms, mea­sured from the ni­tro­gen nu­cleus, should be 120$\POW9,{\circ}$ each. How­ever, as dis­cussed later in chap­ter 5.11.3, va­lence bond the­ory re­quires that the an­gles should be about 90$\POW9,{\circ}$, not 120$\POW9,{\circ}$. (The ac­tual an­gles are about 108$\POW9,{\circ}$ be­cause of rea­sons sim­i­lar to those for wa­ter as dis­cussed in chap­ter 5.11.3.) The key point here is that the ni­tro­gen must sit to the side of the tri­an­gle, and there are two sides, pro­duc­ing once again two dif­fer­ent but equiv­a­lent phys­i­cal states $\psi_1$ and $\psi_2$.

In each case de­scribed above, there are two in­tu­itive phys­i­cal states $\psi_1$ and $\psi_2$. The pe­cu­liar­i­ties of the quan­tum me­chan­ics of two-state sys­tems arise from states that are com­bi­na­tions of these two states, as in

$$\psi=c_1\psi_1+c_2\psi_2$$

Note that ac­cord­ing to the ideas of quan­tum me­chan­ics, the square mag­ni­tude of the first co­ef­fi­cient of the com­bined state, $\vert c_1\vert^2$, rep­re­sents the prob­a­bil­ity of be­ing in state $\psi_1$ and $\vert c_2\vert^2$ the prob­a­bil­ity of be­ing in state $\psi_2$. Of course, the to­tal prob­a­bil­ity of be­ing in one of the states should be one:

$$\vert c_1\vert^2+\vert c_2\vert^2=1$$

(This is only true if the $\psi_1$ and $\psi_2$ states are or­tho­nor­mal. In the hy­dro­gen mol­e­cule cases, or­tho­nor­mal­iz­ing the ba­sic states would change them a bit, but their phys­i­cal na­ture would re­main much the same, es­pe­cially if the pro­tons are not too close.)

The key ques­tion is now what com­bi­na­tion of states has the low­est en­ergy. That will be the ground state $\psi_{\rm {gs}}$ of the two-state sys­tem. The ex­pec­ta­tion value of en­ergy is

$$\left\langle{E}\right\rangle = \langle c_1\psi_1+c_2\psi_2\vert H \vert c_1\psi_1+c_2\psi_2\rangle$$

This can be mul­ti­plied out, tak­ing into ac­count that nu­mer­i­cal fac­tors come out of the left of an in­ner prod­uct as com­plex con­ju­gates. The re­sult is

$$\left\langle{E}\right\rangle = \vert c_1\vert^2 \langle{E... ..._{12} + c_2^*c_1H_{21} + \vert c_2\vert^2 \langle{E}_2\rangle$$

us­ing the short­hand no­ta­tion

$$\langle{E}_1\rangle = \langle\psi_1\vert H\psi_1\rangle, \q... ...\quad \langle{E}_2\rangle = \langle\psi_2\vert H\psi_2\rangle$$

Note that $\langle{E}_1\rangle$ and $\langle{E}_2\rangle$ are real, (2.16). They are the ex­pec­ta­tion en­er­gies of the states $\psi_1$ and $\psi_2$. The states will be or­dered so that $\langle{E}_1\rangle$ is less or equal to $\langle{E}_2\rangle$. (In all the ex­am­ples men­tioned so far, $\langle{E}_1\rangle$ and $\langle{E}_2\rangle$ are equal be­cause the two states are phys­i­cally equiv­a­lent.) Nor­mally, $H_{12}$ and $H_{21}$ are not real but com­plex con­ju­gates, (2.16). How­ever, you can al­ways change the de­f­i­n­i­tion of, say, $\psi_1$ by a com­plex fac­tor of mag­ni­tude one to make $H_{12}$ equal to a real and neg­a­tive num­ber, and then $H_{21}$ will be that same neg­a­tive num­ber.

The above ex­pres­sion for the ex­pec­ta­tion en­ergy con­sists of two kinds of terms, which will be called:

$$\mbox{the averaged energy: } \vert c_1\vert^2 \langle{E}_1\rangle + \vert c_2\vert^2 \langle{E}_2\rangle$$ (5.11)

$$\mbox{the twilight terms: } (c_1^*c_2 + c_2^*c_1) H_{12}%$$ (5.12)

Each of those con­tri­bu­tions will be dis­cussed in turn.

The av­er­aged en­ergy is the en­ergy that you would in­tu­itively ex­pect the com­bined wave func­tion to have. It is a straight­for­ward sum of the ex­pec­ta­tion en­er­gies of the two com­po­nent states $\psi_1$ and $\psi_2$ times the prob­a­bil­i­ties of be­ing in those states. In par­tic­u­lar, in the im­por­tant case that the two states have the same en­ergy, the av­er­aged en­ergy is that en­ergy. What is more log­i­cal than that any mix­ture of two states with the same en­ergy would have that en­ergy too?

But the twi­light terms throw a mon­key wrench in this sim­plis­tic think­ing. It can be seen that they will al­ways make the ground state en­ergy $E_{\rm {gs}}$ lower than the low­est en­ergy of the com­po­nent states $\langle{E}_1\rangle$. (To see that, just take $c_1$ and $c_2$ pos­i­tive real num­bers and $c_2$ small enough that $c_2^2$ can be ne­glected.) This low­er­ing of the en­ergy be­low the low­est com­po­nent state comes out of the math­e­mat­ics of com­bin­ing states; ab­solutely no new phys­i­cal forces are added to pro­duce it. But if you try to de­scribe it in terms of clas­si­cal physics, it re­ally looks like a mys­te­ri­ous new twi­light force is in op­er­a­tion here. It is no new force; it is the weird math­e­mat­ics of quan­tum me­chan­ics.

So, what are these twi­light terms phys­i­cally? If you mean, what are they in terms of clas­si­cal physics, there is sim­ply no an­swer. But if you mean, what are they in terms of nor­mal lan­guage, rather than for­mu­lae, it is easy. Just have an­other look at the de­f­i­n­i­tion of the twi­light terms; they are a mea­sure of the in­ner prod­uct $\langle\psi_1\vert H\psi_2\rangle$. That is the en­ergy you would get if na­ture was in state $\psi_1$ if na­ture was in state $\psi_2$. On quan­tum scales, na­ture can get re­ally, re­ally ethe­real, where it moves be­yond be­ing de­scrib­able by clas­si­cal physics, and the re­sult is very con­crete, but weird, in­ter­ac­tions. For, at these scales twi­light is real, and clas­si­cal physics is not.

For the twi­light terms to be nonzero, there must be a re­gion where the two states over­lap, i.e. there must be a re­gion where both $\psi_1$ and $\psi_2$ are nonzero. In the sim­plest case of the hy­dro­gen mol­e­c­u­lar ion, if the atoms are far apart, the left and right wave func­tions do not over­lap and the twi­light terms will be zero. For the hy­dro­gen mol­e­cule, it gets a bit less in­tu­itive, since the over­lap should re­ally be vi­su­al­ized in the six-di­men­sion­al space of those func­tions. But still, the terms are zero when the atoms are far apart.

The twi­light terms are cus­tom­ar­ily re­ferred to as “ex­change terms,” but every­body seems to have a dif­fer­ent idea of what that is sup­posed to mean. The rea­son may be that these terms pop up all over the place, in all sorts of very dif­fer­ent set­tings. This book prefers to call them twi­light terms, since that most clearly ex­presses what they re­ally are. Na­ture is in a twi­light zone of am­bi­gu­ity.

The low­er­ing of the en­ergy by the twi­light terms pro­duces more sta­ble chem­i­cal bonds than you would ex­pect. Typ­i­cally, the ef­fect of the terms is great­est if the two ba­sic states $\psi_1$ and $\psi_2$ are phys­i­cally equiv­a­lent, like for the men­tioned ex­am­ples. Then the two states have the same ex­pec­ta­tion en­ergy, call it $\langle{E}\rangle_{1,2}$. For such sym­met­ric sys­tems, the ground state will oc­cur for an equal mix­ture of the two states, $c_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $c_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\sqrt{\frac12}$, be­cause then the twi­light terms are most neg­a­tive. (Com­plex co­ef­fi­cients do not re­ally make a phys­i­cal dif­fer­ence, so $c_1$ and $c_2$ can be as­sumed to be real num­bers for con­ve­nience.) In the ground state, the low­est en­ergy is then an amount $\left\vert H_{12}\right\vert$ be­low the en­ergy of the com­po­nent states:

$$\fbox{$\displaystyle \mbox{Symmetric 2-state systems:} \q... ..._{\rm{gs}} = \langle{E}\rangle_{1,2} - \vert H_{12}\vert $} %$$ (5.13)

On the other hand, if the lower en­ergy state $\psi_1$ has sig­nif­i­cantly less en­ergy than state $\psi_2$, then the min­i­mum en­ergy will oc­cur near the lower en­ergy state. That means that $\vert c_1\vert$ $\vphantom0\raisebox{1.1pt}{$\approx$}$ 1 and $\vert c_2\vert$ $\vphantom0\raisebox{1.1pt}{$\approx$}$ 0. (This as­sumes that the twi­light terms are not big enough to dom­i­nate the en­ergy.) In that case $c_1c_2$ $\vphantom0\raisebox{1.1pt}{$\approx$}$ 0 in the twi­light terms (5.12), which pretty much takes the terms out of the pic­ture com­pletely.

This hap­pens for the sin­gle-elec­tron bond of the hy­dro­gen mol­e­c­u­lar ion if the sec­ond pro­ton is re­placed by an­other ion, say a lithium ion. The en­ergy in state $\psi_1$, where the elec­tron is around the pro­ton, will now be sig­nif­i­cantly less than that of state $\psi_2$, where it is around the lithium ion. For such asym­met­ri­cal sin­gle-elec­tron bonds, the twi­light terms are not likely to help much in forg­ing a strong bond. While it turns out that the LiH$\POW9,{+}$ ion is sta­ble, the bind­ing en­ergy is only 0.14 eV or so, com­pared to 2.8 eV for the H$_2^+$ ion. Also, the LiH$\POW9,{+}$ bond seems to be best de­scribed as a Van der Waals at­trac­tion, rather than a true chem­i­cal bond.

In con­trast, for the two-elec­tron bond of the neu­tral hy­dro­gen mol­e­cule, if the sec­ond pro­ton is re­placed by a lithium ion, states $\psi_1$ and $\psi_2$ will still be the same: both states will have one elec­tron around the pro­ton and one around the lithium ion. The two states do have the elec­trons re­versed, but the elec­trons are iden­ti­cal. Thus the twi­light terms are still likely to be ef­fec­tive. In­deed neu­tral LiH lithium hy­dride ex­ists as a sta­ble mol­e­cule with a bind­ing en­ergy of about 2.5 eV at low pres­sures.

(It should be noted that the LiH bond is very ionic, with the shared elec­trons mostly at the hy­dro­gen side, so the ac­tual ground state is quite dif­fer­ent from the co­va­lent hy­dro­gen model. But the model should be bet­ter when the nu­clei are far­ther apart, so the analy­sis can at least jus­tify the ex­is­tence of a sig­nif­i­cant bond.)

For the am­mo­nia mol­e­cule, the two states $\psi_1$ and $\psi_2$ dif­fer only in the side of the hy­dro­gen tri­an­gle that the ni­tro­gen atom is at. Since these two states are phys­i­cally equiv­a­lent, there is again a sig­nif­i­cant low­er­ing of the en­ergy $E_{\rm {gs}}$ for the sym­met­ric com­bi­na­tion $c_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $c_2$. Sim­i­larly, there is a sig­nif­i­cant rais­ing of the en­ergy $E_{\rm {as}}$ for the an­ti­sym­met­ric com­bi­na­tion $c_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-c_2$. Tran­si­tions be­tween these two en­ergy states pro­duce pho­tons of a sin­gle en­ergy in the mi­crowave range. It al­lows a maser (mi­crowave-range laser) to be con­structed. The first maser was in fact an am­mo­nia one. It gave rise to the sub­se­quent de­vel­op­ment of op­ti­cal-range ver­sions. These were ini­tially called op­ti­cal masers, but are now known as lasers. Masers are im­por­tant for pro­vid­ing a sin­gle fre­quency ref­er­ence, like in some atomic clocks. See chap­ter 7.7 for the op­er­at­ing prin­ci­ple of masers and lasers.

The am­mo­nia mol­e­cule may well be the best ex­am­ple of how weird these twi­light ef­fects are. Con­sider, there are two com­mon-sense states in which the ni­tro­gen is at one side of the hy­dro­gen tri­an­gle. What phys­i­cal rea­son could there pos­si­bly be that there is a state of lower en­ergy in which the atom is at both sides at the same time with a 50/50 prob­a­bil­ity? Be­fore you an­swer that, re­call that it only works if you do the 50/50 case right. If you do it wrong, you end up rais­ing the en­ergy. And the only way to fig­ure out whether you do it right is to look at the be­hav­ior of the sign of a phys­i­cally un­ob­serv­able wave func­tion.

It may fi­nally be noted that in the con­text of chem­i­cal bonds, the raised-en­ergy an­ti­sym­met­ric state is of­ten called an “an­ti­bond­ing” state.

Key Points

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In quan­tum me­chan­ics, the en­ergy of dif­fer­ent but phys­i­cally equiv­a­lent states can be low­ered by mix­ing them to­gether.

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This low­er­ing of en­ergy does not come from new phys­i­cal forces, but from the weird math­e­mat­ics of the wave func­tion.

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The ef­fect tends to be much less when the orig­i­nal states are phys­i­cally very dif­fer­ent.

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One im­por­tant place where states are in­deed phys­i­cally the same is in chem­i­cal bonds in­volv­ing pairs of elec­trons. Here the equiv­a­lent states merely have the iden­ti­cal elec­trons in­ter­changed.

5.3 Re­view Ques­tions

1.

The ef­fec­tive­ness of mix­ing states was al­ready shown by the hy­dro­gen mol­e­cule and mol­e­c­u­lar ion ex­am­ples. But the gen­er­al­ized story above re­stricts the ba­sis states to be or­thog­o­nal, and the states used in the hy­dro­gen ex­am­ples were not.

Show that if $\psi_1$ and $\psi_2$ are not or­thog­o­nal states, but are nor­mal­ized and pro­duce a real and pos­i­tive value for $\langle\psi_1\vert\psi_2\rangle$, like in the hy­dro­gen ex­am­ples, then or­thog­o­nal states can be found in the form

$$\bar\psi_1 = \alpha\left(\psi_1 - \varepsilon\psi_2\right) \qquad\bar\psi_2 = \alpha\left(\psi_2 - \varepsilon\psi_1\right).$$

For nor­mal­ized $\psi_1$ and $\psi_2$ the Cauchy-Schwartz in­equal­ity im­plies that $\langle\psi_1\vert\psi_2\rangle$ will be less than one. If the states do not over­lap much, it will be much less than one and $\varepsilon$ will be small.

(If $\psi_1$ and $\psi_2$ do not meet the stated re­quire­ments, you can al­ways re­de­fine them by fac­tors $ae^{{\rm i}{c}}$ and $be^{-{\rm i}{c}}$, with $a$, $b$, and $c$ real, to get states that do.)

So­lu­tion 2state-a

2.

Show that it does not have an ef­fect on the so­lu­tion whether or not the ba­sic states $\psi_1$ and $\psi_2$ are nor­mal­ized, like in the pre­vi­ous ques­tion, be­fore the state of low­est en­ergy is found.

This re­quires no de­tailed analy­sis; just check that the same so­lu­tion can be de­scribed us­ing the nonorthog­o­nal and or­thog­o­nal ba­sis states. It is how­ever an im­por­tant ob­ser­va­tion for var­i­ous nu­mer­i­cal so­lu­tion pro­ce­dures: your set of ba­sis func­tions can be cleaned up and sim­pli­fied with­out af­fect­ing the so­lu­tion you get.

So­lu­tion 2state-b