4.1 The Har­monic Os­cil­la­tor

This sec­tion pro­vides an in-depth dis­cus­sion of a ba­sic quan­tum sys­tem. The case to be an­a­lyzed is a par­ti­cle that is con­strained by some kind of forces to re­main at ap­prox­i­mately the same po­si­tion. This can de­scribe sys­tems such as an atom in a solid or in a mol­e­cule. If the forces push­ing the par­ti­cle back to its nom­i­nal po­si­tion are pro­por­tional to the dis­tance that the par­ti­cle moves away from it, you have what is called an har­monic os­cil­la­tor. Even if the forces vary non­lin­early with po­si­tion, they can of­ten still be ap­prox­i­mated to vary lin­early as long as the dis­tances from the nom­i­nal po­si­tion re­main small.

The par­ti­cle’s dis­place­ment from the nom­i­nal po­si­tion will be in­di­cated by $(x,y,z)$. The forces keep­ing the par­ti­cle con­strained can be mod­eled as springs, as sketched in fig­ure 4.1.

Fig­ure 4.1: Clas­si­cal pic­ture of an har­monic os­cil­la­tor.

The stiff­ness of the springs is char­ac­ter­ized by the so called spring con­stant $c$, giv­ing the ra­tio be­tween force and dis­place­ment. Note that it will be as­sumed that the three spring stiff­nesses are equal.

For a quan­tum pic­ture of a har­monic os­cil­la­tor, imag­ine a light atom like a car­bon atom sur­rounded by much heav­ier atoms. When the car­bon atom tries to move away from its nom­i­nal po­si­tion, the heavy atoms push it back. The har­monic os­cil­la­tor is also the ba­sic rel­a­tivis­tic model for the quan­tum elec­tro­mag­netic field.

Ac­cord­ing to clas­si­cal New­ton­ian physics, the par­ti­cle vi­brates back and forth around its nom­i­nal po­si­tion with a fre­quency

$$\omega=\sqrt{\frac{c}{m}}$$ (4.1)

in ra­di­ans per sec­ond. In quan­tum me­chan­ics, a par­ti­cle does not have a pre­cise po­si­tion. But the nat­ural fre­quency above re­mains a con­ve­nient com­pu­ta­tional quan­tity in the quan­tum so­lu­tion.

Key Points

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The sys­tem to be de­scribed is that of a par­ti­cle held in place by forces that in­crease pro­por­tional to the dis­tance that the par­ti­cle moves away from its equi­lib­rium po­si­tion.

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The re­la­tion be­tween dis­tance and force is as­sumed to be the same in all three co­or­di­nate di­rec­tions.

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Num­ber $c$ is a mea­sure of the strength of the forces and $\omega$ is the fre­quency of vi­bra­tion ac­cord­ing to clas­si­cal physics.

4.1.1 The Hamil­ton­ian

In or­der to find the en­ergy lev­els that the os­cil­lat­ing par­ti­cle can have, you must first write down the to­tal en­ergy Hamil­ton­ian.

As far as the po­ten­tial en­ergy is con­cerned, the spring in the $x$-​di­rec­tion holds an amount of po­ten­tial en­ergy equal to $\frac12cx^2$, and sim­i­larly the ones in the $y$ and $z$ di­rec­tions.

To this to­tal po­ten­tial en­ergy, you need to add the ki­netic en­ergy op­er­a­tor ${\widehat T}$ from sec­tion 3.3 to get the Hamil­ton­ian:

$$H = - \frac{\hbar^2}{2m} \left( \frac{\partial^2}{\parti... ...\right) + {\textstyle\frac{1}{2}}c \left(x^2+y^2+z^2\right) %$$ (4.2)

Key Points

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The Hamil­ton­ian (4.2) has been found.

4.1.2 So­lu­tion us­ing sep­a­ra­tion of vari­ables

This sec­tion finds the en­ergy eigen­func­tions and eigen­val­ues of the har­monic os­cil­la­tor us­ing the Hamil­ton­ian as found in the pre­vi­ous sub­sec­tion. Every en­ergy eigen­func­tion $\psi$ and its eigen­value $E$ must sat­isfy the Hamil­ton­ian eigen­value prob­lem, (or time-in­de­pen­dent Schrö­din­ger equa­tion):

$$\left[- \frac{\hbar^2}{2m} \left( \frac{\partial^2}{\part... ...e\frac{1}{2}}c \left(x^2+y^2+z^2\right) \right] \psi = E \psi$$ (4.3)

The bound­ary con­di­tion is that $\psi$ be­comes zero at large dis­tance from the nom­i­nal po­si­tion. Af­ter all, the mag­ni­tude of $\psi$ tells you the rel­a­tive prob­a­bil­ity of find­ing the par­ti­cle at that po­si­tion, and be­cause of the rapidly in­creas­ing po­ten­tial en­ergy, the chances of find­ing the par­ti­cle very far from the nom­i­nal po­si­tion should be van­ish­ingly small.

Like for the par­ti­cle in the pipe of the pre­vi­ous sec­tion, it will be as­sumed that each eigen­func­tion is a prod­uct of one-di­men­sion­al eigen­func­tions, one in each di­rec­tion:

$$\psi = \psi_x(x) \psi_y(y) \psi_z(z)$$ (4.4)

Find­ing the eigen­func­tions and eigen­val­ues by mak­ing such an as­sump­tion is known in math­e­mat­ics as the “method of sep­a­ra­tion of vari­ables”.

Sub­sti­tut­ing the as­sump­tion in the eigen­value prob­lem above, and di­vid­ing every­thing by $\psi_x(x)\psi_y(y)\psi_z(z)$ re­veals that E con­sists of three parts that will be called $E_x$, $E_y$, and $E_z$:

$$\begin{array}{l} \displaystyle \rule[-1ex]{0pt}{2ex} E = ... ..._z''(z)}{\psi_z(z)} + {\textstyle\frac{1}{2}}c z^2 \end{array}$$ (4.5)

where the primes in­di­cate de­riv­a­tives. The three parts rep­re­sent the $x$, $y$, and $z$ de­pen­dent terms.

By the de­f­i­n­i­tion above, the quan­tity $E_x$ can only de­pend on $x$; vari­ables $y$ and $z$ do not ap­pear in its de­f­i­n­i­tion. But ac­tu­ally, $E_x$ can­not de­pend on $x$ ei­ther, since $E_x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E-E_y-E_z$, and none of those quan­ti­ties de­pends on $x$. The in­escapable con­clu­sion is that $E_x$ must be a con­stant, in­de­pen­dent of all three vari­ables $(x,y,z)$. The same way $E_y$ and $E_z$ must be con­stants.

If now in the de­f­i­n­i­tion of $E_x$ above, both sides are mul­ti­plied by $\psi_x(x)$, a one-di­men­sion­al eigen­value prob­lem re­sults:

$$\left[ - \frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + {\textstyle\frac{1}{2}} c x^2 \right] \psi_x = E_x \psi_x$$ (4.6)

The op­er­a­tor within the square brack­ets here, call it $H_x$, in­volves only the $x$-​re­lated terms in the full Hamil­ton­ian. Sim­i­lar prob­lems can be writ­ten down for $E_y$ and $E_z$. Sep­a­rate prob­lems in each of the three vari­ables $x$, $y$, and $z$ have been ob­tained, ex­plain­ing why this math­e­mat­i­cal method is called sep­a­ra­tion of vari­ables.

Solv­ing the one-di­men­sion­al prob­lem for $\psi_x$ can be done by fairly el­e­men­tary but elab­o­rate means. If you are in­ter­ested, you can find how it is done in de­riva­tion {D.12}, but that is math­e­mat­ics and it will not teach you much about quan­tum me­chan­ics. It turns out that, like for the par­ti­cle in the pipe of the pre­vi­ous sec­tion, there is again an in­fi­nite num­ber of dif­fer­ent so­lu­tions for $E_x$ and $\psi_x$:

$$\begin{array}{ll} \strut^{\strut} E_{x0} = \frac12 \hbar \... ...ega & \psi_{x2}(x)=h_2(x) \\ \;\vdots & \;\vdots \end{array}$$ (4.7)

Un­like for the par­ti­cle in the pipe, here by con­ven­tion the so­lu­tions are num­bered start­ing from 0, rather than from 1. So the first eigen­value is $E_{x0}$ and the first eigen­func­tion $\psi_{x0}$. That is just how peo­ple choose to do it.

Also, the eigen­func­tions are not sines like for the par­ti­cle in the pipe; in­stead, as ta­ble 4.1 shows, they take the form of some poly­no­mial times an ex­po­nen­tial. But you will prob­a­bly re­ally not care much about what kind of func­tions they are any­way un­less you end up writ­ing a text­book on quan­tum me­chan­ics and have to plot them. In that case, you can find a gen­eral ex­pres­sion, (D.4), in de­riva­tion {D.12}.

Ta­ble 4.1: First few one-di­men­sional eigen­func­tions of the har­monic os­cil­la­tor.

But the eigen­val­ues are what you want to re­mem­ber from this so­lu­tion. Ac­cord­ing to the or­tho­dox in­ter­pre­ta­tion, these are the mea­sur­able val­ues of the to­tal en­ergy in the $x$-​di­rec­tion (po­ten­tial en­ergy in the $x$-​di­rec­tion spring plus ki­netic en­ergy of the mo­tion in the $x$-​di­rec­tion.) In­stead of writ­ing them all out as was done above, they can be de­scribed us­ing the generic ex­pres­sion:

$$E_{xn_x} = \frac{2n_x+1}2 \hbar \omega \quad \mbox{for } n_x = 0, 1, 2, 3, \ldots$$ (4.8)

The eigen­value prob­lem has now been solved, be­cause the equa­tions for $Y$ and $Z$ are math­e­mat­i­cally the same and must there­fore have cor­re­spond­ing so­lu­tions:

$$E_{yn_y} = \frac{2n_y+1}2 \hbar \omega \quad \mbox{for } n_y = 0, 1, 2, 3, \ldots$$ (4.9)

$$E_{zn_z} = \frac{2n_z+1}2 \hbar \omega \quad \mbox{for } n_z = 0, 1, 2, 3, \ldots$$ (4.10)

The to­tal en­ergy $E$ of the com­plete sys­tem is the sum of $E_x$, $E_y$, and $E_z$. Any non­neg­a­tive choice for num­ber $n_x$, com­bined with any non­neg­a­tive choice for num­ber $n_y$, and for $n_z$, pro­duces one com­bined to­tal en­ergy value $E_{xn_x}+E_{yn_y}+E_{zn_z}$, which will be in­di­cated by $E_{n_xn_yn_z}$. Putting in the ex­pres­sions for the three par­tial en­er­gies above, these to­tal en­ergy eigen­val­ues be­come:

$$E_{n_xn_yn_z} = \frac{2n_x+2n_y+2n_z+3}2\; \hbar \omega %$$ (4.11)

where the quan­tum num­bers $n_x$, $n_y$, and $n_z$ may each have any value in the range 0, 1, 2, 3, ...

The cor­re­spond­ing eigen­func­tion of the com­plete sys­tem is:

$$\psi_{n_xn_yn_z}=h_{n_x}(x) h_{n_y}(y) h_{n_z}(z) %$$ (4.12)

where the func­tions $h_0$, $h_1$, ...are in ta­ble 4.1 or in (D.4) if you need them.

Note that the $n_x,n_y,n_z$ num­ber­ing sys­tem for the so­lu­tions arose nat­u­rally from the so­lu­tion process; it was not im­posed a pri­ori.

Key Points

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The eigen­val­ues and eigen­func­tions have been found, skip­ping a lot of te­dious math that you can check when the weather is bad dur­ing spring break.

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Generic ex­pres­sions for the eigen­val­ues are above in (4.11) and for the eigen­func­tions in (4.12).

4.1.2 Re­view Ques­tions

1.

Write out the ground state en­ergy.

So­lu­tion harmb-a

2.

Write out the ground state wave func­tion fully.

So­lu­tion harmb-b

3.

Write out the en­ergy $E_{100}$.

So­lu­tion harmb-c

4.

Write out the eigen­state $\psi_{100}$ fully.

So­lu­tion harmb-d

4.1.3 Dis­cus­sion of the eigen­val­ues

As the pre­vi­ous sub­sec­tion showed, for every set of three non­neg­a­tive whole num­bers $n_x,n_y,n_z$, there is one unique en­ergy eigen­func­tion, or eigen­state, (4.12) and a cor­re­spond­ing en­ergy eigen­value (4.11). The “quan­tum num­bers” $n_x$, $n_y$, and $n_z$ cor­re­spond to the num­ber­ing sys­tem of the one-di­men­sion­al so­lu­tions that make up the full so­lu­tion.

This sec­tion will ex­am­ine the en­ergy eigen­val­ues. These are of great phys­i­cal im­por­tance, be­cause ac­cord­ing to the or­tho­dox in­ter­pre­ta­tion, they are the only mea­sur­able val­ues of the to­tal en­ergy, the only en­ergy lev­els that the os­cil­la­tor can ever be found at.

The en­ergy lev­els can be plot­ted in the form of a so-called “en­ergy spec­trum”, as in fig­ure 4.2. The en­ergy val­ues are listed along the ver­ti­cal axis, and the sets of quan­tum num­bers $n_x,n_y,n_z$ for which they oc­cur are shown to the right of the plot.

Fig­ure 4.2: The en­ergy spec­trum of the har­monic os­cil­la­tor.

The first point of in­ter­est il­lus­trated by the en­ergy spec­trum is that the en­ergy of the os­cil­lat­ing par­ti­cle can­not take on any ar­bi­trary value, but only cer­tain dis­crete val­ues. Of course, that is just like for the par­ti­cle in the pipe of the pre­vi­ous sec­tion, but for the har­monic os­cil­la­tor, the en­ergy lev­els are evenly spaced. In par­tic­u­lar the en­ergy value is al­ways an odd mul­ti­ple of $\frac12\hbar\omega$. It con­tra­dicts the New­ton­ian no­tion that a har­monic os­cil­la­tor can have any en­ergy level. But since $\hbar$ is so small, about 10$\POW9,{-34}$ kg m$\POW9,{2}$/s, macro­scop­i­cally the dif­fer­ent en­ergy lev­els are ex­tremely close to­gether. Though the old New­ton­ian the­ory is strictly speak­ing in­cor­rect, it re­mains an ex­cel­lent ap­prox­i­ma­tion for macro­scopic os­cil­la­tors.

Also note that the en­ergy lev­els have no largest value; how­ever high the en­ergy of the par­ti­cle in a true har­monic os­cil­la­tor may be, it will never es­cape. The fur­ther it tries to go, the larger the forces that pull it back. It can’t win.

An­other strik­ing fea­ture of the en­ergy spec­trum is that the low­est pos­si­ble en­ergy is again nonzero. The low­est en­ergy oc­curs for $n_x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n_y$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n_z$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 and has a value:

$$E_{000} = {\textstyle\frac{3}{2}} \hbar \omega$$ (4.13)

So, even at ab­solute zero tem­per­a­ture, the par­ti­cle is not com­pletely at rest at its nom­i­nal po­si­tion; it still has $\frac32\hbar\omega$ worth of ki­netic and po­ten­tial en­ergy left that it can never get rid of. This low­est en­ergy state is the ground state.

The rea­son that the en­ergy can­not be zero can be un­der­stood from the un­cer­tainty prin­ci­ple. To get the po­ten­tial en­ergy to be zero, the par­ti­cle would have to be at its nom­i­nal po­si­tion for cer­tain. But the un­cer­tainty prin­ci­ple does not al­low a pre­cise po­si­tion. Also, to get the ki­netic en­ergy to be zero, the lin­ear mo­men­tum would have to be zero for cer­tain, and the un­cer­tainty prin­ci­ple does not al­low that ei­ther.

The ac­tual ground state is a com­pro­mise be­tween un­cer­tain­ties in mo­men­tum and po­si­tion that make the to­tal en­ergy as small as Heisen­berg's re­la­tion­ship al­lows. There is enough un­cer­tainty in mo­men­tum to keep the par­ti­cle near the nom­i­nal po­si­tion, min­i­miz­ing po­ten­tial en­ergy, but there is still enough un­cer­tainty in po­si­tion to keep the mo­men­tum low, min­i­miz­ing ki­netic en­ergy. In fact, the com­pro­mise re­sults in po­ten­tial and ki­netic en­er­gies that are ex­actly equal, {D.13}.

For en­ergy lev­els above the ground state, fig­ure 4.2 shows that there is a rapidly in­creas­ing num­ber of dif­fer­ent sets of quan­tum num­bers $n_x$, $n_y$, and $n_z$ that all pro­duce that en­ergy. Since each set rep­re­sents one eigen­state, it means that mul­ti­ple states pro­duce the same en­ergy.

Key Points

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En­ergy val­ues can be graph­i­cally rep­re­sented as an en­ergy spec­trum.

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The en­ergy val­ues of the har­monic os­cil­la­tor are equally spaced, with a con­stant en­ergy dif­fer­ence of $\hbar\omega$ be­tween suc­ces­sive lev­els.

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The ground state of low­est en­ergy has nonzero ki­netic and po­ten­tial en­ergy.

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For any en­ergy level above the ground state, there is more than one eigen­state that pro­duces that en­ergy.

4.1.3 Re­view Ques­tions

1.

Ver­ify that the sets of quan­tum num­bers shown in the spec­trum fig­ure 4.2 do in­deed pro­duce the in­di­cated en­ergy lev­els.

So­lu­tion harmc-a

2.

Ver­ify that there are no sets of quan­tum num­bers miss­ing in the spec­trum fig­ure 4.2; the listed ones are the only ones that pro­duce those en­ergy lev­els.

So­lu­tion harmc-b

4.1.4 Dis­cus­sion of the eigen­func­tions

This sec­tion takes a look at the en­ergy eigen­func­tions of the har­monic os­cil­la­tor to see what can be said about the po­si­tion of the par­ti­cle at var­i­ous en­ergy lev­els.

At ab­solute zero tem­per­a­ture, the par­ti­cle will be in the ground state of low­est en­ergy. The eigen­func­tion de­scrib­ing this state has the low­est pos­si­ble num­ber­ing $n_x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n_y$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n_z$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, and is ac­cord­ing to (4.12) of sub­sec­tion 4.1.2 equal to

$$\psi_{000}=h_0(x) h_0(y) h_0(z)$$ (4.14)

where func­tion $h_0$ is in ta­ble 4.1. The wave func­tion in the ground state must be equal to the eigen­func­tion to within a con­stant:

$$\Psi_{\mbox{gs}}=c_{000} h_0(x) h_0(y) h_0(z)$$ (4.15)

where the mag­ni­tude of the con­stant $c_{000}$ must be one. Us­ing the ex­pres­sion for func­tion $h_0$ from ta­ble 4.1, the prop­er­ties of the ground state can be ex­plored.

As noted ear­lier in sec­tion 3.1, it is use­ful to plot the square mag­ni­tude of $\Psi$ as grey tones, be­cause the darker re­gions will be the ones where the par­ti­cle is more likely to be found. Such a plot for the ground state is shown in fig­ure 4.3. It shows that in the ground state, the par­ti­cle is most likely to be found near the nom­i­nal po­si­tion, and that the prob­a­bil­ity of find­ing the par­ti­cle falls off quickly to zero be­yond a cer­tain dis­tance from the nom­i­nal po­si­tion.

Fig­ure 4.3: Ground state of the har­monic os­cil­la­tor

The re­gion in which the par­ti­cle is likely to be found ex­tends, roughly speak­ing, about a dis­tance $\ell$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\sqrt{\hbar/m\omega}$ from the nom­i­nal po­si­tion. For a macro­scopic os­cil­la­tor, this will be a very small dis­tance be­cause of the small­ness of $\hbar$. That is some­what com­fort­ing, be­cause macro­scop­i­cally, you would ex­pect an os­cil­la­tor to be able to be at rest at the nom­i­nal po­si­tion. While quan­tum me­chan­ics does not al­low it, at least the dis­tance $\ell$ from the nom­i­nal po­si­tion, and the en­ergy $\frac32\hbar\omega$ are ex­tremely small.

But ob­vi­ously, the bad news is that the ground state prob­a­bil­ity den­sity of fig­ure 4.3 does not at all re­sem­ble the clas­si­cal New­ton­ian pic­ture of a lo­cal­ized par­ti­cle os­cil­lat­ing back and for­wards. In fact, the prob­a­bil­ity den­sity does not even de­pend on time: the chances of find­ing the par­ti­cle in any given lo­ca­tion are the same for all times. The prob­a­bil­ity den­sity is also spher­i­cally sym­met­ric; it only de­pends on the dis­tance from the nom­i­nal po­si­tion, and is the same at all an­gu­lar ori­en­ta­tions. To get some­thing that can start to re­sem­ble a New­ton­ian spring-mass os­cil­la­tor, one re­quire­ment is that the en­ergy is well above the ground level.

Turn­ing now to the sec­ond low­est en­ergy level, this en­ergy level is achieved by three dif­fer­ent en­ergy eigen­func­tions, $\psi_{100}$, $\psi_{010}$, and $\psi_{001}$. The prob­a­bil­ity dis­tri­b­u­tion of each of the three takes the form of two sep­a­rate blobs; fig­ure 4.4 shows $\psi_{100}$ and $\psi_{010}$ when seen along the $z$-​di­rec­tion. In case of $\psi_{001}$, one blob hides the other, so this eigen­func­tion was not shown.

Fig­ure 4.4: Wave func­tions $\psi_{100}$ and $\psi_{010}$.

Ob­vi­ously, these states too do not re­sem­ble a New­ton­ian os­cil­la­tor at all. The prob­a­bil­ity dis­tri­b­u­tions once again stay the same at all times. (This is a con­se­quence of en­ergy con­ser­va­tion, as dis­cussed later in chap­ter 7.1.) Also, while in each case there are two blobs oc­cu­pied by a sin­gle par­ti­cle, the par­ti­cle will never be be caught on the sym­me­try plane in be­tween the blobs, which naively could be taken as a sign of the par­ti­cle mov­ing from one blob to the other.

The eigen­func­tions for still higher en­ergy lev­els show sim­i­lar lack of re­sem­blance to the clas­si­cal mo­tion. As an ar­bi­trary ex­am­ple, fig­ure 4.5 shows eigen­func­tion $\psi_{213}$ when look­ing along the $z$-​axis. To re­sem­ble a clas­si­cal os­cil­la­tor, the par­ti­cle would need to be re­stricted to, maybe not an ex­act mov­ing point, but at most a very small mov­ing re­gion. In­stead, all en­ergy eigen­func­tions have steady prob­a­bil­ity dis­tri­b­u­tions and the lo­ca­tions where the par­ti­cle may be found ex­tend over large re­gions. It turns out that there is an un­cer­tainty prin­ci­ple in­volved here: in or­der to get some lo­cal­iza­tion of the po­si­tion of the par­ti­cle, you need to al­low some un­cer­tainty in its en­ergy. This will have to wait un­til much later, in chap­ter 7.11.4.

Fig­ure 4.5: En­ergy eigen­func­tion $\psi_{213}$.

The ba­sic rea­son that quan­tum me­chan­ics is so slow is sim­ple. To an­a­lyze, say the $x$ mo­tion, clas­si­cal physics says: “the value of the to­tal en­ergy $E_x$ is

$$E_x = {\textstyle\frac{1}{2}} m \dot x^2 + {\textstyle\frac{1}{2}} c x^2,$$

now go an­a­lyze the mo­tion!”. Quan­tum me­chan­ics says: “the to­tal en­ergy op­er­a­tor $H_x$ is

$$H_x = {\textstyle\frac{1}{2}} m \left(\frac{\hbar}{{\rm i}... ...rtial x}\right)^2 + {\textstyle\frac{1}{2}} c {\widehat x}^2,$$

now first fig­ure out the pos­si­ble en­ergy val­ues $E_{x0},E_{x1},\ldots$ be­fore you can even start think­ing about an­a­lyz­ing the mo­tion.”

Key Points

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The ground state wave func­tion is spher­i­cally sym­met­ric: it looks the same seen from any an­gle.

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In en­ergy eigen­states the par­ti­cle po­si­tion is un­cer­tain.

4.1.4 Re­view Ques­tions

1.

Write out the ground state wave func­tion and show that it is in­deed spher­i­cally sym­met­ric.

So­lu­tion harmd-a

2.

Show that the ground state wave func­tion is max­i­mal at the ori­gin and, like all the other en­ergy eigen­func­tions, be­comes zero at large dis­tances from the ori­gin.

So­lu­tion harmd-b

3.

Write down the ex­plicit ex­pres­sion for the eigen­state $\psi_{213}$ us­ing ta­ble 4.1, then ver­ify that it looks like fig­ure 4.5 when look­ing along the $z$-​axis, with the $x$-​axis hor­i­zon­tal and the $y$-​axis ver­ti­cal.

So­lu­tion harmd-c

4.1.5 De­gen­er­acy

As the en­ergy spec­trum fig­ure 4.2 il­lus­trated, the only en­ergy level for which there is only a sin­gle en­ergy eigen­func­tion is the ground state. All higher en­ergy lev­els are what is called de­gen­er­ate; there is more than one eigen­func­tion that pro­duces that en­ergy. (In other words, more than one set of three quan­tum num­bers $n_x$, $n_y$, and $n_z$.)

It turns out that de­gen­er­acy al­ways re­sults in nonunique­ness of the eigen­func­tions. That is im­por­tant for a va­ri­ety of rea­sons. For ex­am­ple, in the quan­tum me­chan­ics of mol­e­cules, chem­i­cal bonds of­ten se­lect among nonunique the­o­ret­i­cal so­lu­tions those that best fit the given con­di­tions. Also, to find spe­cific math­e­mat­i­cal or nu­mer­i­cal so­lu­tions for the eigen­func­tions of a quan­tum sys­tem, the nonunique­nesses will some­how have to be re­solved.

Nonunique­ness also poses prob­lems for ad­vanced analy­sis. For ex­am­ple, sup­pose you try to an­a­lyze the ef­fect of var­i­ous small per­tur­ba­tions that a har­monic os­cil­la­tor might ex­pe­ri­ence in real life. An­a­lyz­ing the ef­fect of small per­tur­ba­tions is typ­i­cally a rel­a­tively easy math­e­mat­i­cal prob­lem: the per­tur­ba­tion will slightly change an eigen­func­tion, but it can still be ap­prox­i­mated by the un­per­turbed one. So, if you know the un­per­turbed eigen­func­tion you are in busi­ness; un­for­tu­nately, if the un­per­turbed eigen­func­tion is not unique, you may not know which is the right one to use in the analy­sis.

The nonunique­ness arises from the fact that:

Lin­ear com­bi­na­tions of eigen­func­tions at the same en­ergy level pro­duce al­ter­na­tive eigen­func­tions that still have that same en­ergy level.

For ex­am­ple, the eigen­func­tions $\psi_{100}$, and $\psi_{010}$ of the har­monic os­cil­la­tor have the same en­ergy $E_{100}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E_{010}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac52\hbar\omega$ (as does $\psi_{001}$, but this ex­am­ple will be re­stricted to two eigen­func­tions.) Any lin­ear com­bi­na­tion of the two has that en­ergy too, so you could re­place eigen­func­tions $\psi_{100}$ and $\psi_{010}$ by two al­ter­na­tive ones such as:

$$\frac{\psi_{100}+\psi_{010}}{\sqrt2}\quad \mbox {and}\quad \frac{\psi_{010}-\psi_{100}}{\sqrt2}$$

It is read­ily ver­i­fied these lin­ear com­bi­na­tions are in­deed still eigen­func­tions with eigen­value $E_{100}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E_{010}$: ap­ply­ing the Hamil­ton­ian $H$ to ei­ther one will mul­ti­ply each term by $E_{100}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E_{010}$, hence the en­tire com­bi­na­tion by that amount. How do these al­ter­na­tive eigen­func­tions look? Ex­actly like $\psi_{100}$ and $\psi_{010}$ in fig­ure 4.4, ex­cept that they are ro­tated over 45 de­grees. Clearly then, they are just as good as the orig­i­nals, just seen un­der a dif­fer­ent an­gle.

Which raises the ques­tion, how come the analy­sis ended up with the ones that it did in the first place? The an­swer is in the method of sep­a­ra­tion of vari­ables that was used in sub­sec­tion 4.1.2. It pro­duced eigen­func­tions of the form $h_{n_x}(x)h_{n_y}(y)h_{n_z}(z)$ that were not just eigen­func­tions of the full Hamil­ton­ian $H$, but also of the par­tial Hamil­to­ni­ans $H_x$, $H_y$, and $H_z$, be­ing the $x$, $y$, and $z$ parts of it.

For ex­am­ple, $\psi_{100}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $h_1(x)h_0(y)h_0(z)$ is an eigen­func­tion of $H_x$ with eigen­value $E_{x1}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac32\hbar\omega$, of $H_y$ with eigen­value $E_{y0}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12\hbar\omega$, and of $H_z$ with eigen­value $E_{z0}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12\hbar\omega$, as well as of $H$ with eigen­value $E_{100}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac52\hbar\omega$.

The al­ter­na­tive eigen­func­tions are still eigen­func­tions of $H$, but no longer of the par­tial Hamil­to­ni­ans. For ex­am­ple,

$$\frac{\psi_{100}+\psi_{010}}{\sqrt2} = \frac{h_1(x)h_0(y)h_0(z)+h_0(x)h_1(y)h_0(z)}{\sqrt2}$$

is not an eigen­func­tion of $H_x$: tak­ing $H_x$ times this eigen­func­tion would mul­ti­ply the first term by $E_{x1}$ but the sec­ond term by $E_{x0}$.

So, the ob­tained eigen­func­tions were re­ally made de­ter­mi­nate by en­sur­ing that they are si­mul­ta­ne­ously eigen­func­tions of $H$, $H_x$, $H_y$, and $H_z$. The nice thing about them is that they can an­swer ques­tions not just about the to­tal en­ergy of the os­cil­la­tor, but also about how much of that en­ergy is in each of the three di­rec­tions.

Key Points

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De­gen­er­acy oc­curs when dif­fer­ent eigen­func­tions pro­duce the same en­ergy.

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It causes nonunique­ness: al­ter­na­tive eigen­func­tions will ex­ist.

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That can make var­i­ous analy­sis a lot more com­plex.

4.1.5 Re­view Ques­tions

1.

Just to check that this book is not ly­ing, (you can­not be too care­ful), write down the an­a­lyt­i­cal ex­pres­sion for $\psi_{100}$ and $\psi_{010}$ us­ing ta­ble 4.1. Next write down $\left(\psi_{100}+\psi_{010}\right)$$\raisebox{.5pt}{$/$}$​$\sqrt 2$ and $\left(\psi_{010}-\psi_{100}\right)$$\raisebox{.5pt}{$/$}$​$\sqrt 2$. Ver­ify that the lat­ter two are the func­tions $\psi_{100}$ and $\psi_{010}$ in a co­or­di­nate sys­tem $(\bar{x},\bar{y},z)$ that is ro­tated 45 de­grees counter-clock­wise around the $z$-​axis com­pared to the orig­i­nal $(x,y,z)$ co­or­di­nate sys­tem.

So­lu­tion harme-a

4.1.6 Noneigen­states

It should not be thought that the har­monic os­cil­la­tor only ex­ists in en­ergy eigen­states. The op­po­site is more like it. Any­thing that some­what lo­cal­izes the par­ti­cle will pro­duce an un­cer­tainty in en­ergy. This sec­tion ex­plores the pro­ce­dures to deal with states that are not en­ergy eigen­states.

First, even if the wave func­tion is not an en­ergy eigen­func­tion, it can still al­ways be writ­ten as a com­bi­na­tion of the eigen­func­tions:

$$\Psi(x,y,z,t) = \sum_{n_x=0}^\infty\sum_{n_y=0}^\infty\sum_{n_z=0}^\infty c_{n_xn_yn_z} \psi_{n_xn_yn_z}$$ (4.16)

That this is al­ways pos­si­ble is a con­se­quence of the com­plete­ness of the eigen­func­tions of Her­mit­ian op­er­a­tors such as the Hamil­ton­ian. An ar­bi­trary ex­am­ple of such a com­bi­na­tion state is shown in fig­ure 4.6.

Fig­ure 4.6: Ar­bi­trary wave func­tion (not an en­ergy eigen­func­tion).

The co­ef­fi­cients $c_{n_xn_yn_z}$ in the com­bi­na­tion are im­por­tant: ac­cord­ing to the or­tho­dox sta­tis­ti­cal in­ter­pre­ta­tion, their square mag­ni­tude gives the prob­a­bil­ity to find the en­ergy to be the cor­re­spond­ing eigen­value $E_{n_xn_yn_z}$. For ex­am­ple, $\vert c_{000}\vert^2$ gives the prob­a­bil­ity of find­ing that the os­cil­la­tor is in the ground state of low­est en­ergy.

If the wave func­tion $\Psi$ is in a known state, (maybe be­cause the po­si­tion of the par­ti­cle was fairly ac­cu­rately mea­sured), then each co­ef­fi­cient $c_{n_xn_yn_z}$ can be found by com­put­ing an in­ner prod­uct:

$$c_{n_xn_yn_z} = \langle \psi_{n_xn_yn_z} \vert \Psi \rangle$$ (4.17)

The rea­son this works is or­tho­nor­mal­ity of the eigen­func­tions. As an ex­am­ple, con­sider the case of co­ef­fi­cient $c_{100}$:

$$c_{100} = \langle \psi_{100} \vert \Psi \rangle = \langle... ... + c_{001} \psi_{001} + c_{200} \psi_{200} + \ldots \rangle$$

Now proper eigen­func­tions of Her­mit­ian op­er­a­tors are or­tho­nor­mal; the in­ner prod­uct be­tween dif­fer­ent eigen­func­tions is zero, and be­tween iden­ti­cal eigen­func­tions is one:

$$\langle \psi_{100}\vert\psi_{000}\rangle = 0 \quad \langle... ...ad \langle \psi_{100}\vert\psi_{001}\rangle = 0 \quad \ldots$$

So, the in­ner prod­uct above must in­deed pro­duce $c_{100}$.

Chap­ter 7.1 will dis­cuss an­other rea­son why the co­ef­fi­cients are im­por­tant: they de­ter­mine the time evo­lu­tion of the wave func­tion. It may be re­called that the Hamil­ton­ian, and hence the eigen­func­tions de­rived from it, did not in­volve time. How­ever, the co­ef­fi­cients do.

Even if the wave func­tion is ini­tially in a state in­volv­ing many eigen­func­tions, such as the one in fig­ure 4.6, the or­tho­dox in­ter­pre­ta­tion says that en­ergy mea­sure­ment will col­lapse it into a sin­gle eigen­func­tion. For ex­am­ple, as­sume that the en­er­gies in all three co­or­di­nate di­rec­tions are mea­sured and that they re­turn the val­ues:

$$E_{x2}={\textstyle\frac{5}{2}}\hbar\omega \quad E_{y1}={\t... ...}}\hbar\omega \quad E_{z3}={\textstyle\frac{7}{2}}\hbar\omega$$

for a to­tal en­ergy $E$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac{15}2\hbar\omega$. Quan­tum me­chan­ics could not ex­actly pre­dict that this was go­ing to hap­pen, but it did pre­dict that the en­er­gies had to be odd mul­ti­ples of $\frac12\hbar\omega$. Also, quan­tum me­chan­ics gave the prob­a­bil­ity of mea­sur­ing the given val­ues to be what­ever $\vert c_{213}\vert^2$ was. Or in other words, what $\vert\langle\psi_{213}\vert\Psi\rangle\vert^2$ was.

Af­ter the ex­am­ple mea­sure­ment, the pre­dic­tions be­come much more spe­cific, be­cause the wave func­tion is now col­lapsed into the mea­sured one:

$$\Psi^{\mbox{\scriptsize new}} = c^{\mbox{\scriptsize new}}_{213} \psi_{213}$$

This eigen­func­tion was shown ear­lier in fig­ure 4.5.

If an­other mea­sure­ment of the en­er­gies is now done, the only val­ues that can come out are $E_{x2}$, $E_{y1}$, and $E_{z3}$, the same as in the first mea­sure­ment. There is now cer­tainty of get­ting those val­ues; the prob­a­bil­ity $\vert c^{\mbox{\scriptsize {new}}}_{213}\vert^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. This will con­tinue to be true for en­ergy mea­sure­ments un­til the sys­tem is dis­turbed, maybe by a po­si­tion mea­sure­ment.

Key Points

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The ba­sic ideas of quan­tum me­chan­ics were il­lus­trated us­ing an ex­am­ple.

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The en­ergy eigen­func­tions are not the only game in town. Their seem­ingly lowly co­ef­fi­cients are im­por­tant too.

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When the wave func­tion is known, the co­ef­fi­cient of any eigen­func­tion can be found by tak­ing an in­ner prod­uct of the wave func­tion with that eigen­func­tion.