Sub­sec­tions


3.5 A Par­ti­cle Con­fined In­side a Pipe

This sec­tion demon­strates the gen­eral pro­ce­dure for an­a­lyz­ing quan­tum sys­tems us­ing a very el­e­men­tary ex­am­ple. The sys­tem to be stud­ied is that of a par­ti­cle, say an elec­tron, con­fined to the in­side of a nar­row pipe with sealed ends. This ex­am­ple will be stud­ied in some de­tail, since if you un­der­stand it thor­oughly, it be­comes much eas­ier not to get lost in the more ad­vanced ex­am­ples of quan­tum me­chan­ics dis­cussed later. And as the fi­nal sub­sec­tion 3.5.9 shows, as well as much of chap­ter 6, the par­ti­cle in a pipe is re­ally quite in­ter­est­ing de­spite its sim­plic­ity.


3.5.1 The phys­i­cal sys­tem

The sys­tem to be an­a­lyzed is shown in fig­ure 3.4 as it would ap­pear in clas­si­cal non­quan­tum physics.

Fig­ure 3.4: Clas­si­cal pic­ture of a par­ti­cle in a closed pipe.
\begin{figure}\centering
{}%
\epsffile{pipecl.eps}
\end{figure}

A par­ti­cle is bounc­ing around be­tween the two ends of a pipe. It is as­sumed that there is no fric­tion, so the par­ti­cle will keep bounc­ing back and for­ward for­ever. (Fric­tion is a macro­scopic ef­fect that has no place in the sort of quan­tum-scale sys­tems an­a­lyzed here.) Typ­i­cally, clas­si­cal physics draws the par­ti­cles that it de­scribes as lit­tle spheres, so that is what fig­ure 3.4 shows.

The ac­tual quan­tum sys­tem to be an­a­lyzed is shown in fig­ure 3.5.

Fig­ure 3.5: Quan­tum me­chan­ics pic­ture of a par­ti­cle in a closed pipe.
\begin{figure}\centering
{}%
\epsffile{pipeq.eps}
\end{figure}

A par­ti­cle like an elec­tron has no (known) spe­cific shape or size, but it does have a wave func­tion blob. So in quan­tum me­chan­ics the equiv­a­lent of a par­ti­cle bounc­ing around is a wave func­tion blob bounc­ing around be­tween the ends of the pipe.

Please do not ask what this im­pen­e­tra­ble pipe is made off. It is ob­vi­ously a crude ide­al­iza­tion. You could imag­ine that the elec­tron is a va­lence elec­tron in a very tiny bar of cop­per. In that case the pipe walls would cor­re­spond to the sur­face of the cop­per bar, and it is as­sumed that the elec­tron can­not get off the bar.

But of course, a cop­per bar would have nu­clei, and other elec­trons, and the analy­sis here does not con­sider those. So maybe it is bet­ter to think of the par­ti­cle as be­ing a lone he­lium atom stuck in­side a car­bon nan­otube.


Key Points
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An ide­al­ized prob­lem of a par­ti­cle bounc­ing about in a pipe will be con­sid­ered.


3.5.2 Math­e­mat­i­cal no­ta­tions

The first step in the so­lu­tion process is to de­scribe the prob­lem math­e­mat­i­cally. To do so, an $x$-​co­or­di­nate that mea­sures lon­gi­tu­di­nal po­si­tion in­side the pipe will be used, as shown in fig­ure 3.6. Also,the length of the pipe will be called $\ell_x$.

Fig­ure 3.6: De­f­i­n­i­tions for one-di­men­sional mo­tion in a pipe.
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...{$x=0$}}
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To make the prob­lem as easy to solve as pos­si­ble, it will be as­sumed that the only po­si­tion co­or­di­nate that ex­ists is the lon­gi­tu­di­nal po­si­tion $x$ along the pipe. For now, the ex­is­tence of any co­or­di­nates $y$ and $z$ that mea­sure the lo­ca­tion in cross sec­tion will be com­pletely ig­nored.


Key Points
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The only po­si­tion co­or­di­nate to be con­sid­ered for now is $x$.

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The no­ta­tions have been de­fined.


3.5.3 The Hamil­ton­ian

To an­a­lyze a quan­tum sys­tem you must find the Hamil­ton­ian. The Hamil­ton­ian is the to­tal en­ergy op­er­a­tor, equal to the sum of ki­netic plus po­ten­tial en­ergy.

The po­ten­tial en­ergy $V$ is the eas­i­est to find: since it is as­sumed that the par­ti­cle does not ex­pe­ri­ence forces in­side the pipe, (un­til it hits the ends of the pipe, that is), the po­ten­tial en­ergy must be con­stant in­side the pipe:

\begin{displaymath}
V = \mbox{constant}
\end{displaymath}

(The force is the de­riv­a­tive of the po­ten­tial en­ergy, so a con­stant po­ten­tial en­ergy pro­duces zero force.) Fur­ther, since the value of the con­stant does not make any dif­fer­ence phys­i­cally, you may as well as­sume that it is zero and save some writ­ing:

\begin{displaymath}
V = 0
\end{displaymath}

Next, the ki­netic en­ergy op­er­a­tor ${\widehat T}$ is needed. You can just look up its pre­cise form in sec­tion 3.3 and find it is:

\begin{displaymath}
{\widehat T}= - \frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2}
\end{displaymath}

Note that only the $x$ term is used here; the ex­is­tence of the other two co­or­di­nates $y$ and $z$ is com­pletely ig­nored. The con­stant $m$ is the mass of the par­ti­cle, and $\hbar$ is Planck's con­stant.

Since the po­ten­tial en­ergy is zero, the Hamil­ton­ian $H$ is just this ki­netic en­ergy:

\begin{displaymath}
H = - \frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} %
\end{displaymath} (3.13)


Key Points
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The one-di­men­sion­al Hamil­ton­ian (3.13) has been writ­ten down.


3.5.4 The Hamil­ton­ian eigen­value prob­lem

With the Hamil­ton­ian $H$ found, the next step is to for­mu­late the Hamil­ton­ian eigen­value prob­lem, (or “time-in­de­pen­dent Schrö­din­ger equa­tion.”). This prob­lem is al­ways of the form

\begin{displaymath}
H \psi = E \psi
\end{displaymath}

Any nonzero so­lu­tion $\psi$ of this equa­tion is called an en­ergy eigen­func­tion and the cor­re­spond­ing con­stant $E$ is called the en­ergy eigen­value.

Sub­sti­tut­ing the Hamil­ton­ian for the pipe as found in the pre­vi­ous sub­sec­tion, the eigen­value prob­lem is:

\begin{displaymath}
- \frac{\hbar^2}{2m} \frac{\partial^2\psi}{\partial x^2}
= E \psi %
\end{displaymath} (3.14)

The prob­lem is not com­plete yet. These prob­lems also need so called bound­ary con­di­tions, con­di­tions that say what hap­pens at the ends of the $x$ range. In this case, the ends of the $x$ range are the ends of the pipe. Now re­call that the square mag­ni­tude of the wave func­tion gives the prob­a­bil­ity of find­ing the par­ti­cle. So the wave func­tion must be zero wher­ever there is no pos­si­bil­ity of find­ing the par­ti­cle. That is out­side the pipe: it is as­sumed that the par­ti­cle is con­fined to the pipe. So the wave func­tion is zero out­side the pipe. And since the out­side of the pipe starts at the ends of the pipe, that means that the wave func­tion must be zero at the ends {N.5}:

\begin{displaymath}
\psi = 0 \mbox{ at $x=0$}
\qquad\mbox{and}\qquad
\psi = 0 \mbox{ at $x=\ell_x$} %
\end{displaymath} (3.15)


Key Points
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The Hamil­ton­ian eigen­value prob­lem (3.14)has been found.

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It also in­cludes the bound­ary con­di­tions (3.15).


3.5.5 All so­lu­tions of the eigen­value prob­lem

The pre­vi­ous sec­tion found the Hamil­ton­ian eigen­value prob­lem to be:

\begin{displaymath}
- \frac{\hbar^2}{2m} \frac{\partial^2\psi}{\partial x^2}
= E \psi
\end{displaymath}

Now you need to solve this equa­tion. Math­e­mati­cians call an equa­tion of this type an or­di­nary dif­fer­en­tial equa­tion; dif­fer­en­tial be­cause it has a de­riv­a­tive in it, and or­di­nary since there are no de­riv­a­tives with re­spect to vari­ables other than $x$.

If you do not know how to solve or­di­nary dif­fer­en­tial equa­tions, it is no big deal. The best way is usu­ally to look them up any­way. The equa­tion above can be found in most math­e­mat­i­cal ta­ble books, e.g. [40, item 19.7]. Ac­cord­ing to what it says there, (with changes in no­ta­tion), if you as­sume that the en­ergy $E$ is neg­a­tive, the so­lu­tion is

\begin{displaymath}
\psi = C_1 e^{\kappa x} + C_2 e^{-\kappa x}
\qquad \kappa = \frac{\sqrt{-2mE}}{\hbar}
\end{displaymath}

This so­lu­tion may eas­ily by checked by sim­ply sub­sti­tut­ing it into the or­di­nary dif­fer­en­tial equa­tion.

As far as the or­di­nary dif­fer­en­tial equa­tion is con­cerned, the con­stants $C_1$ and $C_2$ could be any two num­bers. But you also need to sat­isfy the two bound­ary con­di­tions given in the pre­vi­ous sub­sec­tion. The bound­ary con­di­tion that $\psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 when $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 pro­duces, if $\psi$ is as above,

\begin{displaymath}
C_1 e^{0} + C_2 e^{0} = 0
\end{displaymath}

and since $e^0$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, this can be used to find an ex­pres­sion for $C_2$:

\begin{displaymath}
C_2=-C_1
\end{displaymath}

The sec­ond bound­ary con­di­tion, that $\psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 at $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\ell_x$, pro­duces

\begin{displaymath}
C_1 e^{\kappa \ell_x} + C_2 e^{-\kappa \ell_x} = 0
\end{displaymath}

or, since you just found out that $C_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-C_1$,

\begin{displaymath}
C_1 \left(e^{\kappa \ell_x} - e^{-\kappa \ell_x}\right) = 0
\end{displaymath}

This equa­tion spells trou­ble be­cause the term be­tween paren­the­ses can­not be zero; the ex­po­nen­tials are not equal. In­stead $C_1$ will have to be zero; that is bad news since it im­plies that $C_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-C_1$ is zero too, and then so is the wave func­tion $\psi$:

\begin{displaymath}
\psi = C_1 e^{\kappa x} + C_2 e^{-\kappa x} = 0
\end{displaymath}

A zero wave func­tion is not ac­cept­able, since there would be no pos­si­bil­ity to find the par­ti­cle any­where!

Every­thing was done right. So the prob­lem must be the ini­tial as­sump­tion that the en­ergy is neg­a­tive. Ap­par­ently, the en­ergy can­not be neg­a­tive. This can be un­der­stood from the fact that for this par­ti­cle, the en­ergy is all ki­netic en­ergy. Clas­si­cal physics would say that the ki­netic en­ergy can­not be neg­a­tive be­cause it is pro­por­tional to the square of the ve­loc­ity. You now see that quan­tum me­chan­ics agrees that the ki­netic en­ergy can­not be neg­a­tive, but says it is be­cause of the bound­ary con­di­tions on the wave func­tion.

Try again, but now as­sume that the en­ergy $E$ is zero in­stead of neg­a­tive. In that case the so­lu­tion of the or­di­nary dif­fer­en­tial equa­tion is ac­cord­ing to [40, item 19.7]

\begin{displaymath}
\psi = C_1 + C_2 x
\end{displaymath}

The bound­ary con­di­tion that $\psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 at $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 now pro­duces:

\begin{displaymath}
C_1 + C_2 0 = C_1 = 0
\end{displaymath}

so $C_1$ must be zero. The bound­ary con­di­tion that $\psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 at $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\ell_x$ gives:

\begin{displaymath}
0 + C_2 \ell_x = 0
\end{displaymath}

so $C_2$ must be zero too. Once again there is no nonzero so­lu­tion, so the as­sump­tion that the en­ergy $E$ can be zero must be wrong too.

Note that clas­si­cally, it is per­fectly OK for the en­ergy to be zero: it would sim­ply mean that the par­ti­cle is sit­ting in the pipe at rest. But in quan­tum me­chan­ics, zero ki­netic en­ergy is not ac­cept­able, and it all has to do with Heisen­berg's un­cer­tainty prin­ci­ple. Since the par­ti­cle is re­stricted to the in­side of the pipe, its po­si­tion is con­strained, and so the un­cer­tainty prin­ci­ple re­quires that the lin­ear mo­men­tum must be un­cer­tain. Un­cer­tain mo­men­tum can­not be zero mo­men­tum; mea­sure­ments will show a range of val­ues for the mo­men­tum of the par­ti­cle, im­ply­ing that it is in mo­tion and there­fore has ki­netic en­ergy.

Try, try again. The only pos­si­bil­ity left is that the en­ergy $E$ is pos­i­tive. In that case, the so­lu­tion of the or­di­nary dif­fer­en­tial equa­tion is ac­cord­ing to [40, item 19.7]:

\begin{displaymath}
\psi = C_1 \cos(k x) + C_2 \sin(k x)
\qquad k = \frac{\sqrt{2mE}}{\hbar}
\end{displaymath}

Here the con­stant $k$ is called the “wave num­ber.”

The bound­ary con­di­tion that $\psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 at $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 is:

\begin{displaymath}
C_1 1 + C_2 0 = C_1 = 0
\end{displaymath}

so $C_1$ must be zero. The bound­ary con­di­tion $\psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 at $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\ell_x$ is then:

\begin{displaymath}
0 + C_2 \sin(k\ell_x) = 0
\end{displaymath}

There fi­nally is pos­si­bil­ity to get a nonzero co­ef­fi­cient $C_2$: this equa­tion can be sat­is­fied if $\sin(k\ell_x)$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 in­stead of $C_2$. In fact, there is not just one pos­si­bil­ity for this to hap­pen: a sine is zero when its ar­gu­ment equals $\pi$, $2\pi$, $3\pi$, .... So there is a nonzero so­lu­tion for each of the fol­low­ing val­ues of the pos­i­tive con­stant $k$:

\begin{displaymath}
k =\frac{\pi}{\ell_x},\;
k =\frac{2\pi}{\ell_x},\;
k =\frac{3\pi}{\ell_x},\;
\ldots
\end{displaymath}

Each of these pos­si­bil­i­ties gives one so­lu­tion $\psi$. Dif­fer­ent so­lu­tions $\psi$ will be dis­tin­guished by giv­ing them a nu­meric sub­script:

\begin{displaymath}
\psi_1 = C_2 \sin\left(\frac{\pi}{\ell_x} x\right),\;
\psi...
...\psi_3 = C_2 \sin\left(\frac{3\pi}{\ell_x} x\right),\;
\ldots
\end{displaymath}

The generic so­lu­tion can be writ­ten more con­cisely us­ing a counter $n$ as:

\begin{displaymath}
\psi_n = C_2 \sin\left(\frac{n\pi}{\ell_x} x\right)
\quad\mbox{for } n = 1,2,3,\ldots
\end{displaymath}

Let’s check the so­lu­tions. Clearly each is zero when $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 and when $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\ell_x$. Also, sub­sti­tu­tion of each of the so­lu­tions into the or­di­nary dif­fer­en­tial equa­tion

\begin{displaymath}
- \frac{\hbar^2}{2m} \frac{\partial^2\psi}{\partial x^2}
= E \psi
\end{displaymath}

shows that they all sat­isfy it, pro­vided that their en­ergy val­ues are, re­spec­tively:

\begin{displaymath}
E_1 = \frac{\hbar^2\pi^2}{2m\ell_x^2},\;
E_2 = \frac{2^2\h...
...l_x^2},\;
E_3 = \frac{3^2\hbar^2\pi^2}{2m\ell_x^2},\;
\ldots
\end{displaymath}

or gener­i­cally:

\begin{displaymath}
E_n = \frac{n^2\hbar^2\pi^2}{2m\ell_x^2}
\quad\mbox{for } n = 1,2,3,\ldots
\end{displaymath}

There is one more con­di­tion that must be sat­is­fied: each so­lu­tion must be nor­mal­ized so that the to­tal prob­a­bil­ity of find­ing the par­ti­cle in­te­grated over all pos­si­ble po­si­tions is 1 (cer­tainty). That re­quires:

\begin{displaymath}
1 = \langle\psi_n\vert\psi_n\rangle =
\int_{x=0}^{\ell_x} ...
...C_2\vert^2 \sin^2\left(\frac{n\pi}{\ell_x} x\right) {\,\rm d}x
\end{displaymath}

which af­ter in­te­gra­tion fixes $C_2$ (as­sum­ing you choose it to be a pos­i­tive real num­ber):

\begin{displaymath}
C_2 = \sqrt{\frac{2}{\ell_x}}
\end{displaymath}

Sum­ma­riz­ing the re­sults of this sub­sec­tion, there is not just one en­ergy eigen­func­tion and cor­re­spond­ing eigen­value, but an in­fi­nite set of them:

\begin{displaymath}
\renewedcommand{arraystretch}{2.9}
\begin{array}{ll}
\psi...
...\ell_x^2}} \\
\qquad \!\vdots & \qquad \vdots
\end{array} %
\end{displaymath} (3.16)

or in generic form:
\begin{displaymath}
\psi_n = \sqrt{\frac{2}{\ell_x}} \sin\left(\frac{n\pi}{\ell...
...r^2\pi^2}{2m\ell_x^2}
\quad\mbox{for } n = 1,2,3,4,5,\ldots %
\end{displaymath} (3.17)

The next thing will be to take a bet­ter look at these re­sults.


Key Points
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Af­ter a lot of grind­ing math­e­mat­ics armed with ta­ble books, the en­ergy eigen­func­tions and eigen­val­ues have fi­nally been found

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There are in­fi­nitely many of them.

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They are as listed in (3.17) above. The first few are also writ­ten out ex­plic­itly in (3.16).

3.5.5 Re­view Ques­tions
1.

Write down eigen­func­tion num­ber 6.

So­lu­tion piped-a

2.

Write down eigen­value num­ber 6.

So­lu­tion piped-b


3.5.6 Dis­cus­sion of the en­ergy val­ues

This sub­sec­tion dis­cusses the en­ergy that the par­ti­cle in the pipe can have. It was al­ready dis­cov­ered in the pre­vi­ous sub­sec­tion that the par­ti­cle can­not have neg­a­tive en­ergy, nor zero en­ergy. In fact, ac­cord­ing to the or­tho­dox in­ter­pre­ta­tion, the only val­ues that the to­tal en­ergy of the par­ti­cle can take are the en­ergy eigen­val­ues

\begin{displaymath}
E_1 = \frac{\hbar^2\pi^2}{2m\ell_x^2},\;
E_2 = \frac{2^2\h...
...l_x^2},\;
E_3 = \frac{3^2\hbar^2\pi^2}{2m\ell_x^2},\;
\ldots
\end{displaymath}

de­rived in the pre­vi­ous sub­sec­tion.

En­ergy val­ues are typ­i­cally shown graph­i­cally in the form of an en­ergy spec­trum, as in fig­ure 3.7.

Fig­ure 3.7: One-di­men­sional en­ergy spec­trum for a par­ti­cle in a pipe.
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\begin{picture}(30...
...,160){\line(1,0){50}}
\put(110,250){\line(1,0){50}}
\end{picture}
\end{figure}

En­ergy is plot­ted up­wards, so the ver­ti­cal height of each en­ergy level in­di­cates the amount of en­ergy it has. To the right of each en­ergy level, the so­lu­tion counter, or quan­tum num­ber, $n$ is listed.

Clas­si­cally, the to­tal en­ergy of the par­ti­cle can have any non­neg­a­tive value. But ac­cord­ing to quan­tum me­chan­ics, that is not true: the to­tal en­ergy must be one of the lev­els shown in the en­ergy spec­trum fig­ure 3.7. It should be noted that for a macro­scopic par­ti­cle, you would not know the dif­fer­ence; the spac­ing be­tween the en­ergy lev­els is macro­scop­i­cally very fine, since Planck's con­stant $\hbar$ is so small. How­ever, for a quan­tum-scale sys­tem, the dis­crete­ness of the en­ergy val­ues can make a ma­jor dif­fer­ence.

An­other point: at ab­solute zero tem­per­a­ture, the par­ti­cle will be stuck in the low­est pos­si­ble en­ergy level, $E_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\hbar^2\pi^2$$\raisebox{.5pt}{$/$}$$2m\ell_x^2$, in the spec­trum fig­ure 3.7. This low­est pos­si­ble en­ergy level is called the ground state. Clas­si­cally you would ex­pect that at ab­solute zero the par­ti­cle has no ki­netic en­ergy, so zero to­tal en­ergy. But quan­tum me­chan­ics does not al­low it. Heisen­berg's prin­ci­ple re­quires some mo­men­tum, hence ki­netic en­ergy to re­main for a con­fined par­ti­cle even at zero tem­per­a­ture.


Key Points
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En­ergy val­ues can be shown as an en­ergy spec­trum.

$\begin{picture}(15,5.5)(0,-3)
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The pos­si­ble en­ergy lev­els are dis­crete.

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But for a macro­scopic par­ti­cle, they are ex­tremely close to­gether.

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The ground state of low­est en­ergy has nonzero ki­netic en­ergy.

3.5.6 Re­view Ques­tions
1.

Plug the mass of an elec­tron, $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 9.109 38 10$\POW9,{-31}$ kg, and the rough size of an hy­dro­gen atom, call it $\ell_x$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 10$\POW9,{-10}$ m, into the ex­pres­sion for the ground state ki­netic en­ergy and see how big it is. Note that $\hbar$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1.054 57 10$\POW9,{-34}$ J s. Ex­press in units of eV, where one eV equals 1.602 18 10$\POW9,{-19}$ J.

So­lu­tion pipee-a

2.

Just for fun, plug macro­scopic val­ues, $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 kg and $\ell_x$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 m, into the ex­pres­sion for the ground state en­ergy and see how big it is. Note that $\hbar$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1.054 57 10$\POW9,{-34}$ J s.

So­lu­tion pipee-b

3.

What is the eigen­func­tion num­ber, or quan­tum num­ber, $n$ that pro­duces a macro­scopic amount of en­ergy, 1 J, for macro­scopic val­ues $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 kg and $\ell_x$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 m? With that many en­ergy lev­els in­volved, would you see the dif­fer­ence be­tween suc­ces­sive ones?

So­lu­tion pipee-c


3.5.7 Dis­cus­sion of the eigen­func­tions

This sub­sec­tion dis­cusses the one-di­men­sion­al en­ergy eigen­func­tions of the par­ti­cle in the pipe. The so­lu­tion of sub­sec­tion 3.5.5 found them to be:

\begin{displaymath}
\psi_1 = \sqrt{\frac{2}{\ell_x}} \sin\left(\frac{\pi}{\ell_...
...{2}{\ell_x}} \sin\left(\frac{3\pi}{\ell_x} x\right),\;
\ldots
\end{displaymath}

The first one to look at is the ground state eigen­func­tion

\begin{displaymath}
\psi_1 = \sqrt{\frac{2}{\ell_x}} \sin\left(\frac{\pi}{\ell_x} x\right).
\end{displaymath}

It is plot­ted at the top of fig­ure 3.8. As noted in sec­tion 3.1, it is the square mag­ni­tude of a wave func­tion that gives the prob­a­bil­ity of find­ing the par­ti­cle. So, the sec­ond graph in fig­ure 3.8 shows the square of the ground state wave func­tion, and the higher val­ues of this func­tion then give the lo­ca­tions where the par­ti­cle is more likely to be found. This book shows re­gions where the par­ti­cle is more likely to be found as darker re­gions, and in those terms the prob­a­bil­ity of find­ing the par­ti­cle is as shown in the bot­tom graphic of fig­ure 3.8.

Fig­ure 3.8: One-di­men­sional ground state of a par­ti­cle in a pipe.
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It is seen that in the ground state, the par­ti­cle is much more likely to be found some­where in the mid­dle of the pipe than close to the ends.

Fig­ure 3.9: Sec­ond and third low­est one-di­men­sional en­ergy states.
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...
\put(0,0){\makebox(0,0)[b]{\epsffile{pipeo3.eps}}}
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\end{figure}

Fig­ure 3.9 shows the two next low­est en­ergy states

\begin{displaymath}
\psi_2 = \sqrt{\frac{2}{\ell_x}} \sin\left(\frac{2\pi}{\ell...
...\sqrt{\frac{2}{\ell_x}} \sin\left(\frac{3\pi}{\ell_x} x\right)
\end{displaymath}

as grey tones. Re­gions where the par­ti­cle is rel­a­tively likely to be found al­ter­nate with ones where it is less likely to be found. And the higher the en­ergy, the more such re­gions there are. Also note that in sharp con­trast to the ground state, for eigen­func­tion $\psi_2$ there is al­most no like­li­hood of find­ing the par­ti­cle close to the cen­ter.

Need­less to say, none of those en­ergy states looks at all like the wave func­tion blob bounc­ing around in fig­ure 3.5. More­over, it turns out that en­ergy eigen­states are sta­tion­ary states: the prob­a­bil­i­ties shown in fig­ures 3.8 and 3.9 do not change with time.

In or­der to de­scribe a lo­cal­ized wave func­tion blob bounc­ing around, states of dif­fer­ent en­ergy must be com­bined. It will take un­til chap­ter 7.11.4 be­fore the an­a­lyt­i­cal tools to do so have been de­scribed. For now, the dis­cus­sion must re­main re­stricted to just find­ing the en­ergy lev­els. And these are im­por­tant enough by them­selves any­way, suf­fi­cient for many prac­ti­cal ap­pli­ca­tions of quan­tum me­chan­ics.


Key Points
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In the en­ergy eigen­func­tions, the par­ti­cle is not lo­cal­ized to within any par­tic­u­lar small re­gion of the pipe.

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In gen­eral there are re­gions where the par­ti­cle may be found sep­a­rated by re­gions in which there is lit­tle chance to find the par­ti­cle.

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The higher the en­ergy level, the more such re­gions there are.

3.5.7 Re­view Ques­tions
1.

So how does, say, the one-di­men­sion­al eigen­state $\psi_6$ look?

So­lu­tion pipef-a

2.

Gen­er­al­iz­ing the re­sults above, for eigen­func­tion $\psi_n$, any $n$, how many dis­tinct re­gions are there where the par­ti­cle may be found?

So­lu­tion pipef-b

3.

If you are up to a trick ques­tion, con­sider the fol­low­ing. There are no forces in­side the pipe, so the par­ti­cle has to keep mov­ing un­til it hits an end of the pipe, then re­flect back­ward un­til it hits the other side and so on. So, it has to cross the cen­ter of the pipe reg­u­larly. But in the en­ergy eigen­state $\psi_2$, the par­ti­cle has zero chance of ever be­ing found at the cen­ter of the pipe. What gives?

So­lu­tion pipef-c


3.5.8 Three-di­men­sional so­lu­tion

The so­lu­tion for the par­ti­cle stuck in a pipe that was ob­tained in the pre­vi­ous sub­sec­tions cheated. It pre­tended that there was only one spa­tial co­or­di­nate $x$. Real life is three-di­men­sion­al. And yes, as a re­sult, the so­lu­tion as ob­tained is sim­ply wrong.

For­tu­nately, it turns out that you can fix up the prob­lem pretty eas­ily if you as­sume that the pipe has a square cross sec­tion. There is a way of com­bin­ing one-di­men­sion­al so­lu­tions for all three co­or­di­nates into full three-di­men­sion­al so­lu­tions. This is called the sep­a­ra­tion of vari­ables idea: Solve each of the three vari­ables $x$, $y$, and $z$ sep­a­rately, then com­bine the re­sults.

The full co­or­di­nate sys­tem for the prob­lem is shown in fig­ure 3.10: in ad­di­tion to the $x$-​co­or­di­nate along the length of the pipe, there is also a $y$-​co­or­di­nate giv­ing the ver­ti­cal po­si­tion in cross sec­tion, and sim­i­larly a $z$-​co­or­di­nate giv­ing the po­si­tion in cross sec­tion to­wards you.

Fig­ure 3.10: De­f­i­n­i­tion of all vari­ables for mo­tion in a pipe.
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...le*{2}}
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Now re­call the one-di­men­sion­al so­lu­tions that were ob­tained as­sum­ing there is just an $x$-​co­or­di­nate, but add sub­scripts $x$ to keep them apart from any so­lu­tions for $y$ and $z$:

\begin{displaymath}
\renewedcommand{arraystretch}{2.9}
\begin{array}{ll}
\psi...
...3^2\hbar^2\pi^2}{2m\ell_x^2}} \\
\vdots & \vdots
\end{array}\end{displaymath} (3.18)

or in generic form:
\begin{displaymath}
\psi_{xn_x} = \sqrt{\frac{2}{\ell_x}}\sin\left(\frac{n_x\pi...
...\hbar^2\pi^2}{2m\ell_x^2}
\quad\mbox{for } n_x = 1,2,3,\ldots
\end{displaymath} (3.19)

Since it is as­sumed that the cross sec­tion of the pipe is square or rec­tan­gu­lar of di­men­sions $\ell_y$ $\times$ $\ell_z$, the $y$ and $z$ di­rec­tions have one-di­men­sion­al so­lu­tions com­pletely equiv­a­lent to the $x$ di­rec­tion:

\begin{displaymath}
\psi_{yn_y} = \sqrt{\frac{2}{\ell_y}}\sin\left(\frac{n_y\pi...
...\hbar^2\pi^2}{2m\ell_y^2}
\quad\mbox{for } n_y = 1,2,3,\ldots
\end{displaymath} (3.20)

and
\begin{displaymath}
\psi_{zn_z} = \sqrt{\frac{2}{\ell_z}}\sin\left(\frac{n_z\pi...
...\hbar^2\pi^2}{2m\ell_z^2}
\quad\mbox{for } n_z = 1,2,3,\ldots
\end{displaymath} (3.21)

Af­ter all, there is no fun­da­men­tal dif­fer­ence be­tween the three co­or­di­nate di­rec­tions; each is along an edge of a rec­tan­gu­lar box.

Now it turns out, {D.11}, that the full three-di­men­sion­al prob­lem has eigen­func­tions $\psi_{n_xn_yn_z}$ that are sim­ply prod­ucts of the one-di­men­sion­al ones:

\begin{displaymath}
\fbox{$\displaystyle
\psi_{n_xn_yn_z} = \sqrt{\frac{8}{\el...
...ll_y} y\right)
\sin\left(\frac{n_z\pi}{\ell_z} z\right)
$} %
\end{displaymath} (3.22)

There is one such three-di­men­sion­al eigen­func­tion for each set of three num­bers $(n_x,n_y,n_z)$. These num­bers are the three quan­tum num­bers of the eigen­func­tion.

Fur­ther, the en­ergy eigen­val­ues $E_{n_xn_yn_z}$ of the three-di­men­sion­al prob­lem are the sum of those of the one-di­men­sion­al prob­lems:

\begin{displaymath}
\fbox{$\displaystyle
E_{n_xn_yn_z} =
\frac{n_x^2\hbar^2\p...
...i^2}{2m\ell_y^2} +
\frac{n_z^2\hbar^2\pi^2}{2m\ell_z^2}
$} %
\end{displaymath} (3.23)

For ex­am­ple, the ground state of low­est en­ergy oc­curs when all three quan­tum num­bers are low­est, $n_x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n_y$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n_z$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. The three-di­men­sion­al ground state wave func­tion is there­fore:

\begin{displaymath}
\psi_{111} = \sqrt{\frac{8}{\ell_x\ell_y\ell_z}}
\sin\left...
...c{\pi}{\ell_y} y\right)
\sin\left(\frac{\pi}{\ell_z} z\right)
\end{displaymath} (3.24)

This ground state is shown in fig­ure 3.11. The $y$ and $z$ fac­tors en­sure that the wave func­tion is now zero at all the sur­faces of the pipe.

Fig­ure 3.11: True ground state of a par­ti­cle in a pipe.
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The ground state en­ergy is:

\begin{displaymath}
E_{111} =
\frac{\hbar^2\pi^2}{2m\ell_x^2} +
\frac{\hbar^2\pi^2}{2m\ell_y^2} +
\frac{\hbar^2\pi^2}{2m\ell_z^2}
\end{displaymath} (3.25)

Since the cross sec­tion di­men­sions $\ell_y$ and $\ell_z$ are small com­pared to the length of the pipe, the last two terms are large com­pared to the first one. They make the true ground state en­ergy much larger than the one-di­men­sion­al value, which was just the first term.

Fig­ure 3.12: True sec­ond and third low­est en­ergy states.
\begin{figure}\centering
{}%
\epsffile{pipet2.eps} \\ [10pt]
\epsffile{pipet3.eps}
\end{figure}

The next two low­est en­ergy lev­els oc­cur for $n_x$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2, $n_y$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n_z$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 re­spec­tively $n_x$ $\vphantom0\raisebox{1.5pt}{$=$}$ 3, $n_y$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n_z$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. (The lat­ter as­sumes that the cross sec­tion di­men­sions are small enough that the al­ter­na­tive pos­si­bil­i­ties $n_y$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2, $n_x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n_z$ $\vphantom0\raisebox{1.5pt}{$=$}$1 and $n_z$ $\vphantom0\raisebox{1.5pt}{$=$}$2, $n_x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n_y$ $\vphantom0\raisebox{1.5pt}{$=$}$1 have more en­ergy.) The en­ergy eigen­func­tions

\begin{displaymath}
\psi_{211} = \sqrt{\frac{8}{\ell_x\ell_y\ell_z}}
\sin\left...
...c{\pi}{\ell_y} y\right)
\sin\left(\frac{\pi}{\ell_z} z\right)
\end{displaymath} (3.26)


\begin{displaymath}
\psi_{311} = \sqrt{\frac{8}{\ell_x\ell_y\ell_z}}
\sin\left...
...c{\pi}{\ell_y} y\right)
\sin\left(\frac{\pi}{\ell_z} z\right)
\end{displaymath} (3.27)

are shown in fig­ure 3.12. They have en­ergy lev­els:
\begin{displaymath}
E_{211} =
\frac{4\hbar^2\pi^2}{2m\ell_x^2} +
\frac{\hbar^...
...c{\hbar^2\pi^2}{2m\ell_y^2} +
\frac{\hbar^2\pi^2}{2m\ell_z^2}
\end{displaymath} (3.28)


Key Points
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Three-di­men­sion­al en­ergy eigen­func­tions can be found as prod­ucts of one-di­men­sion­al ones.

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Three-di­men­sion­al en­er­gies can be found as sums of one-di­men­sion­al ones.

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Ex­am­ple three-di­men­sion­al eigen­states have been shown.

3.5.8 Re­view Ques­tions
1.

If the cross sec­tion di­men­sions $\ell_y$ and $\ell_z$ are one tenth the size of the pipe length, how much big­ger are the en­er­gies $E_{y1}$ and $E_{z1}$ com­pared to $E_{x1}$? So, by what per­cent­age is the one-di­men­sion­al ground state en­ergy $E_{x1}$ as an ap­prox­i­ma­tion to the three-di­men­sion­al one, $E_{111}$, then in er­ror?

So­lu­tion pipeg-a

2.

At what ra­tio of $\ell_y$$\raisebox{.5pt}{$/$}$$\ell_x$ does the en­ergy $E_{121}$ be­come higher than the en­ergy $E_{311}$?

So­lu­tion pipeg-b

3.

Shade the re­gions where the par­ti­cle is likely to be found in the $\psi_{322}$ en­ergy eigen­state.

So­lu­tion pipeg-c


3.5.9 Quan­tum con­fine­ment

Nor­mally, mo­tion in physics oc­curs in three di­men­sions. Even in a nar­row pipe, in clas­si­cal physics a point par­ti­cle of zero size would be able to move in all three di­rec­tions. But in quan­tum me­chan­ics, if the pipe gets very nar­row, the mo­tion be­comes truly one-di­men­sion­al.

To un­der­stand why, the first prob­lem that must be ad­dressed is what mo­tion means in the first place, be­cause nor­mally mo­tion is de­fined as change in po­si­tion, and in quan­tum me­chan­ics par­ti­cles do not have a well-de­fined po­si­tion.

Con­sider the par­ti­cle in the ground state of low­est en­ergy, shown in fig­ure 3.11. This is one bor­ing state; the pic­ture never changes. You might be sur­prised by that; af­ter all, it was found that the ground state has en­ergy, and it is all ki­netic en­ergy. If the par­ti­cle has ki­netic en­ergy, should not the po­si­tions where the par­ti­cle is likely to be found change with time?

The an­swer is no; ki­netic en­ergy is not di­rectly re­lated to changes in likely po­si­tions of a par­ti­cle; that is only an ap­prox­i­ma­tion valid for macro­scopic sys­tems. It is not nec­es­sar­ily true for quan­tum-scale sys­tems, cer­tainly not if they are in the ground state. Like it or not, in quan­tum me­chan­ics ki­netic en­ergy is sec­ond-or­der de­riv­a­tives of the wave func­tion, and noth­ing else.

Next, as al­ready pointed out, all the other en­ergy eigen­states, like those in fig­ure 3.12, have the same bor­ing prop­erty of not chang­ing with time.

Things only be­come some­what in­ter­est­ing when you com­bine states of dif­fer­ent en­ergy. As the sim­plest pos­si­ble ex­am­ple, con­sider the pos­si­bil­ity that the par­ti­cle has the wave func­tion:

\begin{displaymath}
\Psi=\sqrt{{\textstyle\frac{4}{5}}} \psi_{111} + \sqrt{{\textstyle\frac{1}{5}}} \psi_{211}
\end{displaymath}

at some start­ing time, which will be taken as $t$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. Ac­cord­ing to the or­tho­dox in­ter­pre­ta­tion, in an en­ergy mea­sure­ment this par­ti­cle would have a ${\textstyle\frac{4}{5}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 80% chance of be­ing found at the ground state en­ergy $E_{111}$ and a 20% chance of be­ing found at the el­e­vated en­ergy level $E_{211}$. So there is now un­cer­tainty in en­ergy; that is crit­i­cal.

Fig­ure 3.13: A com­bi­na­tion of $\psi_{111}$ and $\psi_{211}$ seen at some typ­i­cal times.
 
a

b

c

d

Move your mouse over any fig­ure to see the an­i­ma­tion. Javascript must be en­abled on your browser. Give it a few sec­onds for the an­i­ma­tion to load, es­pe­cially on a phone line.

In chap­ter 7.1 it will be found that for nonzero times, the wave func­tion of this par­ti­cle is given by

\begin{displaymath}
\Psi=\sqrt{{\textstyle\frac{4}{5}}} e^{-{\rm i}E_{111}t/\hb...
...\textstyle\frac{1}{5}}} e^{-{\rm i}E_{211}t/\hbar} \psi_{211}.
\end{displaymath}

Us­ing this ex­pres­sion, the prob­a­bil­ity of find­ing the par­ti­cle, $\vert\Psi\vert^2$, can be plot­ted for var­i­ous times. That is done in fig­ure 3.13 for four typ­i­cal times. It shows that with un­cer­tainty in en­ergy, the wave func­tion blob does move. It per­forms a pe­ri­odic os­cil­la­tion: af­ter fig­ure 3.13(d), the wave func­tion re­turns to state 3.13(a), and the cy­cle re­peats.

You would not yet want to call the par­ti­cle lo­cal­ized, but at least the lo­ca­tions where the par­ti­cle can be found are now bounc­ing back and for­wards be­tween the ends of the pipe. And if you add ad­di­tional wave func­tions $\psi_{311}$, $\psi_{411}$, ..., you can get closer and closer to a lo­cal­ized wave func­tion blob bounc­ing around.

But if you look closer at fig­ure 3.13, you will note that the wave func­tion blob does not move at all in the $y$-​di­rec­tion; it re­mains at all times cen­tered around the hor­i­zon­tal pipe cen­ter­line. It may seem that this is no big deal; just add one or more wave func­tions with an $n_y$ value greater than one, like $\psi_{121}$, and bingo, there will be in­ter­est­ing mo­tion in the $y$-​di­rec­tion too.

But there is a catch, and it has to do with the re­quired en­ergy. Ac­cord­ing to the pre­vi­ous sec­tion, the ki­netic en­ergy in the $y$-​di­rec­tion takes the val­ues

\begin{displaymath}
E_{y1}=\frac{\hbar^2\pi^2}{2m\ell_y^2} ,\;
E_{y2}=\frac{4\...
..._y^2} ,\;
E_{y3}=\frac{9\hbar^2\pi^2}{2m\ell_y^2} ,\;
\ldots
\end{displaymath}

That will be very large en­er­gies for a nar­row pipe in which $\ell_y$ is small. The par­ti­cle will cer­tainly have the large en­ergy $E_{y1}$ in the $y$-​di­rec­tion; if it is in the pipe at all it has at least that amount of en­ergy. But if the pipe is re­ally nar­row, it will sim­ply not have enough ad­di­tional, say ther­mal, en­ergy to get any­where close to the next level $E_{y2}$. The ki­netic en­ergy in the $y$-​di­rec­tion will there­fore be stuck at the low­est pos­si­ble level $E_{y1}$.

The re­sult is that ab­solutely noth­ing in­ter­est­ing goes on in the $y$-​di­rec­tion. As far as a par­ti­cle in a nar­row pipe is con­cerned, the $y$-​di­rec­tion might just as well not ex­ist. It is ironic that while the ki­netic en­ergy in the $y$-​di­rec­tion, $E_{y1}$, is very large, noth­ing ac­tu­ally hap­pens in that di­rec­tion.

If the pipe is also nar­row in the $z$-​di­rec­tion, the only in­ter­est­ing mo­tion is in the $x$-​di­rec­tion, mak­ing the non­triv­ial physics truly one-di­men­sion­al. It be­comes a “quan­tum wire”. How­ever, if the pipe size in the $z$-​di­rec­tion is rel­a­tively wide, the par­ti­cle will have lots of dif­fer­ent en­ergy states in the $z$-​di­rec­tion avail­able too and the mo­tion will be two-di­men­sion­al, a “quan­tum well”. Con­versely, if the pipe is nar­row in all three di­rec­tions, you get a zero-di­men­sion­al “quan­tum dot” in which the par­ti­cle does noth­ing un­less it gets a siz­able chunk of en­ergy.

An iso­lated atom can be re­garded as an ex­am­ple of a quan­tum dot; the elec­trons are con­fined to a small re­gion around the nu­cleus and will be at a sin­gle en­ergy level un­less they are given a con­sid­er­able amount of en­ergy. But note that when peo­ple talk about quan­tum con­fine­ment, they are nor­mally talk­ing about semi-con­duc­tors, for which sim­i­lar ef­fects oc­cur at sig­nif­i­cantly larger scales, maybe tens of times as large, mak­ing them much eas­ier to man­u­fac­ture. An ac­tual quan­tum dot is of­ten re­ferred to as an “ar­ti­fi­cial atom”, and has sim­i­lar prop­er­ties as a real atom.

It may give you a rough idea of all the in­ter­est­ing things you can do in nan­otech­nol­ogy when you re­strict the mo­tion of par­ti­cles, in par­tic­u­lar of elec­trons, in var­i­ous di­rec­tions. You truly change the di­men­sion­al­ity of the nor­mal three-di­men­sion­al world into a lower di­men­sional one. Only quan­tum me­chan­ics can ex­plain why, by mak­ing the en­ergy lev­els dis­crete in­stead of con­tin­u­ously vary­ing. And the lower di­men­sional worlds can have your choice of topol­ogy (a ring, a let­ter 8, a sphere, a cylin­der, a Möbius strip?, ...,) to make things re­ally ex­cit­ing.


Key Points
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Quan­tum me­chan­ics al­lows you to cre­ate lower-di­men­sion­al worlds for par­ti­cles.