D.13 The har­monic os­cil­la­tor and un­cer­tainty

The given qual­i­ta­tive ex­pla­na­tion of the ground state of the har­monic os­cil­la­tor in terms of the un­cer­tainty prin­ci­ple is ques­tion­able. In par­tic­u­lar, po­si­tion, lin­ear mo­men­tum, po­ten­tial en­ergy, and ki­netic en­ergy are un­cer­tain for the ground state. This note gives a solid ar­gu­ment, but it uses some ad­vanced ideas dis­cussed in chap­ter 4.4 and 4.5.3.

As ex­plained more fully in chap­ter 4.4, the ex­pec­ta­tion value of the ki­netic en­ergy is de­fined as the av­er­age value ex­pected for ki­netic en­ergy mea­sure­ments. Sim­i­larly, the ex­pec­ta­tion value of the po­ten­tial en­ergy is de­fined as the av­er­age value ex­pected for po­ten­tial en­ergy mea­sure­ments.

From the pre­cise form of ex­pec­ta­tion val­ues in quan­tum me­chan­ics, it fol­lows that to­tal en­ergy must be the sum of the ki­netic and po­ten­tial en­ergy ex­pec­ta­tion val­ues. For the har­monic os­cil­la­tor ground state, that gives

$$E_{x0} = {\textstyle\frac{1}{2}}\hbar\omega = \frac{1}{2m}... ...\rangle + \frac{m}{2} \omega^2 \left\langle{x^2}\right\rangle$$

Here $\left\langle{.}\right\rangle$ stands for the av­er­age of the en­closed quan­tity. Only the mo­tion in the $x$-​di­rec­tion will be con­sid­ered here. The $y$ and $z$ di­rec­tions go ex­actly the same way.

Now any value of $p_x$ can be writ­ten as equal to the av­er­age value $\left\langle{p_x}\right\rangle$ plus a de­vi­a­tion from that av­er­age $\Delta{p}_x$. Then

$$\left\langle{p_x^2}\right\rangle = \Big\langle(\left\langle... ...p_x}\right\rangle + \left\langle{(\Delta p_x)^2}\right\rangle$$

Note that an av­er­age is a con­stant that is not af­fected by fur­ther av­er­ag­ing. Next note that the av­er­age of $\Delta{p}_x$ is zero, oth­er­wise the av­er­age of $p_x+\Delta{p}_x$ would not be $\left\langle{p_x}\right\rangle$. So:

$$\left\langle{p_x^2}\right\rangle = \left\langle{p_x}\right\rangle ^2 + \left\langle{(\Delta p_x)^2}\right\rangle$$

Of course, a sim­i­lar ex­pres­sion holds for $\left\langle{x^2}\right\rangle$, so the ground state en­ergy is

$\parbox{400pt}{\hspace{11pt}\hfill$\displaystyle {\textstyle\frac{1}{2}}\hbar\... ... + \frac{m}{2} \omega^2 \left\langle{(\Delta x)^2}\right\rangle $\hfill(1)}$

Con­sider the last two terms. Call them $a^2$ and $b^2$ for now. Note that

$$(a-b)^2 \mathrel{\raisebox{-1pt}{$\geqslant$}}0 \quad\Longr... ...right\rangle } \sqrt{\left\langle{(\Delta x)^2}\right\rangle }$$

as fol­lows from mul­ti­ply­ing out the square. The $\raisebox{-.5pt}{$\geqslant$}$ be­comes $\vphantom0\raisebox{1.5pt}{$=$}$ when $a$ and $b$ are equal.

Now the first square root above is a mea­sure of the un­cer­tainty in $p_x$. If $\Delta{p}_x$ is al­ways zero, then $p_x$ is al­ways its av­er­age value, with­out any un­cer­tainty. Sim­i­larly, the sec­ond square root above is a mea­sure of the un­cer­tainty in $x$. The Heisen­berg un­cer­tainty prin­ci­ple can be made quan­ti­ta­tive as, chap­ter 4.5.3,

$$\sqrt{\left\langle{(\Delta p_x)^2}\right\rangle } \sqrt{\le... ...rel{\raisebox{-1pt}{$\geqslant$}}{\textstyle\frac{1}{2}} \hbar$$

There­fore

$$a^2 + b^2 \mathrel{\raisebox{-1pt}{$\geqslant$}}{\textstyle\frac{1}{2}} \hbar \omega$$

So the min­i­mum value of the fi­nal two terms in the ex­pres­sion (1) for the ground state en­ergy is the com­plete ground state en­ergy. There­fore, in or­der that the right hand side in (1) does not ex­ceed the left hand side, the first two terms must be zero. So the av­er­age par­ti­cle mo­men­tum and po­si­tion are both zero. In ad­di­tion, for the es­ti­mates of the fi­nal two terms, equal­i­ties are needed, not in­equal­i­ties. That means that $a$ must be $b$. That then means that the ex­pec­ta­tion ki­netic en­ergy must be the ex­pec­ta­tion po­ten­tial en­ergy. And the two must be the very min­i­mum al­lowed by the Heisen­berg re­la­tion; oth­er­wise there is still that in­equal­ity.