A.17 The vir­ial the­o­rem

The vir­ial the­o­rem re­lates the ex­pec­ta­tion ki­netic en­ergy of a quan­tum sys­tem to the po­ten­tial. That is of the­o­ret­i­cal in­ter­est, as well as im­por­tant for com­pu­ta­tional meth­ods like “den­sity func­tional the­ory.”

Con­sider a quan­tum sys­tem in a state of def­i­nite en­ergy $E$. In other words, con­sider a quan­tum sys­tem in a sta­tion­ary state. It does not have to be the ground state. The quan­tum sys­tem will be as­sumed to be in in­fi­nite space.

To keep it sim­ple, for now as­sume that there is a sin­gle par­ti­cle with po­si­tion vec­tor ${\skew0\vec r}$ in a po­ten­tial $V({\skew0\vec r})$. That cov­ers our pre­vi­ous ex­am­ples of the har­monic os­cil­la­tor and the hy­dro­gen atom.

Then the vir­ial the­o­rem re­lates the ex­pec­ta­tion ki­netic en­ergy $\left\langle{T}\right\rangle$ to the po­ten­tial $V$ as fol­lows:

$$\fbox{$\displaystyle 2 \left\langle{T}\right\rangle = \left\langle{{\skew0\vec r}\cdot\nabla V}\right\rangle $}$$ (A.75)

(Re­call that nabla, $\nabla$, is just the multi-di­men­sional de­riv­a­tive $\partial/\partial{\skew0\vec r}$.) The above for­mula can be very use­ful.

For ex­am­ple, con­sider the har­monic os­cil­la­tor. There

$$V = \frac12c_xx^2+\frac12c_yy^2+\frac12c_zz^2$$

so in Carte­sian co­or­di­nates

$${\skew0\vec r}\cdot\nabla V = x \frac{\partial V}{\partial... ...rtial V}{\partial y} + z \frac{\partial V}{\partial z} = 2 V$$

Then ac­cord­ing to the vir­ial the­o­rem $2\left\langle{T}\right\rangle$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\left\langle{2V}\right\rangle$. So the ex­pec­ta­tion ki­netic en­ergy and the ex­pec­ta­tion po­ten­tial en­ergy are the same. Com­pute whichever is eas­i­est, or just take half of the to­tal en­ergy $E$ if you know it.

Also con­sider the hy­dro­gen atom. There

$$V = -\frac{e^2}{4\pi\epsilon_0 r}$$

so in po­lar co­or­di­nates

$${\skew0\vec r}\cdot\nabla V = r \frac{\partial V}{\partial r} = - V$$

Then ac­cord­ing to the vir­ial the­o­rem the ex­pec­ta­tion po­ten­tial en­ergy is mi­nus twice the ex­pec­ta­tion ki­netic en­ergy. And their sum, the to­tal en­ergy $E$, is then mi­nus the ex­pec­ta­tion ki­netic en­ergy. In short, $\left\langle{T}\right\rangle =-E$ and $\left\langle{V}\right\rangle =2E$ with $E$ neg­a­tive.

The vir­ial the­o­rem does not ap­ply to the par­ti­cle in a pipe, as that par­ti­cle is in a bounded space. (You can as­sume in­fi­nite space if you take the po­ten­tial in­fi­nite out­side the pipe, but ob­vi­ously by it­self that does not help much. You could as­sume in­fi­nite space with a po­ten­tial

$$V = (x/\ell_x)^p + (y/\ell_y)^p + (z/\ell_z)^p \qquad p \mbox{ even}$$

if you then take the limit $p\to\infty$ to get in­fi­nite po­ten­tial out­side the pipe and zero in­side. That gives the cor­rect but triv­ial re­sult that all the en­ergy is ki­netic.)

But the vir­ial the­o­rem does ap­ply to any num­ber of par­ti­cles, not just to one. Just sum over all the par­ti­cles:

$$2 \left\langle{{\textstyle\sum_i} T_i}\right\rangle = \left... ...extstyle\sum_i} {\skew0\vec r}_i\cdot\nabla_i V}\right\rangle$$

where i is the par­ti­cle num­ber.

For ex­am­ple, con­sider the hy­dro­gen mol­e­cule, where there are four par­ti­cles, two pro­tons and two elec­trons. Here

$$V = \sum_{i \ne j}\frac{q_i q_j}{4\pi\epsilon_0 \vert{\skew0\vec r}_i-{\skew0\vec r}_j\vert}$$

where $q_i$ is $e$ if par­ti­cle $i$ is a pro­ton and $-e$ if it is an elec­tron. Like for the sim­ple hy­dro­gen atom,

$$\left\langle{{\textstyle\sum_i} {\skew0\vec r}_i\cdot\nabla_i V}\right\rangle = - \left\langle{V}\right\rangle$$

so the to­tal ex­pec­ta­tion po­ten­tial en­ergy of the sys­tem is still twice the to­tal en­ergy $E$ and the to­tal ki­netic en­ergy is still mi­nus $E$. And this con­tin­ues to hold for much big­ger sys­tems of nu­clei and elec­trons, which is why it is of in­ter­est for com­pu­ta­tional meth­ods.

In some com­pu­ta­tions you might need to as­sume that the elec­trons are in a state of def­i­nite en­ergy, like in the ground state, but the nu­clei are not. In such com­pu­ta­tions the nu­clei are at an as­sumed po­si­tion and you will only com­pute the state of the elec­trons. So the sum­ma­tion in

$$\left\langle{{\textstyle\sum_i} {\skew0\vec r}_i \cdot \nabla_i V}\right\rangle$$

now ex­tends only over the elec­trons. But this sum­ma­tion does in­cludes po­ten­tials of the elec­trons due to the at­trac­tion by the nu­clei, and those terms are no longer equal to mi­nus the cor­re­spond­ing po­ten­tials. You may need to eval­u­ate these terms ex­plic­itly. But that is not too bad, as these po­ten­tials are now known func­tions of the in­di­vid­ual elec­tron po­si­tions only. The dif­fi­cult term, due to the elec­tron-elec­tron in­ter­ac­tion, is still given by mi­nus the cor­re­spond­ing po­ten­tial.

Fi­nally, you might won­der where the vir­ial the­o­rem comes from. Well, one way to prove the vir­ial the­o­rem, as found in quan­tum text­books and on Wikipedia, is to work out the com­mu­ta­tor in

$$\frac{{\rm d}\langle{\skew0\vec r}\cdot{\skew0\vec p}\rangl... ... i}}{\hbar}\langle[H,{\skew0\vec r}\cdot{\skew0\vec p}]\rangle$$

us­ing the for­mu­lae in chap­ter 4.5.4, to give

$$\frac{{\rm d}\langle{\skew0\vec r}\cdot{\skew0\vec p}\rangl... ...e{T}\right\rangle - \langle{\skew0\vec r}\cdot\nabla V\rangle,$$

and then note that the left hand side above is zero for sta­tion­ary states, (in other words, for states with a pre­cise to­tal en­ergy). This fol­lows the clas­si­cal way of de­riv­ing the clas­si­cal vir­ial the­o­rem, but re­quires a messy purely math­e­mat­i­cal de­riva­tion. The the­o­rem then pops up out of the com­plex math­e­mat­ics with­out any plau­si­ble phys­i­cal rea­son why there would be such a the­o­rem in the first place.

The orig­i­nal de­riva­tion by Fock in 1930 is much more phys­i­cally ap­peal­ing and more in­struc­tive. The idea is to slightly stretch the given quan­tum sys­tem: re­place every po­si­tion co­or­di­nate co­or­di­nate ${\skew0\vec r}$ by a slightly larger one ${\skew0\vec r}_{\rm s}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(1+\varepsilon){\skew0\vec r}$. Here $\varepsilon$ is as­sumed to be a van­ish­ingly small num­ber. We are in­ter­ested in what the ex­pec­ta­tion po­ten­tial and ki­netic en­ergy are in this slightly stretched sys­tem.

First how­ever, re­call that the square mag­ni­tude of the wave func­tion gives the prob­a­bil­ity of that state, and that all prob­a­bil­i­ties must in­te­grate to­gether to 1, cer­tainty. Phrased dif­fer­ently, the ex­pec­ta­tion value of one must be one; $\left\langle{1}\right\rangle =1$, what else? But clearly, if you in­te­grate the same square wave func­tion mag­ni­tude over a slightly larger do­main, you will get a value slightly greater than one. This prob­lem is eas­ily fixed, how­ever, by mul­ti­ply­ing the wave func­tion in the stretched sys­tem by a suit­able con­stant slightly less than one. Then $\left\langle{1}\right\rangle _{\rm {s}}=1$ too. (The pre­cise value of the con­stant de­pends on the num­ber of par­ti­cles and is not im­por­tant.)

Next, the ex­pec­ta­tion ki­netic en­ergy con­sists of terms like $-\hbar^2/2m_i$ times $\left\langle{\partial^2/\partial{}x_{\rm {s,i}}^2}\right\rangle$ be­cause of the form of the ki­netic en­ergy op­er­a­tor. Be­cause of the stretch­ing of the co­or­di­nate in the bot­tom of the de­riv­a­tive, each of these terms changes by a fac­tor $1/(1+\varepsilon)^2$, so

$$\left\langle{T}\right\rangle _{\rm {s}} = \left\langle{T}\r... ...t\rangle - 2 \varepsilon \left\langle{T}\right\rangle + \ldots$$

For the po­ten­tial en­ergy we can use a lin­ear Tay­lor se­ries to fig­ure out how it changes:

$$V({\skew0\vec r}_1+\varepsilon{\skew0\vec r}_1,{\skew0\vec ... ... r}_1 + \nabla_2 V \cdot \varepsilon{\skew0\vec r}_2 + \ldots$$

where in the right hand side $V$ and its de­riv­a­tives are eval­u­ated at $({\skew0\vec r}_1,{\skew0\vec r}_2,\ldots)$. From that

$$\left\langle{V}\right\rangle _s = \left\langle{V}\right\ran... ...um_i} {\skew0\vec r}_i \cdot \nabla_i V}\right\rangle + \ldots$$

From the above ex­pres­sions, it is seen that com­pared to the un­stretched sys­tem, in the stretched sys­tem the sum of ex­pec­ta­tion ki­netic and po­ten­tial en­er­gies is dif­fer­ent by an amount

$$\varepsilon \left( -2 \left\langle{T}\right\rangle + \left... ...ew0\vec r}_i \cdot \nabla_i V}\right\rangle \right) + \ldots$$

But, as de­scribed in {A.7}, if you mess up an en­ergy wave func­tion by an amount of or­der $\varepsilon$, the ex­pec­ta­tion en­ergy should only be messed up by an amount pro­por­tional to $\varepsilon^2$, not $\varepsilon$. (In brief, the amounts of the other en­ergy eigen­func­tions found in the messed up wave func­tion are pro­por­tional to $\varepsilon$. How­ever, the prob­a­bil­i­ties of their en­er­gies are pro­por­tional to the squares of the amounts.) So the fac­tor be­tween paren­the­ses in the ex­pres­sion above must be zero, and that is the vir­ial the­o­rem.