D.32 In­te­gral con­ser­va­tion laws

This sec­tion de­rives the in­te­gral con­ser­va­tion laws given in ad­den­dum {A.14}.

The rules of en­gage­ment are as fol­lows:

• The Carte­sian axes are num­bered us­ing an in­dex $i$, with $i$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, 2, and 3 for $x$, $y$, and $z$ re­spec­tively.
• Also, $r_i$ in­di­cates the co­or­di­nate in the $i$ di­rec­tion, $x$, $y$, or $z$.
• De­riv­a­tives with re­spect to a co­or­di­nate $r_i$ are in­di­cated by a sim­ple sub­script $i$.
• Time de­riv­a­tives are in­di­cated by a sub­script t.
• A bare $\int$ in­te­gral sign is as­sumed to be an in­te­gra­tion over all space, or over the en­tire box for par­ti­cles in a box. The ${\rm d}^3{\skew0\vec r}$ is nor­mally omit­ted for brevity and to be un­der­stood.
• A su­per­script $^*$ in­di­cates a com­plex con­ju­gate.

First it will be shown that ac­cord­ing to the Schrö­din­ger equa­tion $\int\vert\Psi\vert^2$ is con­stant. The Schrö­din­ger equa­tion in free space is

$${\rm i}\hbar \frac{\partial\Psi}{\partial t} = - \frac{\hbar^2}{2m} \nabla^2 \Psi$$

Tak­ing the right hand term to the other side and writ­ing it in in­dex no­ta­tion gives

$${\rm i}\hbar \frac{\partial\Psi}{\partial t} + \sum_i \frac{\hbar^2}{2m} \Psi_{ii} = 0$$

Mul­ti­ply the left hand side by $\Psi^*$$\raisebox{.5pt}{$/$}$​${\rm i}\hbar$ and add the com­plex con­ju­gate of the same equa­tion to get

$$\Psi^* \frac{\partial\Psi}{\partial t} + \Psi \frac{\parti... ...c{\hbar}{2m{\rm i}} (\Psi^* \Psi_{ii} - \Psi \Psi^*_{ii}) = 0$$

To show that the in­te­gral $\int\vert\Psi\vert^2$ is con­stant, it must be shown that its time de­riv­a­tive is zero. Now the first two terms above are the time de­riv­a­tive of $\vert\Psi\vert^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\Psi^*\Psi$. So in­te­grated over all space, they give the time de­riv­a­tive that must be shown to be zero. And the equa­tion above shows that it is in­deed zero pro­vided that the re­main­ing sum in it in­te­grates to zero over all space.

The con­stant is not im­por­tant in show­ing that this is true, so just ex­am­ine for any $i$

$$\int (\Psi^* \Psi_{ii} - \Psi \Psi^*_{ii})$$

This equals

$$\int (\Psi^* \Psi_i - \Psi \Psi^*_i)_i$$

as can be seen by dif­fer­en­ti­at­ing out the par­en­thet­i­cal ex­pres­sion with re­spect to $r_i$. The above in­te­grand can be in­te­grated with re­spect to $r_i$. It will then be zero for a pe­ri­odic box since the ex­pres­sion in paren­the­sis is the same at the up­per and lower lim­its of in­te­gra­tion. It will also be zero for an im­pen­e­tra­ble con­tainer, since $\Psi$ will then be zero on the sur­face of the con­tainer. It will also be zero in an in­fi­nite re­gion pro­vided that $\Psi$ and its de­riv­a­tives van­ish at large dis­tances.

There is an­other way to see that $\int\vert\Psi\vert^2$ is con­stant. First re­call that any so­lu­tion of the Schrö­din­ger equa­tion takes the form

$$\Psi = \sum_n c_n e^{-{\rm i}E_n t} \psi_n({\skew0\vec r})$$

Here the $\psi_n$ are the en­ergy eigen­func­tions. Then

$$\int \vert\Psi\vert^2 = \int \Psi^*\Psi = \int \sum_{\unde... ...0\vec r}) \sum_n c_n e^{-{\rm i}E_n t} \psi_n({\skew0\vec r})$$

Now be­cause of or­tho­nor­mal­ity of the eigen­func­tions, the in­te­gra­tion only pro­duces a nonzero re­sult when ${\underline n}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n$, and then the prod­uct of the eigen­func­tions in­te­grates to 1. So

$$\int \vert\Psi\vert^2 = \sum_n c_n^* c_n$$

That does not de­pend on time, and the nor­mal­iza­tion re­quire­ment makes it 1.

This also clar­i­fies what goes wrong with the Klein-Gor­don equa­tion. For the Klein-Gor­don equa­tion

$$\Psi = \sum_n c_n e^{-{\rm i}E_n t} \psi_n({\skew0\vec r}) + \sum_n d_n e^{{\rm i}E_n t} \psi_n({\skew0\vec r})$$

The first sum are the par­ti­cle states and the sec­ond sum the an­tipar­ti­cle states. That gives:

$$\int\vert\Psi\vert^2 = \sum_n \left( c_n^* c_n + d_n^* d_n ... ...* d_n e^{2{\rm i}E_n t} + d_n^* c_n e^{-2{\rm i}E_n t} \right)$$

The fi­nal two terms in the sum os­cil­late in time. So the in­te­gral is no longer con­stant.

The ex­cep­tion is if there are only par­ti­cle states (no $d_n$) or only an­tipar­ti­cle states (no $c_n$). In those two cases, the in­te­gral is con­stant. In gen­eral

$$\Psi = \Psi_1 + \Psi_2 \qquad \Psi_1 = \sum_n c_n e^{-{\rm... ...ad \Psi_2 = \sum_n d_n e^{{\rm i}E_n t} \psi_n({\skew0\vec r})$$

where the in­te­grated square mag­ni­tudes of $\Psi_1$ and $\Psi_2$ are con­stant.

Next it will be shown that the re­arranged Klein-Gor­don equa­tion

$$\frac{1}{c^2} \frac{\partial^2\Psi}{\partial t^2} - \sum_i \Psi_{ii} + \left(\frac{mc^2}{\hbar c}\right)^2 \Psi = 0$$

pre­serves the sum of in­te­grals

$$\int \left\vert \frac{1}{c} \frac{\partial\Psi}{\partial t}... ...t^2 + \int \left\vert \frac{mc^2}{\hbar c} \Psi \right\vert^2$$

To do so it suf­fices to show that the sum of the time de­riv­a­tives of the three in­te­grals is zero. That can be done by mul­ti­ply­ing the Klein-Gor­don equa­tion by $\partial\Psi^*$$\raisebox{.5pt}{$/$}$​$\partial{t}$, adding the com­plex con­ju­gate of the ob­tained equa­tion, and in­te­grat­ing over all space. Each of the three terms in the Klein-Gor­don equa­tion will then give one of the three needed time de­riv­a­tives. So their sum will in­deed be zero.

To check that, look at what each term in the Klein-Gor­don equa­tion pro­duces sep­a­rately. The first term gives

$$\int \frac{1}{c^2} \left( \frac{\partial \Psi^*}{\partial... ...si}{\partial t} \frac{\partial^2\Psi^*}{\partial t^2} \right)$$

or tak­ing one time de­riv­a­tive out­side the in­te­gral, that is

$$\frac{1}{c^2} \frac{{\rm d}}{{\rm d}t} \int \frac{\partial \Psi^*}{\partial t} \frac{\partial\Psi}{\partial t}$$

That is the first needed time de­riv­a­tive, since a num­ber times its com­plex con­ju­gate is the square mag­ni­tude of that num­ber.

The sec­ond term in the Klein-Gor­don equa­tion pro­duces

$$- \sum_i \int \left( \frac{\partial \Psi^*}{\partial t} \Psi_{ii} + \frac{\partial \Psi}{\partial t} \Psi^*_{ii} \right)$$

That equals

$$- \sum_i \int \left( \frac{\partial \Psi^*}{\partial t} \... ...ght)_i + \sum_i \frac{{\rm d}}{{\rm d}t} \int \Psi_i^* \Psi_i$$

as can be seen by dif­fer­en­ti­at­ing out the par­en­thet­i­cal ex­pres­sion in the first in­te­gral with re­spect to $r_i$ and bring­ing the time de­riv­a­tive in the sec­ond term in­side the in­te­gral. The first in­te­gral above is zero for a pe­ri­odic box, for an im­pen­e­tra­ble con­tainer, and for in­fi­nite space for the same rea­sons as given in the de­riva­tion for the Schrö­din­ger equa­tion. The sec­ond term above is the needed sec­ond time de­riv­a­tive.

The fi­nal of the three terms in the Klein-Gor­don equa­tion pro­duces

$$\left(\frac{mc^2}{\hbar c}\right)^2 \int \left( \frac{\pa... ...al t} \Psi + \frac{\partial \Psi}{\partial t} \Psi^* \right)$$

That equals

$$\left(\frac{mc^2}{\hbar c}\right)^2 \frac{{\rm d}}{{\rm d}t} \int \Psi^* \Psi$$

as can be seen by bring­ing the time de­riv­a­tive in­side the in­te­gral. This is the last of the three needed time de­riv­a­tives.