A.32 The evo­lu­tion of prob­a­bil­ity

This note looks at con­ser­va­tion of prob­a­bil­ity, and the re­sult­ing de­f­i­n­i­tions of the re­flec­tion and trans­mis­sion co­ef­fi­cients in scat­ter­ing. It also ex­plains the con­cept of the “prob­a­bil­ity cur­rent” that you may oc­ca­sion­ally run into.

For the un­steady Schrö­din­ger equa­tion to pro­vide a phys­i­cally cor­rect de­scrip­tion of non­rel­a­tivis­tic quan­tum me­chan­ics, par­ti­cles should not be able to dis­ap­pear into thin air. In par­tic­u­lar, dur­ing the evo­lu­tion of the wave func­tion of a sin­gle par­ti­cle, the to­tal prob­a­bil­ity of find­ing the par­ti­cle if you look every­where should stay one at all times:

$$\int_{x=-\infty}^{\infty} \vert\Psi\vert^2 { \rm d}x = 1 \mbox{ at all times}$$

For­tu­nately, the Schrö­din­ger equa­tion

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

does in­deed con­serve this to­tal prob­a­bil­ity, so all is well.

To ver­ify this, note first that $\vert\Psi\vert^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\Psi^*\Psi$, where the star in­di­cates the com­plex con­ju­gate, so

$$\frac{\partial\vert\Psi\vert^2}{\partial t} = \Psi^*\frac{... ...ial\Psi}{\partial t} + \Psi\frac{\partial\Psi^*}{\partial t}$$

To get an ex­pres­sion for that, take the Schrö­din­ger equa­tion above times $\Psi^*$$\raisebox{.5pt}{$/$}$​${\rm i}\hbar$ and add the com­plex con­ju­gate of the Schrö­din­ger equa­tion,

$$- {\rm i}\hbar \frac{\partial\Psi^*}{\partial t} = - \frac{\hbar^2}{2m} \frac{\partial^2\Psi^*}{\partial x^2} + V \Psi^*,$$

times $\vphantom{0}\raisebox{1.5pt}{$-$}$$\Psi$$\raisebox{.5pt}{$/$}$​${\rm i}\hbar$. The po­ten­tial en­ergy terms drop out, and what is left is

$$\frac{\partial\vert\Psi\vert^2}{\partial t} = \frac{{\rm i... ...al x^2} - \Psi\frac{\partial^2\Psi^*}{\partial x^2} \right).$$

Now it can be ver­i­fied by dif­fer­en­ti­at­ing out that the right hand side can be rewrit­ten as a de­riv­a­tive:

$$\frac{\partial\vert\Psi\vert^2}{\partial t} = - \frac{\part... ...partial x} - \Psi^*\frac{\partial\Psi}{\partial x} \right) %$$ (A.232)

For rea­sons that will be­come ev­i­dent be­low, $J$ is called the “prob­a­bil­ity cur­rent.” Note that $J$, like $\Psi$, will be zero at in­fi­nite $x$ for proper, nor­mal­ized wave func­tions.

If (A.232) is in­te­grated over all $x$, the de­sired re­sult is ob­tained:

$$\frac{{\rm d}}{{\rm d}t} \int_{x=-\infty}^{\infty}\vert\Psi\vert^2 { \rm d}x = - J\Big\vert _{x=-\infty}^{\infty} = 0.$$

There­fore, the to­tal prob­a­bil­ity of find­ing the par­ti­cle does not change with time. If a proper ini­tial con­di­tion is pro­vided to the Schrö­din­ger equa­tion in which the to­tal prob­a­bil­ity of find­ing the par­ti­cle is one, then it stays one for all time.

It gets a lit­tle more in­ter­est­ing to see what hap­pens to the prob­a­bil­ity of find­ing the par­ti­cle in some given fi­nite re­gion $a$ $\raisebox{-.3pt}{$\leqslant$}$ $x$ $\raisebox{-.3pt}{$\leqslant$}$ $b$. That prob­a­bil­ity is given by

$$\int_{x=a}^b \vert\Psi\vert^2 { \rm d}x$$

and it can change with time. A wave packet might en­ter or leave the re­gion. In par­tic­u­lar, in­te­gra­tion of (A.232) gives

$$\frac{{\rm d}}{{\rm d}t} \int_{x=a}^b\vert\Psi\vert^2 { \rm d}x = J_a - J_b$$

This can be un­der­stood as fol­lows: $J_a$ is the prob­a­bil­ity flow­ing out of the re­gion $x$ $\raisebox{.3pt}{$<$}$ $a$ into the in­ter­val $[a,b]$ through the end $a$. That in­creases the prob­a­bil­ity within $[a,b]$. Sim­i­larly, $J_b$ is the prob­a­bil­ity flow­ing out of $[a,b]$ at $b$ into the re­gion $x$ $\raisebox{.3pt}{$>$}$ $b$; it de­creases the prob­a­bil­ity within $[a,b]$. Now you see why $J$ is called prob­a­bil­ity cur­rent; it is equiv­a­lent to a stream of prob­a­bil­ity in the pos­i­tive $x$-​di­rec­tion.

The prob­a­bil­ity cur­rent can be gen­er­al­ized to more di­men­sions us­ing vec­tor cal­cu­lus:

$$\vec J = \frac{{\rm i}\hbar}{2m} \left(\Psi\nabla\Psi^* - \Psi^*\nabla\Psi \right) %$$ (A.233)

and the net prob­a­bil­ity flow­ing out of a re­gion is given by

$$\int \vec J \cdot {\vec n}{ \rm d}A %$$ (A.234)

where $A$ is the out­side sur­face area of the re­gion, and ${\vec n}$ is a unit vec­tor nor­mal to the sur­face. A sur­face in­te­gral like this can of­ten be sim­pli­fied us­ing the di­ver­gence (Gauss or what­ever) the­o­rem of cal­cu­lus.

Re­turn­ing to the one-di­men­sion­al case, it is of­ten de­sir­able to re­late con­ser­va­tion of prob­a­bil­ity to the en­ergy eigen­func­tions of the Hamil­ton­ian,

$$- \frac{\hbar^2}{2m} \frac{{\rm d}^2\psi}{{\rm d}x^2} + V \psi = E \psi$$

be­cause the en­ergy eigen­func­tions are generic, not spe­cific to one par­tic­u­lar ex­am­ple wave func­tion $\Psi$.

To do so, first an im­por­tant quan­tity called the “Wron­skian” must be in­tro­duced. Con­sider any two eigen­func­tions $\psi_1$ and $\psi_2$ of the Hamil­ton­ian:

$$\begin{eqnarray*} & \displaystyle - \frac{\hbar^2}{2m} \frac{{\rm d}^2\psi_1}{... ...2}{2m} \frac{{\rm d}^2\psi_2}{{\rm d}x^2} + V \psi_2 = E \psi_2 \end{eqnarray*}$$

If you mul­ti­ply the first equa­tion above by $\psi_2$, the sec­ond by $\psi_1$ and then sub­tract the two, you get

$$\frac{\hbar^2}{2m} \left( \psi_1\frac{{\rm d}^2\psi_2}{{\... ...}x^2} - \psi_2\frac{{\rm d}^2\psi_1}{{\rm d}x^2} \right) = 0$$

The con­stant $\hbar^2$$\raisebox{.5pt}{$/$}$​$2m$ can be di­vided out, and by dif­fer­en­ti­a­tion it can be ver­i­fied that the re­main­der can be writ­ten as

$$\frac{{\rm d}W}{{\rm d}x} = 0 \qquad \mbox{where } W = \p... ...\rm d}\psi_2}{{\rm d}x} - \psi_2\frac{{\rm d}\psi_1}{{\rm d}x}$$

The quan­tity $W$ is called the Wron­skian. It is the same at all val­ues of $x$.

As an ap­pli­ca­tion, con­sider the ex­am­ple po­ten­tial of fig­ure A.11 in ad­den­dum {A.27} that bounces a par­ti­cle com­ing in from the far left back to where it came from. In the left re­gion, the po­ten­tial $V$ has a con­stant value $V_{\rm {l}}$. In this re­gion, an en­ergy eigen­func­tion is of the form

$$\psi_E = C^{\rm {l}}_{\rm {f}} e^{{\rm i}p_{\rm {c}}^{\rm {... ...ad\mbox{where } p_{\rm {c}}^{\rm {l}}=\sqrt{2m(E-V_{\rm {l}})}$$

At the far right, the po­ten­tial grows with­out bound and the eigen­func­tion be­comes zero rapidly. To make use of the Wron­skian, take the first so­lu­tion $\psi_1$ to be $\psi_E$ it­self, and $\psi_2$ to be its com­plex con­ju­gate $\psi_E^*$. Since at the far right the eigen­func­tion be­comes zero rapidly, the Wron­skian is zero there. And since the Wron­skian is con­stant, that means it must be zero every­where. Next, if you plug the above ex­pres­sion for the eigen­func­tion in the left re­gion into the de­f­i­n­i­tion of the Wron­skian and clean up, you get

$$W = \frac{2{\rm i}p_{\rm {c}}^{\rm {l}}}{\hbar} \left(\ver... ...}}_{\rm {b}}\vert^2-\vert C^{\rm {l}}_{\rm {f}}\vert^2\right).$$

If that is zero, the mag­ni­tude of $C^{\rm {l}}_{\rm {b}}$ must be the same as that of $C^{\rm {l}}_{\rm {f}}$.

This can be un­der­stood as fol­lows: if a wave packet is cre­ated from eigen­func­tions with ap­prox­i­mately the same en­ergy, then the terms $C^{\rm {l}}_{\rm {f}}e^{{{\rm i}}p_{\rm {c}}^{\rm {l}}x/\hbar}$ com­bine for large neg­a­tive times into a wave packet com­ing in from the far left. The prob­a­bil­ity of find­ing the par­ti­cle in that wave packet is pro­por­tional to the in­te­grated square mag­ni­tude of the wave func­tion, hence pro­por­tional to the square mag­ni­tude of $C^{\rm {l}}_{\rm {f}}$. For large pos­i­tive times, the $C^{\rm {l}}_{\rm {b}}e^{-{{\rm i}}p_{\rm {c}}^{\rm {l}}x/\hbar}$ terms com­bine in a sim­i­lar wave packet, but one that re­turns to­wards the far left. The prob­a­bil­ity of find­ing the par­ti­cle in that de­part­ing wave packet must still be the same as that for the in­com­ing packet, so the square mag­ni­tude of $C^{\rm {l}}_{\rm {b}}$ must be the same as that of $C^{\rm {l}}_{\rm {f}}$.

Next con­sider a generic scat­ter­ing po­ten­tial like the one in fig­ure 7.22. To the far left, the eigen­func­tion is again of the form

$$\psi_E = C^{\rm {l}}_{\rm {f}} e^{{\rm i}p_{\rm {c}}^{\rm {... ...ad\mbox{where } p_{\rm {c}}^{\rm {l}}=\sqrt{2m(E-V_{\rm {l}})}$$

while at the far right it is now of the form

$$\psi_E = C^{\rm {r}} e^{{\rm i}p_{\rm {c}}^{\rm {r}}x/\hbar... ...ad\mbox{where } p_{\rm {c}}^{\rm {r}}=\sqrt{2m(E-V_{\rm {r}})}$$

The Wron­skian can be found the same way as be­fore:

$$W = \frac{2{\rm i}p_{\rm {c}}^{\rm {l}}}{\hbar} \left(\ver... ...2{\rm i}p_{\rm {c}}^{\rm {r}}}{\hbar} \vert C^{\rm {r}}\vert^2$$

The frac­tion of the in­com­ing wave packet that ends up be­ing re­flected back to­wards the far left is called the “re­flec­tion co­ef­fi­cient” $R$. Fol­low­ing the same rea­son­ing as above, it can be com­puted from the co­ef­fi­cients in the far left re­gion of con­stant po­ten­tial as:

$$R = \frac{\vert C^{\rm {l}}_{\rm {b}}\vert^2}{\vert C^{\rm {l}}_{\rm {f}}\vert^2}$$

The re­flec­tion co­ef­fi­cient gives the prob­a­bil­ity that the par­ti­cle can be found to the left of the scat­ter­ing re­gion at large times.

Sim­i­larly, the frac­tion of the in­com­ing wave packet that passes through the po­ten­tial bar­rier to­wards the far right is called the “trans­mis­sion co­ef­fi­cient” $T$. It gives the prob­a­bil­ity that the par­ti­cle can be found to the right of the scat­ter­ing re­gion at large times. Be­cause of con­ser­va­tion of prob­a­bil­ity, $T$ $\vphantom0\raisebox{1.5pt}{$=$}$ $1-R$.

Al­ter­na­tively, be­cause of the Wron­skian ex­pres­sion above, the trans­mis­sion co­ef­fi­cient can be ex­plic­itly com­puted from the co­ef­fi­cient of the eigen­func­tion in the far right re­gion as

$$T = \frac{p_{\rm {c}}^{\rm {r}}\vert C^{\rm {r}}\vert^2} {... ...m {l}})} \quad p_{\rm {c}}^{\rm {r}}=\sqrt{2m(E-V_{\rm {r}})}$$

If the po­ten­tial en­ergy is the same at the far right and far left, the two clas­si­cal mo­menta are the same, $p_{\rm {c}}^{\rm {r}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $p_{\rm {c}}^{\rm {l}}$. Oth­er­wise, the rea­son that the ra­tio of clas­si­cal mo­menta ap­pears in the trans­mis­sion co­ef­fi­cient is be­cause the clas­si­cal mo­menta in a wave packet have a dif­fer­ent spac­ing with re­spect to en­ergy if the po­ten­tial en­ergy is dif­fer­ent. (The above ex­pres­sion for the trans­mis­sion co­ef­fi­cient can also be de­rived ex­plic­itly us­ing the Par­se­val equal­ity of Fourier analy­sis, in­stead of in­ferred from con­ser­va­tion of prob­a­bil­ity and the con­stant Wron­skian.)