diff options
Diffstat (limited to 'source/know/concept/selection-rules')
| -rw-r--r-- | source/know/concept/selection-rules/index.md | 155 |
1 files changed, 78 insertions, 77 deletions
diff --git a/source/know/concept/selection-rules/index.md b/source/know/concept/selection-rules/index.md index 84227ba..373486e 100644 --- a/source/know/concept/selection-rules/index.md +++ b/source/know/concept/selection-rules/index.md @@ -10,33 +10,33 @@ layout: "concept" In quantum mechanics, it is often necessary to evaluate matrix elements of the following form, -where $\ell$ and $m$ respectively represent -the total angular momentum and its $z$-component: +where $$\ell$$ and $$m$$ respectively represent +the total angular momentum and its $$z$$-component: $$\begin{aligned} \matrixel{f}{\hat{O}}{i} = \matrixel{\ell_f m_f}{\hat{O}}{\ell_i m_i} \end{aligned}$$ -Where $\hat{O}$ is an operator, $\Ket{i}$ is an initial state, and -$\Ket{f}$ is a final state (usually at least; $\Ket{i}$ and $\Ket{f}$ +Where $$\hat{O}$$ is an operator, $$\Ket{i}$$ is an initial state, and +$$\Ket{f}$$ is a final state (usually at least; $$\Ket{i}$$ and $$\Ket{f}$$ can be any states). **Selection rules** are requirements on the relations -between $\ell_i$, $\ell_f$, $m_i$ and $m_f$, which, if not met, +between $$\ell_i$$, $$\ell_f$$, $$m_i$$ and $$m_f$$, which, if not met, guarantee that the above matrix element is zero. ## Parity rules -Let $\hat{O}$ denote any operator which is odd under spatial inversion +Let $$\hat{O}$$ denote any operator which is odd under spatial inversion (parity): $$\begin{aligned} \hat{\Pi}^\dagger \hat{O} \hat{\Pi} = - \hat{O} \end{aligned}$$ -Where $\hat{\Pi}$ is the parity operator. -We wrap this property of $\hat{O}$ -in the states $\Ket{\ell_f m_f}$ and $\Ket{\ell_i m_i}$: +Where $$\hat{\Pi}$$ is the parity operator. +We wrap this property of $$\hat{O}$$ +in the states $$\Ket{\ell_f m_f}$$ and $$\Ket{\ell_i m_i}$$: $$\begin{aligned} \matrixel{\ell_f m_f}{\hat{O}}{\ell_i m_i} @@ -48,7 +48,7 @@ $$\begin{aligned} \end{aligned}$$ Which clearly can only be true if the exponent is even, -so $\Delta \ell \equiv \ell_f - \ell_i$ must be odd. +so $$\Delta \ell \equiv \ell_f - \ell_i$$ must be odd. This leads to the following selection rule, often referred to as **Laporte's rule**: @@ -60,9 +60,9 @@ $$\begin{aligned} If this is not the case, then the only possible way that the above equation can be satisfied -is if the matrix element vanishes $\matrixel{\ell_f m_f}{\hat{O}}{\ell_i m_i} = 0$. +is if the matrix element vanishes $$\matrixel{\ell_f m_f}{\hat{O}}{\ell_i m_i} = 0$$. We can derive an analogous rule for -any operator $\hat{E}$ which is even under parity: +any operator $$\hat{E}$$ which is even under parity: $$\begin{aligned} \hat{\Pi}^\dagger \hat{E} \hat{\Pi} = \hat{E} @@ -76,10 +76,10 @@ $$\begin{aligned} ## Dipole rules Arguably the most common operator found in such matrix elements -is a position vector operator, like $\vu{r}$ or $\hat{x}$, +is a position vector operator, like $$\vu{r}$$ or $$\hat{x}$$, and the associated selection rules are known as **dipole rules**. -For the $z$-component of angular momentum $m$ we have the following: +For the $$z$$-component of angular momentum $$m$$ we have the following: $$\begin{aligned} \boxed{ @@ -92,7 +92,7 @@ $$\begin{aligned} <label for="proof-dipole-m">Proof</label> <div class="hidden" markdown="1"> <label for="proof-dipole-m">Proof.</label> -We know that the angular momentum $z$-component operator $\hat{L}_z$ satisfies: +We know that the angular momentum $$z$$-component operator $$\hat{L}_z$$ satisfies: $$\begin{aligned} \comm{\hat{L}_z}{\hat{x}} = i \hbar \hat{y} @@ -103,7 +103,7 @@ $$\begin{aligned} \end{aligned}$$ We take the first relation, -and wrap it in $\Bra{\ell_f m_f}$ and $\Ket{\ell_i m_i}$, giving: +and wrap it in $$\Bra{\ell_f m_f}$$ and $$\Ket{\ell_i m_i}$$, giving: $$\begin{aligned} i \hbar \matrixel{\ell_f m_f}{\hat{y}}{\ell_i m_i} @@ -114,7 +114,7 @@ $$\begin{aligned} &= \hbar (m_f - m_i) \matrixel{\ell_f m_f}{\hat{x}}{\ell_i m_i} \end{aligned}$$ -Next, we do the same thing with the second relation, for $[\hat{L}_z, \hat{y}]$, giving: +Next, we do the same thing with the second relation, for $$[\hat{L}_z, \hat{y}]$$, giving: $$\begin{aligned} - i \hbar \matrixel{\ell_f m_f}{\hat{x}}{\ell_i m_i} @@ -125,7 +125,7 @@ $$\begin{aligned} &= \hbar (m_f - m_i) \matrixel{\ell_f m_f}{\hat{y}}{\ell_i m_i} \end{aligned}$$ -Respectively isolating the two above results for $\hat{x}$ and $\hat{y}$, +Respectively isolating the two above results for $$\hat{x}$$ and $$\hat{y}$$, we arrive at these equations: $$\begin{aligned} @@ -144,11 +144,11 @@ $$\begin{aligned} &= (m_f - m_i)^2 \matrixel{\ell_f m_f}{\hat{y}}{\ell_i m_i} \end{aligned}$$ -This can only be true if $\Delta m = \pm 1$, -unless the inner products of $\hat{x}$ and $\hat{y}$ are zero, -in which case we cannot say anything about $\Delta m$ yet. +This can only be true if $$\Delta m = \pm 1$$, +unless the inner products of $$\hat{x}$$ and $$\hat{y}$$ are zero, +in which case we cannot say anything about $$\Delta m$$ yet. Assuming the latter, we take the inner product of -the commutator $\comm{\hat{L}_z}{\hat{z}} = 0$, and find: +the commutator $$\comm{\hat{L}_z}{\hat{z}} = 0$$, and find: $$\begin{aligned} 0 @@ -159,17 +159,17 @@ $$\begin{aligned} &= \hbar (m_f - m_i) \matrixel{\ell_f m_f}{\hat{z}}{\ell_i m_i} \end{aligned}$$ -If $\matrixel{f}{\hat{z}}{i} \neq 0$, we require $\Delta m = 0$. -The previous requirement was $\Delta m = \pm 1$, -implying that $\matrixel{f}{\hat{x}}{i} = \matrixel{f}{\hat{y}}{i} = 0$ -whenever $\matrixel{f}{\hat{z}}{i} \neq 0$. -Only if $\matrixel{f}{\hat{z}}{i} = 0$ -does the previous rule $\Delta m = \pm 1$ hold, -in which case the inner products of $\hat{x}$ and $\hat{y}$ are nonzero. +If $$\matrixel{f}{\hat{z}}{i} \neq 0$$, we require $$\Delta m = 0$$. +The previous requirement was $$\Delta m = \pm 1$$, +implying that $$\matrixel{f}{\hat{x}}{i} = \matrixel{f}{\hat{y}}{i} = 0$$ +whenever $$\matrixel{f}{\hat{z}}{i} \neq 0$$. +Only if $$\matrixel{f}{\hat{z}}{i} = 0$$ +does the previous rule $$\Delta m = \pm 1$$ hold, +in which case the inner products of $$\hat{x}$$ and $$\hat{y}$$ are nonzero. </div> </div> -Meanwhile, for the total angular momentum $\ell$ we have the following: +Meanwhile, for the total angular momentum $$\ell$$ we have the following: $$\begin{aligned} \boxed{ @@ -195,7 +195,7 @@ $$\begin{aligned} <label for="proof-dipole-l-comm">Proof</label> <div class="hidden" markdown="1"> <label for="proof-dipole-l-comm">Proof.</label> -To begin with, we want to find the commutator of $\hat{L}^2$ and $\hat{x}$: +To begin with, we want to find the commutator of $$\hat{L}^2$$ and $$\hat{x}$$: $$\begin{aligned} \comm{\hat{L}^2}{\hat{x}} @@ -263,7 +263,7 @@ $$\begin{aligned} &= 2 i \hbar (\hat{y} \hat{L}_z - \hat{z} \hat{L}_y - i \hbar \hat{x}) \end{aligned}$$ -Repeating this process for $\comm{\hat{L}^2}{\hat{y}}$ and $\comm{\hat{L}^2}{\hat{z}}$, +Repeating this process for $$\comm{\hat{L}^2}{\hat{y}}$$ and $$\comm{\hat{L}^2}{\hat{z}}$$, we find analogous expressions: $$\begin{aligned} @@ -274,7 +274,7 @@ $$\begin{aligned} &= 2 i \hbar (\hat{x} \hat{L}_y - \hat{y} \hat{L}_x - i \hbar \hat{z}) \end{aligned}$$ -Next, we take the commutator with $\hat{L}^2$ of the commutator we just found: +Next, we take the commutator with $$\hat{L}^2$$ of the commutator we just found: $$\begin{aligned} \comm{\hat{L}^2}{\comm{\hat{L}^2}{\hat{x}}} @@ -285,7 +285,7 @@ $$\begin{aligned} - i \hbar \comm{\hat{L}^2}{\hat{x}} \big) \end{aligned}$$ -Where we used that $\comm{\hat{L}^2}{\hat{L}_y} = \comm{\hat{L}^2}{\hat{L}_z} = 0$. +Where we used that $$\comm{\hat{L}^2}{\hat{L}_y} = \comm{\hat{L}^2}{\hat{L}_z} = 0$$. The other commutators look familiar: $$\begin{aligned} @@ -305,8 +305,8 @@ $$\begin{aligned} \end{aligned}$$ Substituting the well-known commutators -$i \hbar \hat{L}_y = \comm{\hat{L}_z}{\hat{L}_x}$ and -$i \hbar \hat{L}_z = \comm{\hat{L}_x}{\hat{L}_y}$: +$$i \hbar \hat{L}_y = \comm{\hat{L}_z}{\hat{L}_x}$$ and +$$i \hbar \hat{L}_z = \comm{\hat{L}_x}{\hat{L}_y}$$: $$\begin{aligned} \comm{\hat{L}^2}{\comm{\hat{L}^2}{\hat{x}}} @@ -320,7 +320,7 @@ $$\begin{aligned} + 2 \hbar^2 \big( \hat{L}^2 \hat{x} - \hat{x} \hat{L}^2 \big) \end{aligned}$$ -By definition, $\hat{L}_x^2 + \hat{L}_y^2 + \hat{L}_z^2 = \hat{L}^2$, +By definition, $$\hat{L}_x^2 + \hat{L}_y^2 + \hat{L}_z^2 = \hat{L}^2$$, which we use to arrive at: $$\begin{aligned} @@ -343,12 +343,12 @@ $$\begin{aligned} = 0 \end{aligned}$$ -Where $\vu{L} = \vu{r} \cross \vu{p}$ by definition, +Where $$\vu{L} = \vu{r} \cross \vu{p}$$ by definition, and the cross product of a vector with itself is zero. This process can be repeated for -$\comm{\hat{L}^2}{\comm{\hat{L}^2}{\hat{y}}}$ and -$\comm{\hat{L}^2}{\comm{\hat{L}^2}{\hat{z}}}$, +$$\comm{\hat{L}^2}{\comm{\hat{L}^2}{\hat{y}}}$$ and +$$\comm{\hat{L}^2}{\comm{\hat{L}^2}{\hat{z}}}$$, leading us to: $$\begin{aligned} @@ -362,13 +362,13 @@ $$\begin{aligned} &= 2 \hbar^2 (\hat{z} \hat{L}^2 + \hat{L}^2 \hat{z}) \end{aligned}$$ -At last, this brings us to the desired equation for $\comm{\hat{L}^2}{\comm{\hat{L}^2}{\vu{r}}}$, -with $\vu{r} = (\hat{x}, \hat{y}, \hat{z})$. +At last, this brings us to the desired equation for $$\comm{\hat{L}^2}{\comm{\hat{L}^2}{\vu{r}}}$$, +with $$\vu{r} = (\hat{x}, \hat{y}, \hat{z})$$. </div> </div> -We then multiply this relation by $\Bra{f} = \Bra{\ell_f m_f}$ on the left -and $\Ket{i} = \Ket{\ell_i m_i}$ on the right, +We then multiply this relation by $$\Bra{f} = \Bra{\ell_f m_f}$$ on the left +and $$\Ket{i} = \Ket{\ell_i m_i}$$ on the right, so the right-hand side becomes: $$\begin{aligned} @@ -450,21 +450,22 @@ $$\begin{aligned} &= \big( (\ell_f + \ell_i + 1)^2 - 1 \big) \big( (\ell_f - \ell_i)^2 - 1 \big) \end{aligned}$$ -The first factor is zero if $\ell_f = \ell_i = 0$, -in which case the matrix element $\matrixel{f}{\vu{r}}{i} = 0$ anyway. +The first factor is zero if $$\ell_f = \ell_i = 0$$, +in which case the matrix element $$\matrixel{f}{\vu{r}}{i} = 0$$ anyway. The other, non-trivial option is therefore: $$\begin{aligned} (\ell_f - \ell_i)^2 = 1 \end{aligned}$$ + </div> </div> ## Rotational rules -Given a general (pseudo)scalar operator $\hat{s}$, +Given a general (pseudo)scalar operator $$\hat{s}$$, which, by nature, must satisfy the following relations with the angular momentum operators: @@ -476,8 +477,8 @@ $$\begin{aligned} \comm{\hat{L}_{\pm}}{\hat{s}} = 0 \end{aligned}$$ -Where $\hat{L}_\pm \equiv \hat{L}_x \pm i \hat{L}_y$. -The inner product of any such $\hat{s}$ must obey these selection rules: +Where $$\hat{L}_\pm \equiv \hat{L}_x \pm i \hat{L}_y$$. +The inner product of any such $$\hat{s}$$ must obey these selection rules: $$\begin{aligned} \boxed{ @@ -490,9 +491,9 @@ $$\begin{aligned} \end{aligned}$$ It is common to write this in the following more complete way, where -$\matrixel{\ell_f}{|\hat{s}|}{\ell_i}$ is the **reduced matrix element**, -which is identical to $\matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i}$, but -with a different notation to say that it does not depend on $m_f$ or $m_i$: +$$\matrixel{\ell_f}{|\hat{s}|}{\ell_i}$$ is the **reduced matrix element**, +which is identical to $$\matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i}$$, but +with a different notation to say that it does not depend on $$m_f$$ or $$m_i$$: $$\begin{aligned} \boxed{ @@ -506,7 +507,7 @@ $$\begin{aligned} <label for="proof-rot-scalar">Proof</label> <div class="hidden" markdown="1"> <label for="proof-rot-scalar">Proof.</label> -Firstly, we look at the commutator of $\hat{s}$ with the $z$-component $\hat{L}_z$: +Firstly, we look at the commutator of $$\hat{s}$$ with the $$z$$-component $$\hat{L}_z$$: $$\begin{aligned} 0 = \matrixel{\ell_f m_f}{\comm{\hat{L}_z}{\hat{s}}}{\ell_i m_i} @@ -515,10 +516,10 @@ $$\begin{aligned} &= \hbar (m_f - m_i) \matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i} \end{aligned}$$ -Which can only be true if $m_f \!-\! m_i = 0$, unless, -of course, $\matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i} = 0$ by itself. +Which can only be true if $$m_f \!-\! m_i = 0$$, unless, +of course, $$\matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i} = 0$$ by itself. -Secondly, we look at the commutator of $\hat{s}$ with the total angular momentum $\hat{L}^2$: +Secondly, we look at the commutator of $$\hat{s}$$ with the total angular momentum $$\hat{L}^2$$: $$\begin{aligned} 0 @@ -528,7 +529,7 @@ $$\begin{aligned} &= \hbar^2 \big( \ell_f (\ell_f \!+\! 1) - \ell_i (\ell_i \!+\! 1) \big) \matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i} \end{aligned}$$ -Assuming $\matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i} \neq 0$, +Assuming $$\matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i} \neq 0$$, this can only be satisfied if the following holds: $$\begin{aligned} @@ -537,11 +538,11 @@ $$\begin{aligned} = (\ell_f + \ell_i) (\ell_f - \ell_i) + (\ell_f - \ell_i) \end{aligned}$$ -If $\ell_f = \ell_i = 0$ this equation is trivially satisfied. -Otherwise, the only option is $\ell_f \!-\! \ell_i = 0$, +If $$\ell_f = \ell_i = 0$$ this equation is trivially satisfied. +Otherwise, the only option is $$\ell_f \!-\! \ell_i = 0$$, which is another part of the selection rule. -Thirdly, we look at the commutator of $\hat{s}$ with the ladder operators $\hat{L}_\pm$: +Thirdly, we look at the commutator of $$\hat{s}$$ with the ladder operators $$\hat{L}_\pm$$: $$\begin{aligned} 0 @@ -551,9 +552,9 @@ $$\begin{aligned} &= C_f \matrixel{\ell_f (m_f\!\mp\!1)}{\hat{s}}{\ell_i m_i} - C_i \matrixel{\ell_f m_f}{\hat{s}}{\ell_i (m_i\!\pm\!1)} \end{aligned}$$ -Where $C_f$ and $C_i$ are constants given below. -We already know that $\Delta \ell = 0$ and $\Delta m = 0$, -so the above matrix elements are only nonzero if $m_f = m_i \pm 1$. +Where $$C_f$$ and $$C_i$$ are constants given below. +We already know that $$\Delta \ell = 0$$ and $$\Delta m = 0$$, +so the above matrix elements are only nonzero if $$m_f = m_i \pm 1$$. Therefore: $$\begin{aligned} @@ -568,7 +569,7 @@ $$\begin{aligned} &= \hbar \sqrt{\ell_f (\ell_f \!+\! 1) - m_i (m_i \!\pm\! 1)} \end{aligned}$$ -In other words, $C_f = C_i$. The above equation therefore reduces to: +In other words, $$C_f = C_i$$. The above equation therefore reduces to: $$\begin{aligned} \matrixel{\ell_f m_i}{\hat{s}}{\ell_i m_i} @@ -576,13 +577,13 @@ $$\begin{aligned} \end{aligned}$$ Which means that the value of the matrix element -does not depend on $m_i$ (or $m_f$) at all. +does not depend on $$m_i$$ (or $$m_f$$) at all. </div> </div> -Similarly, given a general (pseudo)vector operator $\vu{V}$, +Similarly, given a general (pseudo)vector operator $$\vu{V}$$, which, by nature, must satisfy the following commutation relations, -where $\hat{V}_\pm \equiv \hat{V}_x \pm i \hat{V}_y$: +where $$\hat{V}_\pm \equiv \hat{V}_x \pm i \hat{V}_y$$: $$\begin{gathered} \comm{\hat{L}_z}{\hat{V}_z} = 0 @@ -596,7 +597,7 @@ $$\begin{gathered} \comm{\hat{L}_{\pm}}{\hat{V}_{\mp}} = \pm 2 \hbar \hat{V}_z \end{gathered}$$ -The inner product of any such $\vu{V}$ must obey the following selection rules: +The inner product of any such $$\vu{V}$$ must obey the following selection rules: $$\begin{aligned} \boxed{ @@ -637,17 +638,17 @@ Selection rules are not always about atomic electron transitions, or angular mom According to the **principle of indistinguishability**, permuting identical particles never leads to an observable difference. In other words, the particles are fundamentally indistinguishable, -so for any observable $\hat{O}$ and multi-particle state $\Ket{\Psi}$, we can say: +so for any observable $$\hat{O}$$ and multi-particle state $$\Ket{\Psi}$$, we can say: $$\begin{aligned} \matrixel{\Psi}{\hat{O}}{\Psi} = \matrixel{\hat{P} \Psi}{\hat{O}}{\hat{P} \Psi} \end{aligned}$$ -Where $\hat{P}$ is an arbitrary permutation operator. -Indistinguishability implies that $\comm{\hat{P}}{\hat{O}} = 0$ -for all $\hat{O}$ and $\hat{P}$, -which lets us prove the above equation, using that $\hat{P}$ is unitary: +Where $$\hat{P}$$ is an arbitrary permutation operator. +Indistinguishability implies that $$\comm{\hat{P}}{\hat{O}} = 0$$ +for all $$\hat{O}$$ and $$\hat{P}$$, +which lets us prove the above equation, using that $$\hat{P}$$ is unitary: $$\begin{aligned} \matrixel{\hat{P} \Psi}{\hat{O}}{\hat{P} \Psi} @@ -656,9 +657,9 @@ $$\begin{aligned} = \matrixel{\Psi}{\hat{O}}{\Psi} \end{aligned}$$ -Consider a symmetric state $\Ket{s}$ and an antisymmetric state $\Ket{a}$ +Consider a symmetric state $$\Ket{s}$$ and an antisymmetric state $$\Ket{a}$$ (see [Pauli exclusion principle](/know/concept/pauli-exclusion-principle/)), -which obey the following for a permutation $\hat{P}$: +which obey the following for a permutation $$\hat{P}$$: $$\begin{aligned} \hat{P} \Ket{s} @@ -668,8 +669,8 @@ $$\begin{aligned} = - \Ket{a} \end{aligned}$$ -Any obervable $\hat{O}$ then satisfies the equation below, -again thanks to the fact that $\hat{P} = \hat{P}^{-1}$: +Any obervable $$\hat{O}$$ then satisfies the equation below, +again thanks to the fact that $$\hat{P} = \hat{P}^{-1}$$: $$\begin{aligned} \matrixel{s}{\hat{O}}{a} |