Orbital Angular Momentum Operators PDF

# Eigen Functions/Eigen States & Eigen Values of Orbital Angular Momentum Operators $L^2$ & $L_z$ * Eigen functions/ Eigen states & Eigen values of orbital angular momentum operators $L^2$ & $L_z$: * $[L^2, L_x] = 0$, $[L^2, L_y] = 0$, $[L^2, L_z] = 0$ * Since, $L^2$ commute with $L_x$, $L...

# Eigen Functions/Eigen States & Eigen Values of Orbital Angular Momentum Operators $L^2$ & $L_z$ * Eigen functions/ Eigen states & Eigen values of orbital angular momentum operators $L^2$ & $L_z$: * $[L^2, L_x] = 0$, $[L^2, L_y] = 0$, $[L^2, L_z] = 0$ * Since, $L^2$ commute with $L_x$, $L_y$, $L_z$ each component of 'L' can be separately diagonized. Hence it has a simultaneous eigen function with $L^2$. * $[L_x, L_y] \neq [L_y, L_z] \neq [L_z, L_x] \neq 0$ * Since the components of $L_x$, $L_y$ & $L_z$ donot mutually commute we can choose only one of the them to be simultaneously diagonized with $L^2$. Simply we can just aswell as take $[L^2, L_x]$ & $[L^2, L_y]$. * By convention we take $L_z$. # Let $|l,m>$ be common eigen states/eigen vectors/eigen functions of $L^2$ & $L_z$ with corresponding eigen values $ħ^2 l(l+1)$ & $mħ$ respectively. * $L^2 |l, m> = ħ^2 l(l+1) |l, m>$ * $L_z|l, m> = mħ|l, m>$ * Such eigen states constitute a complete orthonormal state. * i.e. $<l',m'|l, m> = δ_{l',l} δ_{m',m}$ * Consider * $<l', m'|L_z^2 |l, m> - <l', m'|L_z^2 |l, m> = <l', m'|L_x^2 + L_y^2 |l, m>$ * $=> <l', m'|L_z^2 |l, m> - <l', m'|L_z^2 |l, m> = <l', m'|L_x^2 |l, m> + <l', m'|L_y^2 |l, m>$ * $=> ħ^2 m^2 <l', m'|l, m> - m'^2 ħ^2 <l', m'|l, m> ≥ 0$ * $=> (ħ^2 (m^2 - m'^2) <l', m'|l, m> ≥ 0$ * $=> (m - m') (m + m') ≥ 0$ # Now, we want to introduce raising and lowering operators $L_+$ & $L_-$$ * $L_+ = \frac{1}{ħ}(L_x + iL_y)$ * $L_- = \frac{1}{ħ}(L_x - iL_y)$ * $[L^2, L±] = [L^2, L_x ± iL_y] = [L^2, L_x] ± i[L^2, L_y]$ * $=>[L^2, L±] = 0$ * Now $[L_z, L±] = [L_z, L_x ± iL_y] = [L_z, L_x] ± i[L_z, L_y]$ * $=> [L_z, L±] = iħL_y ± i(iħL_x) = iħL_y ± ħL_x$ * $=> [L_z, L±] = ħ(L_x ± iL_y)$ * $=> [L_z, L±] = ħL±$ # Now $[L_+, L_-] = [L_x + iL_y; L_x - iL_y]$ * $=> [L_+, L_] = (L_x + iL_y)(L_x - iL_y) - (L_x - iL_y)(L_x + iL_y)$ * $=> [L_+, L-] = L_x^2 - iL_xL_y + iL_yL_x + L_y^2 - L_x^2 - iL_xL_y + iL_yL_x - L_y^2$ * $=> [L_+, L-] = - i[L_x, L_y] - i[L_x, L_y]$ * $=> [L_+, L-] = -2i(iħL_z) = 2ħL_z$ * $=> [L_+, L-] = 2ħL_z$ # Now $(L_+L_-)+(L_-L_+)$= $(L_x + iL_y)(L_x - iL_y) + (L_x - iL_y)(L_x + iL_y)$ * $=> (L_+L_-)+(L_-L_+)$= $L_x^2 + L_y^2 + L_x^2 + L_y^2$ * $=> (L_+L_-)+(L_-L_+)$= $2(L_x^2 + L_y^2)$ * $=> (L_+L_-)+(L_-L_+)$= $2(L^2 - L_z^2) = 2ħ^2 l(l+1) - 2ħ^2 m^2$ * $=> L^2 = \frac{1}{2} [(L_+L_-)+(L_-L_+)] + L_z^2$ # Now multiply $L_+$ in equn (3) * $L_+ L^2 |l, m> = ħ^2 l(l+1) L_+ |l, m>$ * $As [L^2, L_+] = 0 => (L^2 L_+ - L_+ L^2)|l, m> = 0$ * $=> L^2 L_+ |l, m> = ħ^2 l(l+1) L_+ |l, m>$ * $=> L^2 L_+ |l, m> = ħ^2 (l+1)(l+2) L_+ |l, m>$ * $=> L^2 {L_+ |l, m>} = ħ^2 (l+1)(l+2){L_+ |l, m>}$ * $L_+ |l, m>$ is also a eigen vector of $L^2$ corresponding to same eigen value $ħ^2 (l+1)(l+2)$ * Similarly operating $L_+$ on equn (4) * $L_z L_+= mħ L_+$. * $=> L_z L_+ |l, m> = mħ L_+ |l, m>$ * $=> L_z L_+ |l, m> + ħ L_+ |l, m> = mħ L_+ |l, m> + ħ L_+ |l, m>$ * $=> L_z L_+ |l, m> = (m+1)ħ L_+ |l, m>$ * $=> L_z {L_+ |l, m>} = (m+1)ħ {L_+ |l, m>}$ * Thes equr (15) shows that $L_+ |l, m>$ is an eigen vector / eigen function of $L_z$ corresponding to eigen value $(m+1)ħ$. # Now operating $L_-$ on equn (4) * $L_- L_z |l, m> = mħ L_- |l, m>$ * $As [L_z, L_-] = -ħL_-$ * $=> (L_z L_- - L_- L_z) = -ħL_-$ * $=> (L_zL_-) = (L_-L_z) + ħL_-$ * Using these in equn (16) * $L_zL_-|l, m> + ħL_-|l, m> = mħL_-|l, m>$ * $=> L_zL_-|l, m> = mħ L_-|l, m> - ħL_-|l, m>$ * $=> L_z {L_- |l, m>} = (m-1)ħ {L_- |l, m>}$ * Thes equr (17) show that $L_- |l, m>$ is an eigen vector/ eigen function of $L_z$ corresponding to eigen value $(m-1)ħ$. * We may write * $L_+ |l, m> = C_+ |l, m + 1>$ * $L_- |l, m> = C_- |l, m - 1>$ # Since, $L_+|l,m>$ and $L_-|l,m>$ are raising and lowering the second quantum no. 'm' by one unity, $C_+$ & $(-)$ are normalization constant. * Equ (8) and (19) show why $L_±$ are called ladder operator. That $L_+$ raises eigen ket to the higher value of 'm' & $L_-$ is an operator that lowers eigen ket to the lower value of 'm'. * $L_+ |l, m +1> = C_+ |l, m + 2>$ * $L_+ |l, m +2>= C_+ |l, m + 3>$ * $...$ * $L_+ |l, m+l> = C_+ |l, m+l+1>$ * $L_+ |l, m+l+1> = C_+ |l, m+l+2> = 0$ * $L_- |l, m-1> = C_- |l, m - 2>$ * $L_- |l, m - 2> = C_- |l, m - 3>$ * $...$ * $L_- |l, m -l> = C_- |l, m-l-1>$ * $L_- |l, m -l-1> = C_- |l, m-l-2> = 0$ * Equ (20) & (21) shows that we can not go indefinitely otherwise we violate $|l^2| ≥ m^2$. * Highest Value m = m+ * Lowest Value m = m- * m+ - m- = +ve number # In similar manner if we argue that in ladder operator L not change. * Since, $|l, m + >$ and $|l, m - >$ are obtained from $|l, m>$ by repeated application of $L_+$ & $L_-$ respectively with adjacent eigen value 'm' differing by one unity. * We have $(m +) - (m - ) = +ve integer | zero. * Operating equr (20) by $L_-$ * $L_-L_+ |l, m + > = 0$ * $=>(L^2 - L_z^2 - ħL_z)|l, m + > = 0$ * $=> L^2 |l, m + > - L_z^2 |l, m + > -ħL_z |l, m + > = 0$ * $=> ħ²l(l+1) |l, m + > - m^2ħ^2 |l, m + > - m+ħ^2 |l, m + > = 0$ * $=> (ħ^2 (l(l+1) - m^2 - m+) |l, m + > = 0$ * $=> l(l+1) - m^2 - m+ = 0 => l(l+1) = m^2 + m+$ # Operating equn (14) by $L_+$ * $L+L_-|l, m-> = 0$ * $=> (L^2 - L_z^2 + ħL_z)|l, m-> = 0$ * $=> L^2 |l, m-> - L_z^2 |l, m-> + ħL_z |l, m-> = 0$ * $=> ħ²l(l+1) |l, m-> - m^2ħ^2 |l, m-> + m-ħ^2 |l, m-> = 0$ * $=> (ħ^2 (l(l+1) - m^2 + m-) |l, m-> = 0$ * $=> l(l+1) - m^2 + m- = 0 => l(l+1) = m^2 - m-$ # Comparing equ (33) & (23) * m + (m + + 1) = m-(m - 1). * $=> m^2 + m + = m^2 - m - $ * $=> m^2 + m^+ - m^2 + m_- = 0$ * $=> (m^+ + m-)(m^+ - m-) + (m^+ + m-) = 0$ * $=> (m^+ + m-)(m^+ - m- + 1) = 0$ * $=> (m^+ + m-) = 0$ 'on' $(m^+ - m- + 1) = 0$ * $=> m^+ = -m-$ 'or' $m^+ = m- - 1$ * or $m^+ = m- - 1$ : not acceptable because m+ - m- > 0. * Now λħ² be the eigen value of $L^2$ & λħ² depends only on $l$. So, that according to equn (22) & (23) m+ & m- should be function of $l$ only. * Let us choose the maximum value of 'm' should be equal to 'm'. # According to equ (24) $m + = l$ * $m + - m - = (l - (-l)) = 2l$ = (ve integer including 0) * ... m+ - m - allow value λ = 0, 1, 2, 3, .... * Therefore allow value λ = 0, 1, 2, 3, .... # For each value of 'l' the 'm' values goes from -l to +l in states of unity. * Using equation (26) in each of equn (22) & (23) * λ = l(l+1) & λ = -l(l - 1) = l(l + 1) * Therefore $|l, m> = λ²|l, m> = l(l+1)²|l, m>$ * The eigen value of $L²=l(l+1)ħ$ * But the eigen states of $L_z$ are different * $L_z|l, m> = mħ|l, m>$ * m = -l to +l through zero. * No. of state = ₂C^l+1 # Let us now determine $C_+$ & $C_-$. * $|P> = L_+|l, m> = C_+|l, m + 1>$ * $<P| = <l, m|L_- = <l, m+1|$ * According to normalization condition * $<P|P> = <l, m|L_-L_+|l, m> = C_+C_+^*<l, m + 1|l, m + 1>$ * $=> <l, m|L^2 - L_z^2 - ħL_z|l, m> = |C_+|^2$ * $=> ħ^2l(l+1)<l, m|l, m> - m^2ħ^2<l, m|l, m> - m ħ^2<l, m|l, m>= |C_+|^2$ * $=> (ħ^2 (l(l+1) - m^2 - m) |l, m|l, m> = |C_+|^2$ * $=> (ħ^2(l + 1 - m)(l - m))|l, m|l, m> = |C_+|^2$ * $=> (l + 1 - m)(l - m)|l, m|l, m> = |C_+|^2$ * $=> (l + 1 - m)(l - m) = |C_+|^2$ * $=> C_+ = \sqrt{(l + 1 - m)(l - m)}\hbar$ * $.. L_+|l, m>= \sqrt{(l + 1 - m)(l -m)}\hbar |l, m+1>$ * Again $L_-|l, m> = C_-|l, m-1>$ * According to normalisation condition * $<l, m|L_+L_-|l, m> = |C_-|^2$ * $=> <l, m|L^2 - L_z^2 + ħL_z|l, m> = |C_-|^2$ * $=> ħ^2l(l+1)<l, m|l, m> - m^2ħ^2<l, m|l, m> + m ħ^2<l, m|l, m>= |C_-|^2$ * $=> (ħ ^2 (l(l+1) - m^2 + m) |l, m|l, m> = |C_-|^2$ * $=> (ħ^2(l + 1 + m)(l - m))|l, m|l, m> = |C_-|^2$ * $=> (l + 1 + m)(l - m) = |C_-|^2$ * $=> C_- = \sqrt{(l + 1 + m)(l - m)}ħ$ * .. $L_-|l,m>= \sqrt{(l + 1 + m)(l - m)}\hbar |l, m -1>$ * For $m=l= (m +)$, then equn (30) becomes $L_+|l, l> = 0$ * For $m=-l= (m - )$, then equn (34) becomes $L_-|l, -l> = \sqrt{(l + 1 +(-l))(l - (-l))}\hbar |l, -l-1>$ * $=> \sqrt{(l + 1 -l)(l + l)}\hbar |l, -l-1>$ * $= \sqrt{(l + 1)} \hbar|l, -l-1>$ * Therefore our assumption $m + = l$ & $m - = -l$ is satisfied. * ... Total no. of 'm' states are $(2l + 1)$ * Let us now find eigen value and eigen vector of orbital angular momentum. * $L^2 |l, m> = ħ^2 l(l+1) |l, m>$ * $L_z |l, m> = mħ |l, m>$ * Where $l = 0, \frac{1}{2}, 1, \frac{3}{2}, 2,....$ & $m = -l$ to $+l$ through zero. # Schwarz Inequality:- * For any two states $|ψ>$ & $|\phi>$ of the Hilbert space 'H' we can show that * $|<ψ|\phi>|^2 ≤ <ψ|ψ><\phi|\phi>$ * Proof: * Let us consider a vector $|ψ + λ\phi>$ where λ is a complex no. Since, a scalar product of ψ + λ\phi with itself should always be greater than or equal to zero. * <ψ + λ*φ|ψ + λφ> ≥ 0 * Let $f(λ) = <ψ + λ\phi|ψ + λφ>$ * $=> <ψ|ψ> + <\phi|ψ>*λ + λ*<ψ|\phi> + λλ*<φ|\phi> ≥ 0$ * Differentiating equation (1) w.r.t. λ * $\frac{df(λ)}{dλ}$ = [$<ψ|ψ>+<φ|ψ>*λ + λ*<ψ|\phi>+ λλ*<φ|\phi>$] ≥ 0 * $=> \frac{df(λ)}{dλ} = <φ|ψ>*λ + λ*<ψ|\phi> ≥ 0$ * $=> \frac{df(λ)}{dλ} $ = $λ*<φ|ψ> + λ*<ψ|\phi> ≥ 0$ * For minimum $\frac{df(λ)}{dλ} = 0$ * $=> <φ|ψ>*λ + λ*<ψ|\phi> = 0$ * $=> λ = - \frac{<φ|ψ>*}{<ψ|\phi>}$ * Using equation (1) in equation (2) * $[f(λ)]_{min} = <ψ|ψ> - <φ|ψ>* \frac{<ψ|\phi>}{<φ|\phi>} + \frac{<φ|ψ>*}{<φ|\phi>}$ <ψ|\phi> ≥ 0 * $=> <ψ|ψ> - \frac{<φ|ψ>*}{<φ|\phi>}$ <ψ|\phi> ≥ 0 * $=> <ψ|ψ> <φ|\phi> - <φ|ψ>*<ψ|\phi> ≥ 0$ * $=> <ψ|ψ> <φ|\phi> ≥ <φ|ψ>*<ψ|\phi>$ * $=> <ψ|ψ> <φ|\phi> ≥ |<φ|ψ>|^2$ * $=> 1|<ψ|\phi>|^2 ≤ <ψ|ψ><\phi|\phi>$ (proved) # Q1. Show that $<ψ|ψ> + <φ|φ> ≥ 2Re <ψ|φ>$ * Let $|ψ> = |ψ_1 - ψ_2>$ & $|φ> = |ψ_1 - ψ_2>$ * The scalar product of $|ψ>$ & $<φ|$ * $<ψ|φ> = (|ψ_1>-|ψ_2>)(|ψ_1>-|ψ_2>)$ * $=> <ψ_1|ψ_1> - <ψ_1|ψ_2> - <ψ_2|ψ_1> + <ψ_2|ψ_2>$ * $=> <ψ_1|ψ_1> - <ψ_2|ψ_2> ≥ <ψ_1|ψ_2> + <ψ_2|ψ_1>$ * $=> <ψ_1|ψ_1> + <ψ_2|ψ_2> ≥ <ψ_1|ψ_2> + <ψ_2|ψ_1> + <ψ_1|ψ_2> + <ψ_2|ψ_1>$ * $=> <ψ_1|ψ_1> + <ψ_2|ψ_2> ≥ 2Re <ψ_1|ψ_2>$ (proved) # Triangle Inequality:- * Show that $Kψ + φ|ψ + φ> ≤ \sqrt{<ψ|ψ>} + \sqrt{<φ|φ>}$. * Proof: * If states ψ & φ are two states that are linearly independent then they satisfy above equation which is called triangle inequality. * $<ψ + φ|ψ + φ> = <ψ|ψ> + <ψ|φ> + <φ|ψ> + <φ|φ>$ * $=> <ψ + φ|ψ + φ> = <ψ|ψ> + <ψ|φ> + <φ|ψ>* + <φ|φ>$ * $=> <ψ + φ|ψ + φ> = <ψ|ψ> + <ψ|φ> + 2Re <ψ|φ>$ * If '$z$' is a complex no. , $Re~ z ≤ |z|$ * $=> Re <ψ|φ> ≤ |<ψ|φ>|$ * Equation (1) can be written as * $<ψ + φ|ψ + φ> <= |<ψ|ψ> + <φ|φ> + 2 <ψ|φ>|$ * According to Schwartz inequality * $<ψ|ψ><φ|φ> ≥ |<ψ|φ>|^2$ * $=> 1 <ψ|φ> | ≤ \sqrt{<ψ|ψ><φ|φ>}$ * Using equation (3) in equn (2) * $ <ψ + φ|ψ + φ> ≤ \sqrt{<ψ|ψ>} + \sqrt{<φ|φ>} + 2\sqrt{<ψ|ψ><φ|φ>}$ * $=> <ψ + φ|ψ + φ> ≤ (\sqrt{<ψ|ψ>} + \sqrt{<φ|φ>})^2$ * $=> \sqrt{<ψ + φ|ψ + φ>} ≤ \sqrt{<ψ|ψ>} + \sqrt{<φ|φ>}$ (proved) # If $|ψ>$ & $ φ>$ are linearly dependent, $|ψ>= α|φ> $ & if the proportionality scalar 'α' is real and positive, the triangle inequality becomes an equality. * The counter part of this inequality in Euclidean space is given by $|A + B| ≤ |A| + |B|$. * Uncertainty relation between two operators: # Let $<A>$ and $<B>$ denote the expectation values of two hermitian operators A & B with respect to normalized state vector $|ψ>$. * $<A> = <ψ|A|ψ>$ * <B> = <ψ|B|ψ> * And At A & Bt B = B # Now introducing the operator * ΔA = A - <A > => (ΔA)² = (A - <A>)² * $=> (ΔA)² = (A)² - 2 A <A> + <A>)²$ * ΔB = B - <B > => (ΔB)² = (B - <B>)² * $=> (ΔB)² = (B)² - 2B <B> + <B>)²$ * $<(ΔA)²> = <A²> - 2<A><A> + <A²>$ * $=> <(ΔA)²> = <A²> - 2<A²> + <A²>$ * $=> <(ΔA)²> = <A²> - <A²>$ * $<(ΔB)²> = <B²> - 2<B><B> + <B²>$ * $=> <(ΔB)²>= <B²> - 2 <B²> + <B²>$ * $=> <(ΔB)²> = <B²> - <B²>$ # The uncertainty ΔA & ΔB are defined by * ΔA = $\sqrt{<(ΔA)²>} = \sqrt{<A²> - <A>²}$ * ΔB = $\sqrt{<(ΔB)²>} = \sqrt{<B²> - <B>²}$ # Let us write the action of the operators of any state ψ as follows * $|χ> = ΔA|ψ> = (A - <A>)|ψ>$ * $|$〉 = $ΔB|ψ> = (B - <B>)|ψ>$ * The Schwartz inequality of two states $χ>$ & $〉$ * <χ|χ><〉|〉> $|<χ|〉|^2$ * Since, A & B are hermitian, so ΔA & ΔB must be hermitian. * $(ΔA)^+ = (A - <A>)^+ = A^t - <A>^*I = A - <A> = ΔA$ * $(ΔB)^+ = (B - <B>)^+ = B^t - <B>^*I = B - <B> = ΔB$ * Now, <χ|χ> = <ψ|ΔA^†ΔA|ψ> = <ψ|ΔAΔA|ψ> * $=> <χ|χ> = <ψ|(ΔA)^2 |ψ>$ * <ψ|ψ> = <ψ|ΔB^†ΔΒ|ψ> = <ψ|ΔΒΔΒ|ψ> * $=> <ψ|ψ> = <ψ|(ΔΒ)^2 |ψ>$ * <χ|ψ> = <ψ|ΔA^†ΔΒ|ψ> = <ψ|ΔΑΔΒ|ψ> * $=> <χ|ψ> = <ψ|ΔΑΔΒ|ψ>$ * Using equation (13), (14) in equation (11) * <ψ|(ΔA)²|ψ><ψ|(ΔΒ)²|ψ> ≥ $ |<ψ|ΔΑΔΒ|ψ>|^2$ * Using completeness property * <ψ|(ΔA)²(ΔΒ)²|ψ> ≥ $ |<ψ|ΔΑΔΒ|ψ>|^2$ * ≥ $|<ψ|ΔΑΔΒ|ψ>|<ψ|ΔΑ^†ΔΒ^†|ψ>$ * ≥ $<ψ|ΔΑΔΒ ΔΑ^†ΔΒ^†|ψ>$ * ≥ $<ψ|ΔΑΔΒ ΔΑ^†ΔΒ^†|ψ>$ * ≥ $<ψ|(ΔΑΔΒ)^2|ψ>$ * ≥ $<ψ|(ΔΑΔΒ)^2|ψ>> |<ψ|(ΔΑΔΒ)|ψ>|$ * $=> <ψ|(ΔA)²(AB)²|ψ> ≥ |<ψ|(ΔΑΔΒ)^2|ψ>$ * Now, ΔΑΔΒ = [ΔΑ, ΔΒ] + $\frac{1}{2}$ {ΔΑ, ΔΒ} * $=> ΔΑΔΒ = \frac{1}{2} [ΔΑ, ΔΒ] + \frac{1}{2}$ {ΔΑ, ΔΒ} * .. [ΔΑ, ΔΒ] = ΔΑΔΒ - ΔΒΔΑ * = (A - <A>)(B - <B>) - (B - <B>)(A - <A>) * = AB - A<B> - <A>B + <A><B> - BA + B<A + <B>A - <B><A> * $=> <[ΔΑ, ΔΒ]> = <AB> - <A><B>- <A><B> + <A><B> - <BA> + <B><A>+<B><A>-<B><A>$ * $=> <[ΔΑ, ΔΒ]> = <AB> - <BA> = <[A,B]>$ * Now, $KΔΑ, ΔΒ>^2 = \frac{1}{4}K[A,B]>^2 + \frac{1}{4} K{ΔΑ, ΔΒ}>^2$ * $=> KΔΑ, ΔΒ>^2 ≥ \frac{1}{4}K[A,B]>^2$ * Now put A = x : B = P_x: position operator * Δx.ΔP_x ≥ $\frac{1}{2}$ : momentum operator * $=> Δx.ΔP_x ≥ \frac{ħ}{2}$ $=>$ Δx.ΔP_x ≥ $\frac{ħ}{2}$ (proved) * Similarly, Δy.ΔP_y ≥ $\frac{ħ}{2}$ , Δz.ΔP_z ≥ $\frac{ħ}{2}$ , Δθ.Δ$L$ ≥ $\frac{ħ}{2}$ * The minimum uncertainty wave packet - (Gaussian wave packet) * According to exact statement of uncertainty wave packet we have. * The minimum uncertainty product is, * Δx.ΔP_x = $\frac{ħ}{2}$ * The minimum uncertainty product will be obtained if we choose the function in schwartz inequality, such that there is a sign of equality . * The sign of equality can be obtained if the function χ & ψ are proportional to each other. i.e. χ & ψ are linearly independent. * So, we can write χ = αψ. * Let choose X = pψ * $x=αψ$ * $=> \frac{dx}{dt} = α\frac{dψ}{dt}$. * $=> -\frac{ħ}{i} \frac{dψ}{dx} = αψ$ * $=> -iħ \frac{dψ}{dx} = αψ$ * Put -ħ/i = β, so β $\frac{dψ}{dx}$ = αψ * Integrating both sides of above equation. # $β \int_{}^{} \frac{dψ}{ψ} = α \int_{}^{} dx => \int_{}^{} \frac{dψ}{ψ} = \frac{α}{β} \int_{}^{} dx => lnψ = \frac{α}{β} x + c$ * $=> lnψ = \frac{α}{β} x + lnC$ * $=> ψ = Ce^{\frac{α}{β} x}$ * $=> ψ = Ne^{\frac{-x^2}{2β}}$ * Where N → normalization const. # As we want to represent a wave packet for which the integral ∫ ψ* ψ dx converges. β must be positive. * The magnitude of 'N' in equation (6) can be calculated by normalizing ψ* ψ dx = 1. * i.e . $\int_{}^{-\infty} ψ* ψ dx = 1 => \int_{}^{-\infty} N^2 e^{\frac{-x^2}{β}} dx = 1 $ * $=> N^2 \int_{}^{-\infty} e^{\frac{-x^2}{β}} dx = 1 => N^2 \int_{}^{-\infty} e^{\frac{-x^2}{β}} dx = 1 $ * Let put x²/β = t => x=√βt * $=> dx = 1β \frac{1}{2}t^{\frac{-1}{2}} dt = \frac{1}{2} √β t^{\frac{-1}{2}} dt$ # $N^2 \sqrtβ \int_{}^{} e^{-t} t^{\frac{-1}{2}} dt = 1 => N^2 \sqrtβ \int_{}^{} e^{-t} t^{\frac{-1}{2}} dt = 1 $ * $=> N^2 \sqrtβ Γ(\frac{1}{2}) = 1 => N^2√βπ = 1 => N^2 = \frac{1}{√βπ}$ * $=> N = \frac{1}{(βπ)^{1/4}}$ * Now the wave function * $ψ = (βπ)^{1/4}e^{\frac{-x^2}{2β}}$ *