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Wavefunctions of Harmonic oscillator are expressed as
\[
\phi_n(x) = A_n H_n\left(x\sqrt{\frac{\omega m}{2\hbar}}\right) e^{-\frac{m\omega x^2}{2\hbar}}
\]
[Let \(x_0 = \sqrt{\frac{2\hbar}{m\omega}}\)]
\[
= A_n\, e^{-\frac{x^2}{x_0^2}}\, H_n\!\left(\frac{x}{x_0}\right)
\]
where,
\[
\left\{
\begin{aligned}
H_n(x) &= \text{Hermite polynomial of order 'n'.}\\
H_0(x) &= 1\\
H_1(x) &= 2x\\
\vdots\\
H_n(x) &= \begin{cases} \text{odd powers of 'x'} & n = \text{odd}\\ \text{even powers of 'x'}, & n = \text{even}\end{cases}
\end{aligned}
\right.
\]
It is family of orthogonal functions. (Mutually orthogonal)
\(\{\)cond\(^{\text{n}}\) for orthogonality
\[
\int_{-\infty}^{\infty} dx\ e^{-x^2}\ H_n(x)\, H_m(x) = (\underline{\qquad})\ \delta_{nm}
\]
\(\hookrightarrow\) weight factor.
\(\{\)Rodrigue's formula\(\}\) where,
\[
H_n(x) = e^{+x^2}\cdot\left(\frac{-d}{dx}\right)^{n} e^{-x^2}
\]
\[
H_0(x) = e^{+x^2}\, e^{-x^2} = 1
\]
\[
H_1(x) = e^{+x^2}\left(\frac{-d}{dx}\right)e^{-x^2} = 2x\, e^{0} = 2x
\]
\[
H_2(x) = e^{+x^2}\left(\frac{-d}{dx}\right)\left(\frac{-d}{dx}\right)e^{-x^2} = e^{+x^2}\left[\frac{-d}{dx}\left(e^{-x^2}\cdot 2x\right)\right]
\]
\[
= e^{+x^2} e^{-x^2}\cdot 2x\,(2x + x)
\]
\[
= 4x(x+1) \qquad 4x^2 + 2
\]
\[
4x^2 - 2
\]
[the expressions \(4x(x+1)\) and \(4x^2+2\) are struck out in the notebook; the answer written below them is \(4x^2-2\)]
Generating function :
\[
e^{2tx - t^2} = \sum_{n=0}^{\infty} \frac{H_n(x)\, t^n}{n!}
\]
Generating functions generates power of 't' whose coefficients are precisely the \(\left(\frac{\text{Hermite polynomials}}{n!}\right)\).
One more thing \(\rightarrow\)
Hermite polynomials are regular solutions of the below differential equations.
Recurrence formula \(\}\)
\[
\frac{d^2H_n}{dx^2} - 2x\,\frac{dH_n}{dx} + 2n\,H_n = 0
\]
we can solve it using power series solution.
Note :-
The Hamiltonian of Harmonic oscillator
\[
H = \frac{p^2}{2m} + \frac{1}{2}m\omega^2 x^2
\]
This equation is symmetric in both \(\hat x\) & \(\hat p\).
In position basis :- \(\left(\hat p = -i\hbar\frac{\partial}{\partial x}\right)\) , \(\hat x = \hat x\)
\[
\hat H = \left(\frac{-\hbar^2}{2m}\frac{d^2}{dx^2} + \frac{1}{2}m\omega^2\hat x^2\right)
\]
In momentum basis :- \(\hat x = i\hbar\frac{\partial}{\partial p}\) , \(\hat p = \hat p\)
\[
\hat H = \left(\frac{\hat p^2}{2m} - \frac{1}{2}m\omega^2\hbar^2\frac{d^2}{dp^2}\right)
\]
So the solution of wave fun./eigen states in momentum basis are also going to be \(\left(H_n(p)\cdot e^{-p^2/2m}\right)\).
\[
\boxed{\tilde\phi_n(p) = B_n\, e^{-\frac{p^2}{2m\omega\hbar}}\cdot H_n\!\left(\frac{p}{\sqrt{2m\omega\hbar}}\right)}
\]
The ground state of Harmonic oscillator is gaussian, both in \(\{|x\rangle\}\) basis and \(\{|p\rangle\}\) basis.
\[
\tilde\phi_0(p) = B_0\, e^{-\frac{p^2}{2m\omega\hbar}}
\]
\[
\phi_0(x) = A_0\, e^{-\frac{x^2 m\omega}{2\hbar}}
\]
also we can see both are fourier transform of each other.
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\draw[dashed] (1.0,0) -- (1.0,0.9);
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\draw[->] (0,-0.3) -- (0,2.5) node[above] {$\tilde\phi_0(p)$};
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Note :-- classically particle does to and fro motion b/w \((A, -A)\).
But q.mechanics says \(\exists\) non zero finite probability to find it anywhere on \(x\)-axis.
letus find \((\Delta x)_n = \left\langle (x - \langle x\rangle)^2\right\rangle_n^{1/2}\)
first, let me find out, \(\langle n|a|n\rangle\)
\(\{|n\rangle = n^{\text{th}}\) eigen state of \(H\).
where
\[
a = x\sqrt{\frac{m\omega}{2\hbar}} + ip\left(\frac{1}{\sqrt{2m\omega\hbar}}\right)
\]
\[
a^+ = x\sqrt{\frac{m\omega}{2\hbar}} - i\frac{p}{\sqrt{2m\omega\hbar}}
\]
\[
E_n = \hbar\omega\left(n+\tfrac{1}{2}\right)\ , \qquad H = \hbar\omega\left(a^+a + \frac{I}{2}\right)
\]
also we know \(a^+a|n\rangle = n|n\rangle\) ; \(a^+a = N\) (Number operator)
we also know that,
\[
a|n\rangle = \sqrt{n}\,|n-1\rangle
\]
\[
a^+|n\rangle = \sqrt{n+1}\,|n+1\rangle
\]
\(\{\)and all \(|n\rangle\) are orthonormal.
Given that, what is \(\langle n|a|n\rangle\).
\[
\langle n|\sqrt{n}|n-1\rangle = \sqrt{n}\,\langle n|n-1\rangle = 0
\]
similarly, \(\langle n|a^+|n\rangle = 0\)
we can represent \(a, a^+, \hat x, \hat p\) as \(\infty\)-diamensional matrices.
\(\{\)So what looks like \(-i\hbar\frac{d}{dx}\) operator in one basis, same operator looks simply \(p\) in another basis, & same operator in energy basis / no. operator basis looks like \(\infty\)-diamensional matrices. So These are abstract operator which wear many clothes.
since matrix element of \(a\) in \(\{|n\rangle\}\) basis can be written as \(a_{ij} = \langle i|a|j\rangle\) or \(\langle n|a|m\rangle = a_{nm}\)
since
\[
\left.\begin{aligned}\langle n|a|n\rangle &= 0\\ \langle n|a^+|n\rangle &= 0\end{aligned}\right\} \Rightarrow \ \text{diagonal elements of } a,\ a^+ \text{ are zero so is the true for } \hat x,\ \hat p \text{ too}
\]
as
\[
a = \sqrt{\frac{m\omega}{2\hbar}}\,\hat x + \frac{i\,\hat p}{\sqrt{2m\omega\hbar}}
\]
\[
a^+ = \underline{\qquad} \ - \frac{i\,\hat p}{\sqrt{\ }}
\]
\[
\hat x = \frac{a+a^+}{2\sqrt{\frac{m\omega}{2\hbar}}}\ , \qquad \hat p = \left(\frac{a-a^+}{i\,2}\right)\sqrt{2m\omega\hbar}
\]
also don't have diagonal elements.
lets choose units such that \((m,\omega,\hbar = 1)\)
\[
a = \frac{x+ip}{\sqrt2}
\]
\[
a^+ = \frac{x-ip}{\sqrt2}
\]
\[
\Rightarrow \quad x = \frac{a+a^+}{\sqrt2}
\]
\[
p = \frac{a-a^+}{i\sqrt2}
\]
since \(a, a^+\) are not hermitian.
but, \(x, p\) are hermitian operators, \(\left.\begin{aligned}x^+ &= x,\\ p^+ &= p\end{aligned}\right\}\)
\(\downarrow\)
(physical operators)
lets find
\[
\left.\begin{aligned}\langle n|x|n\rangle &= 0 = \langle x\rangle\\ \langle n|p|n\rangle &= 0 = \langle p\rangle\end{aligned}\right\}\ \text{diagonal elements are zero.}
\]
\[
\langle x^2\rangle \Rightarrow \langle n|x^2|n\rangle = \left\langle n\left|\frac{(a+a^+)^2}{2}\right|n\right\rangle
\]
\[
= \left\langle n\left|\frac{a^2 + aa^+ + a^+a + a^{+2}}{2}\right|n\right\rangle
\]
\[
= \left\langle n\left|\frac{a^2}{2}\right|n\right\rangle + \left\langle n\left|\frac{aa^+}{2}\right|n\right\rangle + \left\langle n\left|\frac{a^+a}{2}\right|n\right\rangle + \left\langle n\left|\frac{a^{+2}}{2}\right|n\right\rangle
\]
\[
= \frac{n}{2}\langle n|n\rangle + \left\langle n\left|\frac{aa^+}{2}\right|n\right\rangle
\]
\[
= \frac{n}{2} + \left\langle n\left|\frac{I + a^+a}{2}\right|n\right\rangle
\]
\[
= \frac{n}{2} + \frac{n}{2} + \frac{1}{2} = n + \frac{1}{2}
\]
Now putting \(m,\omega,\hbar\) ,
\[
\langle x^2\rangle = \left(n+\tfrac{1}{2}\right)\frac{\hbar}{m\omega}
\]
\[
\langle p^2\rangle = \langle n|p^2|n\rangle
\]
\[
= \left\langle n\left|\frac{(a-a^+)^2}{2(i)^2}\right|n\right\rangle
\]
\[
= \left\langle n\left|\frac{a^2 - aa^+ - a^+a + a^{+2}}{(-2)}\right|n\right\rangle
\]
\[
= \left(\frac{-1}{2}\right)\left\langle n\left|a^2\right|n\right\rangle + \frac{1}{2}\left\langle n\left|aa^+\right|n\right\rangle + \frac{1}{2}\left\langle n\left|a^+a\right|n\right\rangle + \frac{1}{2}\left\langle n\left|a^{+2}\right|n\right\rangle
\]
\[
= \left(n+\tfrac{1}{2}\right)
\]
\(\left\{\begin{aligned}\langle x\rangle &= \langle n|\hat x|n\rangle = 0\\ \langle p\rangle &= \langle n|\hat p|n\rangle = 0\\ \hat x &= \frac{a+a^+}{\sqrt2}\\ \hat p &= \frac{a-a^+}{\sqrt2\,i}\end{aligned}\right.\)
putting back \((m,\omega,\hbar)\)
\[
\langle p^2\rangle = \left(n+\tfrac{1}{2}\right)\cdot m\omega\hbar
\]
\[
\Delta x = \left\langle (x - \langle x\rangle)^2\right\rangle^{1/2} = \sqrt{\left\langle x^2 - 2x\langle x\rangle + \langle x\rangle^2\right\rangle}
\]
\[
= \sqrt{\langle x^2\rangle - 2\langle x\rangle^2 + \langle x\rangle^2}
\]
\[
= \sqrt{\langle x^2\rangle - \langle x\rangle^2}
\]
\[
(\Delta x)_n(\Delta p)_n = \sqrt{\left(n+\tfrac{1}{2}\right)\frac{\hbar}{m\omega}}\ \cdot\ \sqrt{\left(n+\tfrac{1}{2}\right)m\omega\hbar}\ = \left(n+\tfrac{1}{2}\right)\hbar
\]
In ground state, \(n = 0\) ;
\[
(\Delta x)_0(\Delta p)_0 = \left(0+\tfrac{1}{2}\right)\hbar = \frac{\hbar}{2}
\]
also, uncertainty principle says. \(\Delta x\,\Delta p \geq \frac{\hbar}{2}\)
"which implies ground state of Harmonic oscillator is minimum uncertainty state."
If we plot \(\left(\frac{\Delta x}{\sqrt{\hbar/m\omega}} \rightarrow x\right)\left(\frac{\Delta p}{\sqrt{m\omega\hbar}} \rightarrow y\right)\)
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\begin{tikzpicture}[scale=1.0]
\draw[->] (-0.4,0) -- (4.3,0) node[right] {$x$};
\draw[->] (0,-0.4) -- (0,3.7) node[above] {$y$};
\draw[domain=0.62:4.1,smooth,variable=\x] plot ({\x},{2/\x});
\draw[domain=1.75:4.1,smooth,variable=\x] plot ({\x},{6/\x});
\fill (1.414,1.414) circle (1.3pt);
\fill (2.449,2.449) circle (1.3pt);
\draw[dashed] (0,1.414) -- (1.414,1.414) -- (1.414,0);
\node[left] at (-0.05,1.414) {$1/\sqrt2$};
\node[below] at (1.414,-0.05) {$1/\sqrt2$};
\node[anchor=west] at (1.55,1.5) {Ground state $|0\rangle$};
\node[anchor=west] at (2.55,2.7) {$|1\rangle$};
\end{tikzpicture}
Number operator states \(|n\rangle\) are called \(-\) folk space states.
"Important thing in Harmonic oscillator is that energy levels are equally spaced. Therefore when we quantize fields (including E-m field), it turns out every time we add some energy, it is added in quantum / const amt of energy.
Question
What are eigen-states of \(a\), \(a^+\). (?).
As these operators are not hermitian, we do not expect real eigen values. (may be complex).
\[
a^+ = \frac{x-ip}{\sqrt2} \quad \xrightarrow{\ \text{in }\{|x\rangle\ \text{basis}\}\ } \quad \frac{1}{\sqrt2}\left(x + i^2\frac{d}{dx}\right)
\]
\[
\frac{1}{\sqrt2}\left(x - \frac{d}{dx}\right)
\]
lets say \(\chi(x)\) be eigen state of \(a^+\).
\[
a^+|\chi(x)\rangle = \lambda\,|\chi(x)\rangle
\]
\[
\frac{1}{\sqrt2}\left(x - \frac{d}{dx}\right)|\chi(x)\rangle = \lambda\,|\chi(x)\rangle
\]
Exercise \(\{\)Find \(\chi(x)\}\)
eigen states of \(a^+\) :-
\[
a^+|\chi\rangle = \lambda|\chi\rangle
\]
we know that, \(|\chi\rangle = \sum_{n=0}^{\infty} c_n|n\rangle\) uniquely in \(\{|n\rangle\}\) basis
\[
|\chi\rangle = c_0|0\rangle + c_1|1\rangle + c_2|2\rangle + \cdots
\]
\[
a^+|\chi\rangle = c_0 a^+|0\rangle + c_1 a^+|1\rangle + c_2 a^+|2\rangle + \cdots
\]
\[
a^+|\chi\rangle = c_0\sqrt{1}\,|1\rangle + c_1\sqrt{2}\,|2\rangle + c_2\sqrt{3}\,|3\rangle + \cdots \qquad -\ \text{(1)}
\]
But
\[
a^+|\chi\rangle = \lambda|\chi\rangle = \lambda\left(c_0|0\rangle + c_1|1\rangle + c_2|2\rangle + \cdots\right) \qquad \text{(2)}
\]
Comparing coefficients of \(|0\rangle\), \(|1\rangle\), \(\cdots\)
\[
\left.\begin{aligned}
c_0 &= 0\\
\lambda c_1 &= c_0 = 0\\
\lambda c_2 &= c_1\sqrt{2} = 0\\
\lambda c_3 &= c_2\sqrt{3} = 0
\end{aligned}\right\} \quad \text{so solution is there are no eigen functions.}
\]
But there are solutions to the differential eq\(^{\text{n}}\)
\[
\frac{1}{\sqrt2}\left(x - \frac{d}{dx}\right)\chi(x) = \lambda\,\chi(x)
\]
\[
\frac{d\chi}{dx} = \frac{(x - \lambda\sqrt2)\,\chi(x)}{\sqrt2}
\]
\[
\Rightarrow \quad \chi(x) = A\,e^{\frac{1}{\sqrt2}\left(\frac{x^2}{2} - \lambda x\sqrt2\right)}
\]
So what is resolution to this ?
" \(\chi(x)\) from above diff. eq\(^{\text{n}}\) is not normalizable ".
or, \(\left(\int_{-\infty}^{\infty} A^2\, e^{\sqrt2\left(\frac{x^2}{2} - \lambda x\sqrt2\right)}\,dx \neq \text{finite}\right)\)
eigen -states of \(a\) :-
\[
a = \frac{x+ip}{\sqrt2}
\]
\[
a|\alpha\rangle = \alpha|\alpha\rangle \qquad \Rightarrow \qquad \frac{1}{\sqrt2}\left(x + \frac{d}{dx}\right)\alpha(x) = \alpha\cdot\alpha(x)
\]
\(\uparrow\) \(\langle x|\alpha\rangle\)
\[
\frac{1}{\sqrt2}\frac{d(\alpha(x))}{dx} = \left(\lambda - \frac{x}{\sqrt2}\right)\alpha(x)
\]
\[
\int\frac{d\alpha(x)}{\alpha(x)} = \int\left(\sqrt2\lambda - x\right)dx
\]
\[
\boxed{\alpha(x) = A\,e^{\sqrt2\lambda x - \frac{x^2}{2}}}
\]
\[
|\alpha\rangle = \sum_{n=0}^{\infty} c_n|n\rangle
\]
\[
= c_0|0\rangle + c_1|1\rangle + c_2|2\rangle + \cdots
\]
\[
a|\alpha\rangle = c_0 a|0\rangle + c_1 a|1\rangle + c_2 a|2\rangle + \cdots
\]
\[
= 0 + c_1
\]
\[
= 0 + c_1\sqrt{1}\,|0\rangle + c_2\sqrt{2}\,|1\rangle + c_3\sqrt{3}\,|2\rangle + \cdots \qquad \text{(1)}
\]
\[
a|\alpha\rangle = \alpha|\alpha\rangle = \alpha c_0|0\rangle + \alpha c_1|1\rangle + \alpha c_2|2\rangle + \cdots \qquad \text{(2)}
\]
comparing coefficients from eq\(^{\text{n}}\) (1) & (2).
\[
\alpha c_0 = c_1 \qquad \Rightarrow \qquad c_1 = \frac{\alpha c_0}{\sqrt1}
\]
\[
\alpha c_1 = c_2\sqrt2 \qquad \Rightarrow \qquad c_2 = \frac{\alpha^2}{\sqrt{1\cdot2}}\cdot c_0
\]
\[
\alpha c_2 = c_3\sqrt3 \qquad \Rightarrow \qquad c_3 = \frac{\alpha^3}{\sqrt{1\cdot2\cdot3}}\, c_0
\]
\[
\boxed{c_n = \frac{\alpha^n}{\sqrt{n!}}\cdot c_0}
\]
So
\[
|\alpha\rangle = c_0\sum_{n=0}^{\infty}\frac{\alpha^n}{\sqrt{n!}}\,|n\rangle
\]
\((\alpha = \) any complex no.\()\)
all no. in complex planes are eigen value of annihilation operator \((\hat a)\). & They have coorespondng normalized states.
\[
\langle\alpha|\alpha\rangle = 1 = |c_0|^2\sum\frac{|\alpha|^{2n}}{n!}\,\langle n|n\rangle
\]
provided \(\langle n|m\rangle = \delta_{mn}\)
\[
1 = |c_0|^2\, e^{|\alpha|^2}
\]
\[
\boxed{c_0 = e^{-\frac{|\alpha|^2}{2}}}
\]
therefore, we can say
\(\hat a\) has normalizable eigen states \(\{|\alpha\rangle\}\).
\[
a|\alpha\rangle = \alpha|\alpha\rangle \qquad\qquad \alpha\in\mathbb{C}
\]
where,
\[
|\alpha\rangle = e^{-\frac{1}{2}|\alpha|^2}\sum_{n=0}^{\infty}\frac{\alpha^n}{\sqrt{n!}}\,|n\rangle
\]
\(\Downarrow\)
\[
|n\rangle = \frac{(a^+)^n}{\sqrt{n!}}\,|0\rangle \qquad (?\ \text{How}).
\]
\[
\left\{
\begin{aligned}
|1\rangle &= \frac{1}{\sqrt1}\,a^+|0\rangle\\
|2\rangle &= \frac{1}{\sqrt2}\,a^+|1\rangle = \frac{(a^+)^2|0\rangle}{\sqrt{2\cdot1}}\\
\vdots\\
|n\rangle &= \frac{(a^+)^n}{\sqrt{n!}}\,|0\rangle
\end{aligned}
\right.
\]
\[
|\alpha\rangle = e^{-\frac{1}{2}|\alpha|^2}\sum_{n=0}^{\infty}\frac{(\alpha a^+)^n}{n!}\,|0\rangle
\]
\[
|\alpha\rangle = e^{-\frac{1}{2}|\alpha|^2}\cdot e^{\alpha a^+}\cdot|0\rangle
\]
\(\left[\langle\alpha|\alpha\rangle = 1\right.\)
These eigen states \(\{|\alpha\rangle\}\) are called coherent states. (Fundamental role in quantum optics).
Every normalizable \(|\Psi\rangle\) can be written as.
\[
|\Psi\rangle = \sum_{n=0}^{\infty} c_n|n\rangle \qquad \{c_0, c_1, c_2 \cdots\}
\]
(Q) Are those coherent states orthogonal to each other?
are eigen states of \(a\). i.e. \(|\alpha\rangle\) orthonormal to each other. ?
\[
\langle\alpha|\beta\rangle = ?
\]
\[
\langle\beta|\alpha\rangle = \left(\sum_n\sum_m \frac{(\beta^*)^m}{\sqrt{m!}}\,\frac{\alpha^n}{\sqrt{n!}}\ \underbrace{\langle m|n\rangle}_{\delta_{nm}}\right)\cdot e^{-\frac{1}{2}\left(|\alpha|^2+|\beta|^2\right)}
\]
\((n=m)\)
\[
= e^{-\frac{1}{2}\left(|\alpha|^2+|\beta|^2\right)}\cdot \sum_n \frac{\alpha\beta^*}{n!}
\]
\[
= e^{-\frac{1}{2}\left(|\alpha|^2+|\beta|^2\right)}\ e^{\alpha\beta^*}
\]
\[
\langle\beta|\alpha\rangle = e^{-\frac{1}{2}\left(|\alpha|^2+|\beta|^2\right)+\alpha\beta^*} \neq 0.
\]
(check for \((\alpha=\beta)\) , \(\left(\langle\alpha|\alpha\rangle = 1\right)\))
So, states of \(a\), are not orthonormal .
(over \(-\) completeness \(= \int\frac{d^2\alpha}{\pi}\ |\alpha\rangle\langle\alpha| = I\) )
set \(\{|\alpha\rangle\}\) is an overcomplete set of states
Q.) How does \(|\alpha\rangle\) looks like in \(|x\rangle\) basis.
\[
\alpha(x) = \langle x|\alpha\rangle = \sum_{n=0}^{\infty}\frac{\alpha^n}{\sqrt{n!}}\,\langle x|n\rangle
\]
\[
= \sum_{n=0}^{\infty}\frac{\alpha^n}{\sqrt{n!}}\,\phi_n(x)
\]
\[
= \sum_{n=0}^{\infty}\frac{\alpha^n}{\sqrt{n!}}\left(A_n\, e^{-\frac{x^2}{2}}\cdot H_n(x)\right)
\]
Using Dirac approach we factorized the \(\hat H\) (which was 2\(^{\text{nd}}\) order & non-linear) into two linear first order operators \((a, a^+)\).
\[
a = \frac{x+ip}{\sqrt2}
\]
\[
a^+ = \frac{x-ip}{\sqrt2}
\]
\(\}\) in suitable units , \(p = -i\hbar\frac{\partial}{\partial x}\)
\[
\hat H = \hbar\omega\left(a^+a + \frac{I}{2}\right)
\]
\(\left[\text{which was earlier } \frac{p^2}{2m} + \frac{1}{2}m\omega^2q^2\right]\)
[Factorization]
This can't be done with any potential, it can be done with only few potentials. This method is called the factorization method / method of inter-twining operators .
we know that,
\[
a|n\rangle = \sqrt{n}\,|n-1\rangle
\]
\[
a^+|n\rangle = \sqrt{n+1}\,|n+1\rangle
\]
matrix elements :-
\[
\langle \ell|a|n\rangle = a_{\ell n} = \sqrt{n}\,\langle \ell|n-1\rangle = \sqrt{n}\ \delta_{\ell,\,n-1}
\]
\[
\hat a = \begin{pmatrix}
0 & \sqrt1 & 0 & 0 & 0 & \cdots\\
0 & 0 & \sqrt2 & 0 & 0 & \cdots\\
0 & 0 & 0 & \sqrt3 & 0 & \cdots\\
0 & 0 & 0 & 0 & \sqrt4 & \cdots\\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots
\end{pmatrix}
\]
\((\infty\times\infty)\) diamensional. implies.
All diagonal elements has to be zero.
\(\left(a_{\ell n} = \sqrt{n}\ \text{for}\ \ell = n-1\right)\)
\(a^+ = \) simply hermitian conjugate of \(\hat a\).
\[
a = x\sqrt{\frac{m\omega}{2\hbar}} + \frac{i\,p}{\sqrt{2m\omega\hbar}}
\]
\[
\hat a^+ = x\sqrt{\frac{m\omega}{2\hbar}} - \frac{i\,p}{\sqrt{2m\omega\hbar}}
\]
\[
x = \frac{a+a^+}{2}\left(\sqrt{\frac{2\hbar}{m\omega}}\right) = \frac{(a+a^+)}{\sqrt2}\sqrt{\frac{\hbar}{m\omega}}
\]
\[
p = \frac{(a-a^+)}{i\sqrt2}\sqrt{m\omega\hbar}
\]
\((x, p, a^+)\) can be represented in matrix form in same way
since
\[
[a, a^+] = I = aa^+ - a^+a
\]
\[
\text{Trace}(I) = \text{Trace}\left(aa^+ - a^+a\right)
\]
\(\left\{\text{Trace}(AB) = \text{Trace}(BA)\right.\) even if \([A,B]\neq0\).
(nonzero \(\neq\) zero) so we can't represent them in terms of finite diamensional matrices.
\(x, p\) are not bounded operators, their eigen values goes upto \(\infty\).
\(\{\)for two operators \(\hat A\) & \(\hat B\}\)
\(A\) and \(B\) are compatible if \(-\)
\[
[A,B] = 0
\]
(or )
common basis set of eigen vectors such that simultaneous states exist.
(or )
(Implication) \(\downarrow\) \([A,B] = 0\)
\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0]
\node at (0,0) {$A$};
\node at (1.1,0) {$B$};
\node at (2.2,0) {$A$};
\node at (3.4,0) {$|\Psi\rangle$};
\draw[->] (0,-0.35) -- (0,-0.9);
\draw[->] (1.1,-0.35) -- (1.1,-0.9);
\draw[->] (2.2,-0.35) -- (2.2,-0.9);
\node at (0,-1.3) {$a_n$};
\draw (0,-1.3) circle (0.3);
\node at (1.1,-1.3) {$b_\ell$};
\node at (2.2,-1.3) {$a_n$};
\draw (2.2,-1.3) circle (0.3);
\end{tikzpicture}
getting same state / state does not change by measurment of different measurable in midde.
\(a_n - \) eigen value of \(\hat A\)
\(b_\ell - \) eigen value of \(\hat B\)
what if \([A,B] \neq 0\) , what is \((\Delta A)\cdot(\Delta B)\)?
\[
\Delta A\cdot\Delta B \geq \frac{1}{2}\left|\left\langle [A,B]\right\rangle\right|
\]
uncertainty principle.
\[
\Delta A = \langle A^2\rangle - \langle A\rangle^2 = \left\langle (A - \langle A\rangle)^2\right\rangle
\]
lets call
\[
\begin{aligned}
A - \langle A\rangle &= A'\\
B - \langle B\rangle &= B'
\end{aligned}
\]
then, what is \(\left\langle[A,B]\right\rangle\) ?
\[
\left\langle [A,B]\right\rangle = \left\langle [A',B']\right\rangle
\]
as \(\langle A\rangle\) is a no which is subtructed from \(A\). \(\left(A - \langle A\rangle = A'\right)\)
\[
\Rightarrow \quad \left\langle\Psi\left|[A'B']\right|\Psi\right\rangle = \left\langle[A,B]\right\rangle
\]
\[
\Rightarrow \quad \left\langle\Psi|A'B'|\Psi\right\rangle - \left\langle\Psi|B'A'|\Psi\right\rangle = \left\langle[A,B]\right\rangle
\]
since \(A', B', A, B\) are hermitian
\(\left(\begin{aligned}&A,B \text{ are hermitian}\\ &\text{so} \Rightarrow A', B' \text{ are too.}\end{aligned}\right)\)
\[
[A,B] = AB - BA
\]
\[
= \left(A' + \langle A\rangle\right)\left(B' + \langle B\rangle\right) - \left(B' + \langle B\rangle\right)\left(A' + \langle A\rangle\right)
\]
\[
= A'B' - B'A' + A'\langle B\rangle + B'\langle A\rangle + \langle A\rangle\langle B\rangle
\]
\[
= A'B' - B'A' = [A',B']
\]
[the cross terms \(A'\langle B\rangle\), \(B'\langle A\rangle\) and \(\langle A\rangle\langle B\rangle\) are struck out]
\[
\Rightarrow \quad \left\langle A'\Psi|B'\Psi\right\rangle - \left\langle B'\Psi|A'\Psi\right\rangle = \left\langle[A,B]\right\rangle
\]
\[
\Rightarrow \quad \left\langle A'\Psi|B'\Psi\right\rangle - \left\langle A'\Psi|B'\Psi\right\rangle^{*} = \left\langle[A,B]\right\rangle
\]
\(\left\{\begin{aligned} x &= a+ib\\ x^* &= a-ib\\ x - x^* &= 2ib\end{aligned}\right.\)
\[
2i\,\text{Im}\left(\left\langle A'\Psi|B'\Psi\right\rangle\right) = \left\langle[A,B]\right\rangle
\]
\[
\left|\text{Img}\left(\left\langle A'\Psi|B'\Psi\right\rangle\right)\right| = \frac{1}{2}\left|\left\langle[A,B]\right\rangle\right|
\]
also.
\[
\left|\text{Im}\left(\left\langle A'\Psi|B'\Psi\right\rangle\right)\right| \leq \left|\left\langle A'\Psi|B'\Psi\right\rangle\right|
\]
\[
\frac{1}{2}\left|\left\langle[A,B]\right\rangle\right| \leq \left|\left\langle A'\Psi|B'\Psi\right\rangle\right|
\]
\(\left\{\langle v|w\rangle \leq |v|\,|w|\right.\) cauchy inequality \(\}\)
\[
\frac{1}{2}\left|\left\langle[A,B]\right\rangle\right| \leq \left\langle A'\Psi|A'\Psi\right\rangle^{1/2}\left\langle B'\Psi|B'\Psi\right\rangle^{1/2}
\]
\[
\frac{1}{2}\left|\left\langle[A,B]\right\rangle\right| \leq \left\langle\Psi|A'A'|\Psi\right\rangle^{1/2}\left\langle\Psi|B'B'|\Psi\right\rangle^{1/2}
\]
\[
\leq \Delta A\cdot\Delta B
\]
\(\left\{\left\langle\Psi|A'^2|\Psi\right\rangle = \left\langle (A - \langle A\rangle)^2\right\rangle = \Delta A^2\right.\)
so
\[
\boxed{\Delta A\cdot\Delta B \geq \frac{1}{2}\left|\left\langle[A,B]\right\rangle\right|}
\]
Question ?
even if \([A,B]\neq0\) we can have \(\left\langle[A,B]\right\rangle = 0\) (? How)
if,
\[
\left.\begin{aligned}A &= x\\ B &= p\end{aligned}\right. \qquad [A,B] = [x,p] = i\hbar I
\]
\(\{\langle I\rangle = 1\)
\[
\Delta A\,\Delta B \geq \frac{\hbar}{2}
\]
\[
\boxed{\Delta x\,\Delta p \geq \frac{\hbar}{2}}
\]
[the notebook writes the heading as "The shrödinger and Heisenberg pictures :-"]
what we did so far is shrödinger's picture. In which state of any system can be represented by state vector \(|\Psi(t)\rangle\) in some hilbert space.
\(|\Psi(t)\rangle\) is explicitly time dependent, and the time dependence is specified by the shrödinger's equation.
\[
i\hbar\frac{d}{dt}|\Psi(t)\rangle = \hat H|\Psi(t)\rangle
\]
\[
|\Psi(t)\rangle = e^{\frac{-iH(t-t_0)}{\hbar}}\cdot|\Psi(t_0)\rangle
\]
where \(|\Psi(t)\rangle\) is obtained from action of unitary operator
\[
U(t,t_0) = e^{-\frac{iH(t-t_0)}{\hbar}}
\]
which takes \(|\Psi(t_0)\rangle\) to \(|\Psi(t)\rangle\)
It is unitary operator because it preserves norm of vector just like unitary operator does.
\[
\left\langle\Psi(t)|\Psi(t)\right\rangle = \left\langle\Psi(t_0)|\Psi(t_0)\right\rangle
\]
\[
UU^{+} = U^{+}U = I
\]
Note :-
\[
U(t,t_0) = e^{-\frac{iH(t-t_0)}{\hbar}}
\]
only in case of autonomous Hamiltonian
Since Norm of state vector is preserved, it looks like state vector is rotating in Hilbert space.
remember \(|\Psi\rangle\) and \(e^{i\alpha}|\Psi\rangle\) have same norm.
It is as if \(|\Psi\rangle\) is rotating on surface of sphere, except it is happening in \(\infty\)-diamensional hilbert space. It is strictly not a rotation but a unitary transformation.
unitary transformation is counter part of canonical transformation which preserves phase-space volume.
\[
\{q_0, p_0\} = 1
\]
\[
\{q_t, p_t\} = 1
\]
In same way just as classical Hamiltonian dynamics is gradual unfolding of sequence of infinitesimal canonical transformation, in exactly same way the Quantum evolution is gradual unfolding of sequence of unitary transformation on state vector.
\[
\langle A\rangle(t) = \left\langle\Psi(t)|A|\Psi(t)\right\rangle
\]
In shrodinger picture \(\hat A\) are time independent but \(|\Psi(t)\rangle\) is time dependant, lets use subscript 's' for shrödinger picture.
\[
\langle A_s\rangle(t) = \left\langle\Psi_s(t)\left|A_s\right|\Psi_s(t)\right\rangle
\]
\[
|\Psi_s(t)\rangle = e^{\frac{-iHt}{\hbar}}\,|\Psi_s(0)\rangle
\]
\[
\langle\Psi_s(t)| = \langle\Psi_s(0)|\,e^{\frac{+iHt}{\hbar}}
\]
\[
\langle A_s\rangle(t) = \left\langle\Psi_s(0)\left|e^{\frac{iHt}{\hbar}}\,A\,e^{\frac{-iHt}{\hbar}}\right|\Psi_s(0)\right\rangle
\]
"Now Heisenberg comes and says, operator itself is time dependent, where as wave function / state vector does not have dependence."
\[
\langle A\rangle(t) = \left\langle\Phi\left|A(t)\right|\Phi\right\rangle
\]
\[
A(t) = e^{\frac{+iHt}{\hbar}}\, A\, e^{\frac{-iHt}{\hbar}}
\]
\[
\boxed{\langle A\rangle(t) = \left\langle\Psi_H\left|A_H(t)\right|\Psi_H\right\rangle}
\]
It must be true for every observable \(\hat A\) and every state vector \(|\Psi_H\rangle\).
These two are completely consistent with each other, provided,
\[
\left.\begin{aligned}
|\Psi_H\rangle &= |\Psi_s(0)\rangle\\
A_H(t) &= e^{\frac{iH_st}{\hbar}}\,A_s\,e^{\frac{-iH_st}{\hbar}}
\end{aligned}\right\}
\]
therefore,
\[
A_H(0) = A_s \qquad (\text{by def}^{\text{n}}).
\]
so, our \(A_H(t)\) becomes
\[
A_H(t) = e^{\frac{iH_st}{\hbar}}\,A_H(0)\,e^{\frac{-iH_st}{\hbar}}
\]
lets \(A = H\), find \(H_H(t) = ?\)
\[
H(t) = e^{\frac{iH_st}{\hbar}}\,H_s\,e^{\frac{-iH_st}{\hbar}}
\]
since \(\left[e^{\frac{iH_st}{\hbar}}, H_s\right] = 0\)
\[
\boxed{H(t) = H_s} = H \ , \ \text{say}
\]
So, Hamiltonian operator is same in both pictures so let us call it \(H\).
\[
A_H(t) = e^{\frac{iHt}{\hbar}}\,A_H(0)\,e^{\frac{-iHt}{\hbar}}
\]
Hamiltonian operator is a special operator which does not change with time.
\[
H(t) = e^{\frac{iHt}{\hbar}}\,H_H(0)\,e^{\frac{-iHt}{\hbar}} = H(t=0).
\]
letus go backward and find,
\[
\frac{d}{dt}A_H(t) = ? \qquad\qquad A_H(t) = e^{\frac{iHt}{\hbar}}\,A_H(0)\,e^{\frac{-iHt}{\hbar}}
\]
\[
\frac{dA_H(t)}{dt} = \frac{iH}{\hbar}A_H(t) + e^{\frac{iHt}{\hbar}}A_H(0)e^{\frac{-iHt}{\hbar}}\left(\frac{-iH}{\hbar}\right)
\]
\[
= \frac{iH}{\hbar}A_H(t) + A_H(t)\left(\frac{-iH}{\hbar}\right)
\]
\[
= \frac{i}{\hbar}\left(H\,A_H(t) - A_H(t)\,H\right)
\]
\[
\frac{dA_H(t)}{dt} = \frac{i}{\hbar}\left[H,\ A_H(t)\right]
\]
\[
i\hbar\,\frac{dA_H(t)}{dt} = -\left[H,\ A_H(t)\right]
\]
\[
\boxed{i\hbar\,\frac{d}{dt}A_H(t) = \left[A_H(t),\ H\right]} \qquad \text{or} \qquad \frac{dA_H(t)}{dt} = \frac{\left[A_H(t), H\right]}{i\hbar}
\]
[\(\leftarrow\) Heisenberg equation of motion.]
solution to this equation is simply
\[
A_H(t) = e^{\frac{+iHt}{\hbar}}\,A_H(0)\,e^{\frac{-iHt}{\hbar}}
\]
classically if \(H = H(q,p)\)
\[
\frac{dA(q,p)}{dt} = \{A, H\}
\]
So indeed the poisson brackett has been replaced by \(\frac{[A,H]}{i\hbar}\)
\[
\begin{array}{c|c|c}
& \text{Shrodinger picture} & \text{Heisenberg picture}\\
\hline
\hat A & \text{Time-independent }(A_s) & \text{Time dependent } A_H(t)\\
\hline
|\Psi\rangle & \text{time dependent } |\Psi(t)\rangle & \text{Time independent } |\Psi_H\rangle\\
\hline
\text{E.O.M.} & i\hbar\dfrac{d}{dt}|\Psi(t)\rangle = \hat H|\Psi(t)\rangle & i\hbar\dfrac{dA_H(t)}{dt} = [A,H]
\end{array}
\]
\[
\langle A\rangle(t) = \left\langle\Psi_s(t)\left|A_s\right|\Psi_s(t)\right\rangle = \left\langle\Psi_H\left|A_H(t)\right|\Psi_H\right\rangle
\]
Note :- That is like saying when average value changes either probability distribution changes and variable itself is not time dependent, or I say no, distribution is fixed for all, where as variables has undergone dignamical evolution.
Heisenberg and shrodinger pictures are connected by unitary transformation.
Also, prove that
\[
\frac{d}{dt}\langle A_s\rangle \equiv \left\langle\frac{dA_H}{dt}\right\rangle
\]
\[
\langle A_s\rangle = \left\langle\Psi_s(t)\left|A_s\right|\Psi_s(t)\right\rangle \Big/ \underbrace{\left\langle\Psi(t)|\Psi(t)\right\rangle}_{1\ (\text{normalized assumptn})}
\]
\[
\frac{d}{dt}\langle A_s\rangle = \frac{d}{dt}\left\langle\Psi_s(t)\left|A\right|\Psi(t)\right\rangle + \left\langle\Psi(t)\left|A\right|\frac{d}{dt}|\Psi(t)\right\rangle
\]
using,
\[
i\hbar\frac{d}{dt}|\Psi(t)\rangle = H|\Psi(t)\rangle
\]
\[
\frac{d}{dt}\langle\Psi(t)| = \frac{\langle\Psi(t)|H}{-i\hbar}
\]
\[
\frac{d}{dt}\langle A_s\rangle = \left[-1\left\langle\Psi(t)\left|\frac{HA}{i\hbar}\right|\Psi(t)\right\rangle + \left\langle\Psi(t)\left|\frac{AH}{i\hbar}\right|\Psi(t)\right\rangle\right]
\]
\[
= \left\langle\Psi(t)\left|\frac{AH - HA}{i\hbar}\right|\Psi(t)\right\rangle
\]
\[
= \left\langle\Psi(t)\left|\frac{[A,H]}{i\hbar}\right|\Psi(t)\right\rangle
\]
\[
\frac{d}{dt}\langle A_s\rangle = \left\langle\Psi(t)\left|\frac{dA_H(t)}{dt}\right|\Psi(t)\right\rangle = \left\langle\frac{dA_H}{dt}\right\rangle
\]
In general if shrödinger equation has operators which are explicit dependent on time.
\[
A = x^2 t + p t^3
\]
then
\[
\boxed{\frac{dA_H(t)}{dt} = \frac{[A,H]}{i\hbar} + \frac{\partial A}{\partial t}}
\]
\(\left(\frac{\partial A}{\partial t} = x^2 + 3pt^2\ \text{in above case}\right)\)
exactly like \(\frac{dA}{dt} = \{A, H\} + \frac{\partial A}{\partial t}\) from classical mechanics.
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