• #Lecture 3
  • #Linear independence :-
  • #Span :-
  • #Basis :-
  • #Diamentionality :-
  • #Normalization :-
  • #Gram -- Schmidt orthonormalization :-
  • #(projection operator)
• #Lecture 4
  • #Redefining normalization condition for legendre polynomials.
  • #Fourier inversion formula

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if \(|\phi_1\rangle, |\phi_2\rangle, \cdots |\phi_n\rangle\) are linearly independent

then if

\[ \sum_i a_i |\phi_i\rangle = 0 \]

so sol\(^{\text{n}}\) becomes \(a_i = 0 = a_1 = a_2\)

A set of vectors \(|\phi_1\rangle, |\phi_2\rangle \cdots |\phi_n\rangle\) are said to span the LVS, iff any vector can be written as linear combination of these vectors. \(\left(|\phi_1\rangle -, |\phi_n\rangle\right)\).

example

(1) \(\hat e_x\) & \(\hat e_y\) \(\longrightarrow\) L.I , do not span \(\mathbb{R}^3\)

(2) \(\hat e_x, \hat e_y, \hat e_z\) \(\rightarrow\) L.I, span \(\mathbb{R}^3\) (orthogonal set) (unit vectors)

(3) \(\hat e_x, \hat e_y, \hat e_z, \hat e_{x} + \hat y\) \(\longrightarrow\) span, not L.I

(4) \(\hat e_x,\ \hat e_y + \hat e_x,\ \hat e_z + \hat e_y + \hat e_x\) \(\longrightarrow\) L.I, span \(\mathbb{R}^3\) (oblique set of coordinate axis vectors) (not-unit vector)

A set of vectors which are linearly independent & span the LVS forms a basis set in the LVS.

\[ \langle\phi_i|\phi_j\rangle = \delta_{ij} \]

For orthogonal basis \(\delta_{ij} = 0\) if \(i \neq j\)

The no. of basis vectors defines the diamentionality of LVS.

\(\therefore\) dim \(V = \#\) of vectors in basis set.

\(\longrightarrow\) Infinite dimentional LVS does not have finite basis

\(\mathbb{R}^2\) is 2-d ; \(\mathbb{R}^3\) is 3-D LVS, \(\cdots\) \(\mathbb{R}^n\) is \(n\)-dimensional LVS.

Here is what can go wrong for infinite-diamensional LVS.

for \(\mathbb{R}^n\):

\[ \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_N \end{pmatrix} = |\phi\rangle \] \[ \langle\phi|\phi\rangle = \sum_{i=1}^{N} |x_i|^2 = \|\phi\|^2 \]

[we could have defined \(\|\phi\|\) as \(\left(\sum_{i=1}^{N}|x_i|^p\right)^{1/p}\), but for \(p = 2\) only \(l_p\) is self dual. dual of \(l_p\) \(= l_q\) such that \(\frac{1}{p} + \frac{1}{q} = 1\) (for \(p = q = 2\), \(l_p = l_q\))]

for \(N \to \infty\) there is no gaurentee that \(\langle\phi|\phi\rangle\) converges / have finite value.

so we have to make this piece (length of vector) to be finite.

\[ \sum_{i=1}^{N} |x_i|^2 < \infty \]

[\(|x_i|^2\) should vanish before \(n \to \infty\) (i.e. \(n^{-1} \to 0\)) \(\Rightarrow\) \(|x_i|\) should vanish faster than \(\frac{1}{\sqrt n}\)]

\[ |f\rangle = \begin{pmatrix} f_1\\ f_2\\ f_3 \end{pmatrix} = \sum_{i=1}^{N} f_i |i\rangle \;\Rightarrow\; \langle f|f\rangle < \infty \;\Rightarrow\; \sum_{i=1}^{N}|f_i|^2 < \infty \]

linear vector space of square summable sequences. , \(l_2\)

ex.,

(1)

\[ (x_1, x_2, x_3 \cdots x_\infty) = \left(1, \tfrac{1}{\sqrt2}, \tfrac{1}{\sqrt3}, \tfrac{1}{\sqrt4} \cdots \right) \]

\(\hookrightarrow\) diverges. not an element of \(l_2\).

(2)

\[ x_r = \frac{1}{r^\epsilon} \qquad \text{for, } \epsilon > \tfrac{1}{2} \text{ it is in } l_2 \]

[\(|\phi\rangle = (x_1, x_2 \cdots x_r)\) is in \(l_2\) if \(x_r = \frac{1}{r^\epsilon}\)

\(\langle\phi|\phi\rangle = (x_1^2 + x_2^2 + \cdots x_r^2)\) , \(\sum_{r=1}^{\infty}(x_r)^2 \to\) converges, \(\sum_r \frac{1}{r^{2\epsilon}} \to\) converges, \(2\epsilon > 1 \Rightarrow (\epsilon > \tfrac{1}{2})\)]

(3)

\[ x_n = \frac{\ln n}{n^{0.6}} \quad \text{in } l_2 \] \[ |x_n|^2 = \frac{(\ln n)^2}{n^{1.2}} \in l_2 \]

also \(\frac{(\ln n)^{100}}{n^{0.6}} \in l_2\)

(log is weaker than power & power is weaker than exponential)

\[ \frac{|\psi\rangle}{\sqrt{\langle\psi|\psi\rangle}} = \frac{|\psi\rangle}{\|\psi\|} \]

finding two orthonormal basis vectors.

let us say we have any \(|\psi_1\rangle, |\psi_2\rangle\) two basis vectors

we can produce \(|\phi_1\rangle\) & \(|\phi_2\rangle\) such that \(\langle\phi_i|\phi_j\rangle = \delta_{ij}\).

\(|\phi_1\rangle = \) one of orthonormal vector would be simply \(\dfrac{|\psi_1\rangle}{\|\psi_1\|}\)

\[ |\phi_2\rangle = \frac{|\psi_2\rangle - \langle\phi_1|\psi_2\rangle|\phi_1\rangle}{\|\xi\|} = |\phi_2\rangle \]

This will gaurantee that

\[ \langle\phi_1|\phi_2\rangle = \boxed{0} \]

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0]
  \draw[->] (0,0) -- (2.6,2.1) node[above left,pos=0.95] {$|\psi_2\rangle$};
  \draw[->] (0,0) -- (1.7,0) node[below,pos=0.6] {$|\phi_1\rangle$};
  \draw[->] (0,0) -- (4.0,0) node[above right] {$|\psi_1\rangle$};
  \draw[dashed] (2.6,0) -- (2.6,2.1);
  \draw[->] (2.6,0.0) -- (2.6,1.6) node[right,pos=0.7] {$|\phi_2\rangle$};
  \draw[<->] (0,-0.75) -- (2.6,-0.75);
  \node[below] at (1.3,-0.78) {$\langle\phi_1|\psi_2\rangle|\phi_1\rangle$};
\end{tikzpicture}

\(\hookrightarrow\) or

\[ \Big\{|\phi_1\rangle\langle\phi_1|\Big\}|\psi\rangle = \langle\phi_1|\psi\rangle\,|\phi_1\rangle \] \[ (n\times 1)(1\times n) \;\longrightarrow\; (n \times n) \]

(projection operator)

let us say

\[ |\psi\rangle = v_1|\phi_1\rangle + v_2|\phi_2\rangle + v_3|\phi_3\rangle \]

magnitude projection along \(|\phi_1\rangle\) \(\hookrightarrow\) \(\left(\langle\phi_1|\psi\rangle = v_1\right)\).

projection of \(|\psi\rangle\) along \(|\phi_1\rangle\) \(=\)

\[ \Big(|\phi_1\rangle\langle\phi_1|\Big)\psi\rangle = |\phi_1\rangle v_1 = v_1|\phi_1\rangle \]

\(\hookrightarrow\) projection operator. (of \((n\times n)\) matrix identification)

[In LVS in addition to linear vectors we have linear operators & these operators acts upon these vectors to produce another vector.]

\[ \vec v = v_1 \hat i + v_2 \hat j + v_3 \hat k \qquad \boxed{P = \hat i\hat i\ (\hat j\hat j)\ \hat k\hat k} \] \[ (\hat i\hat i)\vec v = \vec v \cdot \hat i\,\hat i = v_1 \hat i \quad \text{or} \quad \hat i v_1 \]

\(\hookrightarrow\) dyadic in tensor algebra / or tensor product

\[ P_n = |\phi_n\rangle\langle\phi_n| \] \[ P_n^2 = |\phi_n\rangle\langle\phi_n|\phi_n\rangle\langle\phi_n| \]

(\(\textcircled{1}\) unity iff \(\|\phi_n\| = 1\))

\[ = |\phi_n\rangle\langle\phi_n| \] \[ \left(\sum_n |\phi_n\rangle\langle\phi_n|\right)|\psi\rangle = 1\cdot|\psi\rangle \]

so,

\[ \sum_n |\phi_n\rangle\langle\phi_n| = \text{unit operator.} \]

we know that,

\[ P_n^2 = P_n \qquad \text{(Idempote matrix) i.e. for matrix } M;\ M^{\alpha} = M \text{ for any } \alpha \in Z \] \[ P_n^2 - P_n = 0 \] \[ P_n(P_n - \mathbb{1}) = 0 \qquad \text{eigen value} = (0,1) \]

\(\hookrightarrow\) caushy relation \(\lambda(\lambda-1) = 0\) will be satisfied by matrix itself.

These are two very important properties of LVS.

1. orthonormality \[ \langle\phi_i|\phi_j\rangle = \delta_{ij} \]
2. completeness. \[ \sum_{n=1}^{n} |\phi_n\rangle\langle\phi_n| = \mathbb{1} \] ex. \[ |\phi_1\rangle = \begin{pmatrix} 1\\ 0\end{pmatrix} \qquad |\phi_2\rangle = \begin{pmatrix} 0\\ 1\end{pmatrix} \] [\(|\phi_1\rangle\) & \(|\phi_2\rangle\) also provides basis for operators] \[ \langle\phi_i|\phi_j\rangle = \delta_{ij} \] \[ |\phi_1\rangle\langle\phi_1| = \begin{pmatrix}1\\0\end{pmatrix}\begin{pmatrix}1 & 0\end{pmatrix} = \begin{pmatrix}1 & 0\\ 0 & 0\end{pmatrix} \] \[ |\phi_2\rangle\langle\phi_2| = \begin{pmatrix}0\\1\end{pmatrix}\begin{pmatrix}0 & 1\end{pmatrix} = \begin{pmatrix}0 & 0\\ 0 & 1\end{pmatrix} \] \[ \sum_{i=1}^{2}|\phi_i\rangle\langle\phi_i| = |\phi_1\rangle\langle\phi_1| + \langle\phi_2|\phi_2\rangle = \begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix} = I \] \[ |\phi_1\rangle\langle\phi_2| = \text{operator} = \begin{pmatrix}1\\0\end{pmatrix}\begin{pmatrix}0 & 1\end{pmatrix} = \begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix} \] \[ |\phi_2\rangle\langle\phi_1| = \begin{pmatrix}0\\1\end{pmatrix}\begin{pmatrix}1 & 0\end{pmatrix} = \begin{pmatrix}0 & 0\\ 1 & 0\end{pmatrix} \]

we know that,

\[ \begin{pmatrix} a & b\\ c & d\end{pmatrix} = a\begin{pmatrix}1 & 0\\ 0 & 0\end{pmatrix} + b\begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix} + c\begin{pmatrix}0 & 0\\ 1 & 0\end{pmatrix} + d\begin{pmatrix}0 & 0\\ 0 & 1\end{pmatrix} \] \[ \left\{ \begin{aligned} a &= a_{11}\\ b &= a_{12}\\ c &= a_{21}\\ d &= a_{22} \end{aligned} \right. \] \[ = a_{11}|\phi_1\rangle\langle\phi_1| + a_{12}|\phi_1\rangle\langle\phi_2| + a_{21}|\phi_2\rangle\langle\phi_1| + a_{22}|\phi_2\rangle\langle\phi_2| \] \[ \begin{aligned} a_{11} &= \langle\phi_1|A|\phi_1\rangle\\ a_{12} &= \langle\phi_1|A|\phi_2\rangle\\ a_{21} &= \langle\phi_2|A|\phi_1\rangle\\ a_{22} &= \langle\phi_2|A|\phi_2\rangle \end{aligned} \qquad A_{nm} = \langle\phi_n|A|\phi_m\rangle \]

Any vector \(|\psi\rangle\) can be uniqly expanded in form

\[ |\psi\rangle = \sum_n c_n |\phi_n\rangle \]

Any operator \(A\) can be expanded in form ; where \(\{|\phi_n\rangle\langle\phi_m|\}\) forms basis.

\[ A = \sum_{n,m} A_{nm}\,|\phi_n\rangle\langle\phi_m| \]

\(l_2\) : space of square summable sequence

\[ \sum_{i=1}^{\infty} |x_i|^2 < \infty \] \[ \Big(l_2(a,b)\Big) = \int_a^b |f(x)|^2\,dx \quad (< \infty) \]

necessary cond\(^{\text{n}}\) for \(f(x)\) to be \(l_2\) is that

\[ \left\{ \lim_{x \to \pm\infty} f(x) \longrightarrow 0 \right\} \]

why not \((x \to b, a)\)?

\(\hookrightarrow\) it should die to zero sufficiently rapidly

\(\lim_{|x| \to \infty} f(x) \longrightarrow 0\) should go faster than \(\frac{1}{\sqrt{|x|}}\) (?) How.

\(\rightarrow\) we will require wave function to be bounded.

Let's look at space of square integrable functions. \(\rightarrow\) \(l_2(-1,1)\)

where,

\(L_1 \rightarrow\) space of integrable functions.

\(l_2 \rightarrow\) space of square integrable fun. (self dual)

\[ l_2(-1,1) = \int_{-1}^{1} dx\,|f(x)|^2 < \infty \]

[let us find basis set for any function. so that we can expand it in the given basis]

let \(|\psi_0\rangle, |\psi_1\rangle, |\psi_2\rangle, |\psi_3\rangle \cdots\) \(\to\) \(x^0, x^1, x^2, \cdots\) forms basis. (orthonormal)

\(f_0(x) = c_0\)

whats

\[ \int_{-1}^{1} dx\ c_0^2 = 1 \qquad \Rightarrow \qquad c_0 = \frac{1}{\sqrt2} \] \[ f_1(x) = ax + b \qquad \text{(in general).} \]

find \(a, b\) such that

\[ \left\{ \begin{aligned} \langle\phi_0|\phi_0\rangle &= 1\\ \langle\phi_1|\phi_1\rangle &= 1\\ \langle\phi_0|\phi_1\rangle &= 0 \end{aligned} \right. \] \[ \Rightarrow \int_{-1}^{1} dx\ f_1(x) f_0(x) = 0 \] \[ \Rightarrow \int_{-1}^{1} (ax+b)\,\frac{1}{\sqrt2}\,dx = 0 \] \[ \Rightarrow \left[\frac{ax^2}{2} + bx\right]_{-1}^{1} = 0 \] \[ \left(\frac{a}{2} + b\right) - \left(\frac{a}{2} - b\right) = 0 \] \[ 2b = 0 \quad \Rightarrow \quad \boxed{b = 0} \]

So

\[ f_1(x) = ax \]

so normalizing

\[ \int_{-1}^{1} f_1(x)^2\,dx = 1 \] \[ \int a^2 x^2\,dx = 1 \] \[ \frac{a^2}{3}\left(x^3\right)\Big|_{-1}^{1} = 1 \] \[ \frac{a^2}{3}(1+1) = 1 \] \[ \boxed{a = \left(\frac{3}{2}\right)^{1/2}} \] \[ f_2(x) = ax^2 + bx + c \] \[ \begin{aligned} \int_{-1}^{1} dx\ f_2(x) f_0(x) &= 0\\ \int dx\ f_2(x) f_1(x) &= 0\\ \int dx\ \big(f_2(x)\big)^2 &= 1 \end{aligned} \]

three cond\(^{\text{n}}\) are sufficent to find \(a, b, c\).

Let us Redefine Normalization condition to get rid of weird constants.

lets call these functions as \(P_n(x)\) (Legendre polynomials)

we have,

\[ \int_{-1}^{1} dx\ P_n(x) P_m(x) = \frac{2}{2n+1}\delta_{nm} = \frac{2}{2m+1}\delta_{mn} = \frac{2\ \delta_{mn}}{\sqrt{(2n+1)(2m+1)}} \]

if

\[ P_0(x) = 1 \qquad P_1(x) = x, \qquad P_2(x) = \frac{3x^2-1}{2} \]

Even the fourier transform (series) is really writing a function in some basis set. Any periodic fun. can be written as (expanded in fourier series). where '\(\cos nx\)' and '\(\sin nx\)' are basis set for state vector \(f(x)\).

\[ f(x+\beta) = f(x) \] \[ f(x) = \sum_{n=0}^{\infty}\big(a_n \cos nx + b_n \sin nx\big) \qquad (n = 0, 1, 2, 3 \cdots) \]

\(a_n, b_n\) are coefficients of expansion

\[ f(x) = \sum_{n=-\infty}^{\infty} c_n e^{inx} \]

[Generalized normalized condition

\[ \int_{-\infty}^{\infty} d\mu(x)\ \phi_n^{*}(x)\phi_m(x) = A_n \delta_{nm} \qquad (A_n \to \text{weight factor}) \]

\(d\mu(x) \to\) measure

ex.

\[ \int_{-\infty}^{\infty} dx\, e^{-x^2}\phi_n^{*}(x)\phi_m(x) = A_n \delta_{nm} \]

\(\hookrightarrow\) \(\phi_n(x) = \) Hermite polynomials

\(d\mu(x) \to\) measure is used to make sure that \(\int_{-\infty}^{\infty} d\mu(x)\phi_n^{*}(x)\phi_n(x) < \infty\) remains finite.

for \(d\mu(x) = dx\,e^{-x}\), \(\phi(x)\) forms Laguerre polynomial.]

If \(f(x)\) is not periodic, we can expand \(f(x)\) in fourier integral

\[ f(x) = \int_{-\infty}^{\infty} dk\ e^{ikx}\,\tilde f(k) \]

for orthonormality

\[ \langle\phi_n|\phi_m\rangle = 0 \ , \quad \Longleftrightarrow \quad \int_a^b f_n^{*}(x) f_m(x)\,dx = \delta_{mn} \]

for completeness

\[ \sum_{n=1}^{\infty} |\phi_n\rangle\langle\phi_n| = 1 \] \[ \sum_n f_n^{*}(x) f_n(x') = \delta(x-x') \] \[ \int dx'\ \delta(x-x')\,\psi(x') = \psi(x) \]

\(\rightarrow\) unit operator

[so in this way we can define our own polynomials. let, \(d\mu(x) = dx\cdot x^2\), \(\phi_n(x)\) forms Vishal's polynomial. \(\smile\)]

If \(\{|\phi_n\rangle\}\) is an orthonormal basis in any LVS, any ket \(|\psi\rangle \in V\) can be uniquely expanded as

\[ |\psi\rangle = \sum_{n=1}^{N} c_n |\phi_n\rangle \qquad (n \to \infty) \text{ for LVS of infinite diamensions.} \]

where,

\[ c_m = \langle\phi_m|\psi\rangle \]

If we have different basis say \(|\chi_n\rangle\)

\[ |\psi\rangle = \sum_{n=1}^{N} d_n |\chi_n\rangle = \sum_i d_i |\chi_i\rangle \qquad \cdots \textcircled{1} \]

where

\[ d_j = \langle\chi_j|\psi\rangle \qquad \cdots \textcircled{2} \]

To find relation between \(c_m\) and \(d_i\)

we can expand \(|\chi_i\rangle\) in \(|\phi_n\rangle\) basis

\[ |\chi_i\rangle = \sum_m h_{mi}|\phi_m\rangle \] \[ \Rightarrow \quad h_{mi} = \langle\phi_m|\chi_i\rangle \]

So, eq\(^{\text{n}}\) \(\textcircled{1}\) becomes

\[ |\psi\rangle = \sum_i^{N} d_i|\chi_i\rangle = \sum_i \sum_m d_i\,h_{mi}\,|\phi_m\rangle \]

or

\[ \sum_n \left(\sum_i h_{ni} d_i\right)|\phi_n\rangle \]

So

\[ \boxed{c_n = \sum_i h_{ni} d_i} \] \[ c_n = \langle\phi_n|\psi\rangle = \sum_i \langle\phi_n|\chi_i\rangle\langle\chi_i|\psi\rangle \] \[ = \langle\phi_n|\left(\sum_i |\chi_i\rangle\langle\chi_i|\right)|\psi\rangle \]

(\(\to\) unit operator)

\[ = \langle\phi_n|\psi\rangle \]

So the whole point is any time we take new basis we automatically insert the identity (unit operator).

\[ \langle\psi| = \sum_{n=1}^{N} c_n^{*}\langle\phi_n| \ , \qquad |\psi\rangle = \sum_{n=1}^{N} c_n|\phi_n\rangle \] \[ \langle\psi|\psi\rangle = \sum_{n=1}^{N} c_n^{*} c_n \langle\phi_n|\phi_n\rangle = \sum_n |c_n|^2 < \infty \]

as, \(\langle\phi_n|\phi_n\rangle = \underline{1}\)

we can expand any function \(f(x)\) on basis set formed by polynomials (Legendre, Laguerre polynomials, hermite polynomials and fourier series (if \(f(x)\) is periodic). If \(f(x)\) is not periodic then we can expand it in continuous basis \(\{e^{ikx}\}\) labelled by '\(k\)'.

\[ f(x) = \int_{-\infty}^{\infty} dk\ e^{ikx}\,\tilde f(k) \]

\(\tilde f(k) = \) fourier transform of \(f(x)\).

[\(e^{ikx} \sim\) unit vectors / basis set. \(\tilde f(k)\) are like components in \(e^{ikx}\) direction / basis.]

for \(l_2(-1,1)\)

\[ f(\theta) = \sum_{\ell=0}^{\infty} c_\ell\,P_\ell(\cos\theta) \] \[ c_\ell = \tfrac{1}{2}\int_{-1}^{1} d(\cos\theta)\ P_\ell^{*}(\cos\theta) f(\theta) \] \[ \left\{ P_\ell(\cos\theta) \text{ is real so, } P_\ell^{*}(\cos\theta) = P_\ell(\cos\theta)\right. \]

\(\rightarrow\) comes from orthogonality relation

\[ \frac{1}{2\pi}\int_{-\infty}^{\infty} dk\ e^{ik(x-x')} = \delta(x-x') \] \[ f(x) = \int_{-\infty}^{\infty} dk\ \tilde f(k)\,e^{ikx} \ , \qquad \tilde f(k) = \frac{1}{2\pi}\int_{-\infty}^{\infty} dx\ e^{-ikx} f(x) \]

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