• #Lecture 7 :-
  • #Simultaneous Eigen states :-
  • #The uncertainty in an observable :-
  • #IN Momentum basis :-
• #Lecture 8
  • #The shrödinger equation For a particle :-
  • #Stationary states (eigen states) :-
  • #Particle in 1-D BOX :-
  • #Solving problem in momentum space wave function
  • #Particle in a ring :- (circle).
  • #Particle in 2-D box :-

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operators in Q.M are repesented as \(\hat A\).

\(\ast\) A particle moving in one diamension with no potential.

1. Analog of \(H\) in Q.M \[ \hat H = \frac{\hat p^2}{2m} \]

Shrodinger eq\(^{\text{n}}\) becomes.

\[ i\hbar\,\frac{d}{dt}|\Psi(t)\rangle = \hat H|\psi(t)\rangle = \frac{\hat p^2}{2m}|\Psi(t)\rangle \] \[ |\Psi(t)\rangle = e^{-\frac{i\hat H t}{\hbar}}\,|\Psi(0)\rangle \] \[ [\hat x, \hat p] = i\hbar\,\mathbb{1} \]

Just as we take coordinate system / axis for \((x, p)\), In Q.M. we must choose some basis. (in terms of which every other vector is expressed).

\(\ast\) Most basic physical measurable quantity is \(\hat x, \hat p\) everything else is derived.

For convenience we should choose the basis set formed from eigen vector of operator (which is associated with each some physical measurable of system).

lets use some basis called as position basis.

\[ \{|x\rangle\} \ \text{ such that } \] \[ \hat x |x_0\rangle = x_0 |x_0\rangle \]

\(\hat x = \) operator

\(|x_0\rangle = \) eigen state (\(|x_0\rangle\) is ket corresponding to a particle with precise location \(x_0\)).

\(x_0 = \) eigen value

\(x_0\) can take any no. from \(-\infty\) to \(+\infty\)

\[ x_0 \in (-\infty, \infty). \]

So it is a contineous basis.

lets choose basis orthonormal to each other.

\[ \langle x|x'\rangle = \delta(x - x') = \left\{ \begin{aligned} 1 & \qquad x = x'\\ 0 & \qquad x \neq x' \end{aligned} \right. \]

The completeness relation looks like

\[ \int dx\ |x\rangle\langle x| = \mathbb{1} \qquad \text{completeness.} \]

Any state vector can be written as

\[ |\Psi(t)\rangle = \sum_{n=0}^{\infty} c_n|x_n\rangle \qquad c_n = \langle x_n|\Psi(t)\rangle \]

but basis are contineous, so we should replace it with integral. (as \(|x\rangle\) is contineous basis).

\[ |\Psi(t)\rangle = \int dx\ \langle x_0|\Psi(t)\rangle\,|x_0\rangle \]

[we could hv written as, \(|\Psi(t)\rangle = \mathbb{1}|\Psi(t)\rangle = \int dx\,|x\rangle\langle x|\Psi(t)\rangle\)

\(|\Psi(t)\rangle = \int dx\ \langle x|\Psi(t)\rangle\,|x\rangle\)]

coefficients are now labled by contineous variable '\(x\)'.

\[ \langle x_n|\Psi(t)\rangle = c_x(t) = \Psi(x,t) \]

\(=\) probability amplitude that position is '\(x\)' at time \(t\).

\(|\psi(x,t)|^2 = \) probability density that position lies b/w \(x\), \(x + dx\) at time \(t\).

lets drop subscript '\(n\)' as \(x\) is contineous.

\[ \langle x|\Psi(t)\rangle = \Psi(x,t) = \text{wave function} \]

since it is in position basis, \(\Psi(x,t)\) is called as position wave function.

lets choose momentum basis / momentum space set \(\{|p_0\rangle\}\)

\[ \hat p|p_0\rangle = p_0|p_0\rangle \]

(\(\hat p \to\) operator, \(p_0 \to\) eigen value, \(|p_0\rangle = \) eigen vector.)

\[ \langle p|p'\rangle = \delta(p - p') \qquad \text{orthogonal } \{|p_0\rangle\} \text{ basis set} \] \[ \sum |p\rangle\langle p| \quad \Rightarrow \quad \int dp\ |p\rangle\langle p| = \mathbb{1} \qquad \text{completeness.} \]

so state vector can be written as.

\[ |\Psi(t)\rangle = \int dp\ \langle p|\psi(t)\rangle\,|p\rangle \] \[ \langle p|\Psi(t)\rangle = \tilde\Psi(p,t) \]

(\(\to\) to distinguish from \(\Psi(x,t)\))

\(\hookrightarrow\) wave function in momentum basis. or momentum space wave function

we can write

\[ \tilde\Psi(p,t) = \langle p|\Psi(t)\rangle \] \[ = \langle p|\left(\int dx\ |x\rangle\langle x|\right)|\Psi(t)\rangle \qquad (\to \text{unit operator}) \] \[ \tilde\Psi(p,t) = \int dx\ \langle p|x\rangle\,\langle x|\Psi(t)\rangle \] \[ \tilde\Psi(p,t) = \int dx\ \langle p|x\rangle\,\Psi(x,t) \]

where \(\langle p|x\rangle\) is the probability amplitude that when the position of particle is '\(x\)' its momentum is '\(p\)'.

where, \(\langle p|x\rangle\) will turn out to be \(e^{\frac{ipx}{\hbar}}\)

\[ \psi(x,t) = \int dp\ \langle x|p\rangle\,\tilde\Psi(p,t) \]

Q) which Basis is more convenient out of \(|x\rangle\) & \(|p\rangle\)?

Ans It is easier to work in position basis. because of following reason.

we know that,

\[ \hat H = \frac{\hat p^2}{2m} + V(\hat x) \]

if we use position basis

\[ \hat H|x_0\rangle = \frac{\hat p^2|x_0\rangle}{2m} + V(\hat x)|x_0\rangle \] \[ = \frac{\hat p^2|x_0\rangle}{2m} + V(x_0)|x_0\rangle \]

(\(V(\hat x) \to\) operator ; \(V(x_0) \to\) eigen value)

once we know how \(\hat p\) acts on \(\{|x_0\rangle\}\) we can easily compute \(\hat H|x_0\rangle\), but in case of momentum basis \(\{|p\rangle\}\).

\[ \hat H|p_0\rangle = \frac{\hat p^2}{2m}|p_0\rangle + V(\hat x)|p_0\rangle \] \[ = \frac{p_0^2}{2m} + V(\hat x)|p_0\rangle \]

\(V(\hat x)\) can be complicated function of \(x\), thus will complicate the solution.

for \(\hat H = \frac{p^2}{2m} + \frac{1}{2}k\hat x^2\) simple - Harmonic oscillator.

we can use \(\{|x\rangle\}\) and \(\{|p\rangle\}\) with same priority.

Q) what is \(\hat p|x_0\rangle\)?

Suppose we have two operators \(A, B\) such that

\[ \begin{aligned} A^{\dagger} &= A\\ B^{\dagger} &= B \end{aligned} \qquad \text{i.e. Hermitian operators.} \]

Q.) can I find simultaneous eigen states of \(A\) and \(B\). can I find eigen states / eigen functions which are common to both of these operators.

Ans) In general there is no reason we can find such eigen states. But we can expect some eigen states can be common.

If \([A, A^{\dagger}] = 0\) i.e. if \(A\) commutes with its hermitian

such a matrix can always be diagonalised by similarity transformation.

for \(A^{\dagger} = A\) (In all physical operator \(A^{\dagger} = A\))

\[ \Rightarrow \quad [A, A^{\dagger}] = 0 \]

we can simultaneously diagonalise \(A, B\) if

\[ [A, B] = 0 \ ; \quad AB - BA = 0 \] \[ \left\{ \begin{aligned} \text{Such that } SAS^{\dagger} &= \text{diagonal } M\\ SBS^{\dagger} &= \text{Diagonal } N \end{aligned} \right. \]

Here "simultaneously" means we can diagonalise with the same similarity transformation \((S)\).

\(\rightarrow\) If \([A, B] \neq 0\)

we can not find simultaneous eigen states.

i.e. we can't find complete common set of eigen states. But they may share one or more common eigen-states.

Note :-- This does not mean that every eigen-state of \(A\) is also the eigen-state of \(B\). if \([A,B] = 0\).

ex. take fun. on a line.

\[ \mathbb{1}\,f(x) = f(x) \]

Parity operator

\[ \mathbb{P}\,f(x) = f(-x) \] \[ \mathbb{P}^2 f(x) = P\big(P(f(x))\big) = P\big(f(-x)\big) = f(x) \] \[ P^2 = \mathbb{1} \]

we can see that

\[ [P, \mathbb{1}] = P\cdot 1 - 1\cdot P = 0 \]

also

\[ [P^2, P] = [\mathbb{1}, P] = 0 \]

we know that every function is an eigen-state of unit operator '\(\mathbb{1}\)', but every fun. is not eigen-state for parity operator.

\[ \mathbb{P}\,\phi(x) = \left\{ \begin{aligned} &\phi(-x)\\ &= \alpha\,\phi(x) \end{aligned} \right. \qquad P^2\phi(x) = \alpha^2\phi(x) = \phi(x) \] \[ \alpha = \pm 1 \]

\(\alpha = 1\) even function

odd fun, even functions are eigen states of parity operator

\[ P f_e(x) = f(-x) = f(x) \qquad \alpha = 1 \] \[ P f_o(x) = f(-x) = -f(x) \qquad \alpha = -1 \ \text{ odd function.} \]

So set of eigen states of \(P^2\) is much bigger than set of eigen functions of parity operator. / eigen state set of \(\mathbb{P}\) is subset of spectrum of \(P^2\) or \(\mathbb{1}\).

since \([x, p] \neq 0\) they do not share common complete eigen value set. (But they can share few eigen values. even if \([A,B] \neq 0\)).

we know that

\[ \langle \hat A\rangle = \langle\Psi|\hat A|\Psi\rangle \qquad \text{assuming } \langle\Psi|\Psi\rangle = 1 \] \[ \langle \hat A^2\rangle = \langle\Psi|A^2|\Psi\rangle \] \[ (\Delta\hat A)^2 = \left\langle\left(\hat A - \langle A\rangle\right)^2\right\rangle \] \[ (\Delta\hat A)^2 = \langle\hat A^2\rangle - \langle\hat A\rangle^2 \]

for position operator \(\hat x\)

\[ \langle\hat x\rangle = \langle\Psi|\hat x|\Psi\rangle \] \[ |\Psi\rangle = \int dx\ \psi(x)|x\rangle \] \[ \langle\Psi| = \int dx'\ \langle x'|\,\psi^{*}(x') \] \[ \langle\hat x\rangle = \int dx \int dx'\ \psi^{*}(x')\,\psi(x)\ \langle x'|\hat A|x\rangle \]

(\(\to\) element of \(A\) in position basis.)

\[ = \int dx \int dx'\ \psi^{*}(x')\,\psi(x)\ A(x', x) \]

Note :- we know that

\[ \psi(x,t) = \int dp\ \langle x|p\rangle\,\tilde\Psi(p,t) \]

find \(\langle x|p\rangle\)?

we know that,

\[ \hat x\hat p - \hat p\hat x = i\hbar\,\mathbb{1} \]

let us find matrix element of \([\hat x, \hat p]\) in position basis.

\[ \langle x|\hat x\hat p - \hat p\hat x|x'\rangle = i\hbar\,\langle x|x'\rangle = i\hbar\,\delta(x-x') \] \[ x^{*}\langle x|\hat p|x'\rangle - x'\langle x|\hat p|x'\rangle = i\hbar\,\delta(x-x') \]

(\(x\) is real, so \(x^{*} = x\) ; \(\downarrow\) Position has to be real)

\[ \langle x|\hat p|x'\rangle\,(x - x') = i\hbar\,\delta(x-x') \] \[ \Rightarrow \quad \langle x|\hat p|x'\rangle = i\hbar\,\frac{\delta(x-x')}{(x-x')} \]

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defined :-

\[ \left(\delta'(x) = \frac{-\delta(x)}{x}\right) \qquad \text{(How?)} \]

so

\[ \delta'(x-x') = \frac{-\delta(x-x')}{(x-x')} \] \[ \Rightarrow \quad \langle x|\hat p|x'\rangle = -i\hbar\,\frac{\partial}{\partial x}\big(\delta(x-x')\big) = -i\hbar\,\frac{\partial}{\partial x}\big(\langle x|x'\rangle\big) \]

for any arbitrary state \(\Psi(t)\)

\[ \langle x|\hat p|\Psi(t)\rangle = -i\hbar\,\frac{\partial}{\partial x}\langle x|\Psi(t)\rangle \]

So it \(\Rightarrow\)

\[ \left\{\left(\hat p \longrightarrow -i\hbar\frac{\partial}{\partial x}\right) \ \text{IN THE POSITION BASIS}\right. \] \[ \hat x|x\rangle = x|x\rangle \]

but \(\hat p|x\rangle\) is totally different \(\longrightarrow\) \(= -i\hbar\frac{\partial}{\partial x}|x\rangle\)

So,

\[ \langle x|p|x'\rangle = -i\hbar\,\frac{\partial}{\partial x}\langle x|x'\rangle \]

likewise,

\[ \langle x|\hat x|x'\rangle = x\,\langle x|x'\rangle = x'\,\langle x|x'\rangle \] \[ \left\{ \begin{aligned} \langle x'|x|x\rangle &= 1 \quad \text{if } x = x'\\ &= 0 \quad \text{if } x \neq x' \end{aligned} \right. \qquad \langle x|x'\rangle = \delta(x-x') \] \[ \hat p|p_0\rangle = p_0|p_0\rangle \]

what is \(\hat x|p_0\rangle = ?\)

\[ \left\{\hat x = +i\hbar\frac{\partial}{\partial p} \quad \text{IN momentum Basis}\right. \]

In contineous basis rather than kronical delta we use \(\delta(x - x')\).

\[ \int f(x')\,\delta(x-x')\,dx' = f(x) \] \[ I = \int \delta(x-x')\,dx' \qquad ? \]

To Find the inner product \(\langle x|p\rangle\)?

(\(\langle x| \to\) position eigen state ; \(|p\rangle \to\) momentum eigen state)

given \([\hat x, \hat p] = i\hbar\,\mathbb{1}\)

what is overlap b/w position eigen states & momentum eigen states. since \([x, p] \neq 0\) we can't find complete common eigen -- states. values.

we know that

\[ \langle \hat x|\hat p|\Psi(t)\rangle = -i\hbar\,\frac{\partial}{\partial x}\langle x|\Psi(t)\rangle \]

(\(\hat x \to\) not a operator)

instead of \(\Psi(t)\), we replace it with any momentum eigen -state.

\[ \langle \hat x|\hat p|p\rangle = p\,\langle \hat x|p\rangle \]

(\(\to\) eigen value) as \(\hat p|p\rangle = \hat p|p\rangle\) \(\hookrightarrow\) abstract eigen state (regardless of any basis)

also

\[ \langle x|\hat p|p\rangle = -i\hbar\,\frac{\partial}{\partial x}\langle x|p\rangle = p\,\langle x|p\rangle \] \[ \langle x|p\rangle = \text{some fun. of } x \text{ labelled by } 'p' = f_p(x) \] \[ -i\hbar\,\frac{\partial}{\partial x} f_p(x) = p\,f_p(x) \] \[ \int \frac{d f(x)}{f(x)} = \int \frac{ip}{\hbar}\,dx \] \[ f_p(x) \propto e^{\frac{ipx}{\hbar}} \]

(\(p \to\) eigen value corresponding to \(\langle x|\))

\[ \boxed{\langle x|p\rangle \propto e^{\frac{ipx}{\hbar}}} \]

(\(\to\) eigen value corresponding to \(|p\rangle\))

\(x, p\) are both no. (eigen values).

[\(\langle x|p\rangle \propto e^{\frac{ip\cdot x}{\hbar}}\) , \(\langle p|x\rangle \propto e^{-\frac{ip}{\hbar}x}\)

here \(p\) & \(x\) are eigen values of \(|p\rangle\) & \(|x\rangle\) respectively.]

\[ \Rightarrow \quad \langle p|x\rangle \sim e^{-\frac{ipx}{\hbar}} \qquad \text{(complex conjugate of } \langle x|p\rangle) \]

we know that

\[ \langle x|\Psi(t)\rangle = \Psi(x,t) = \int dp\ \langle x|p\rangle\langle p|\Psi(t)\rangle = \int dp\ e^{\frac{ipx}{\hbar}}\,\tilde\Psi(p,t) \] \[ \langle p|\Psi(t)\rangle = \tilde\Psi(p,t) = \int dx\ \langle p|x\rangle\langle x|\Psi(t)\rangle = \int dx\ e^{-\frac{ipx}{\hbar}}\,\Psi(x,t) \]

"So we can see that position space wave fun. & momentum space wave functions are fourier transforms of each other."

In 3-D,

\[ \langle \vec r|\vec p\rangle \propto e^{\frac{i\vec p\cdot\vec r}{\hbar}} \] \[ \langle \vec p|\vec r\rangle \propto e^{-\frac{i\vec p\cdot\vec r}{\hbar}} \]

also

\[ \langle \vec r|\hat{\vec p}|\Psi(t)\rangle = -i\hbar\,\nabla_{\vec r}\,\langle\vec r|\Psi(t)\rangle \]

(\(\vec r, \hat{\vec p}\) means 3-D)

\[ \hat H(\hat r, \hat p) = \hat H = \frac{\hat p^2}{2m} + \hat V(\hat r) \] \[ i\hbar\,\frac{d}{dt}|\Psi(t)\rangle = \hat H|\Psi(t)\rangle \]

so, we want to know how above eq\(^{\text{n}}\) looks like in position basis.

taking inner product with \(\langle\vec r|\cdots\rangle\) both sides.

\[ \left\langle \vec r\left| i\hbar\frac{d}{dt}\right|\Psi(t)\right\rangle = \langle\vec r|\hat H|\Psi(t)\rangle \]

\(\left\{\right.\) Here \(|\vec r\rangle\) or \(\langle\vec r|\) is any particular eigen state of position basis.

above eq\(^{\text{n}}\) can be written as,

\[ i\hbar\,\frac{d}{dt}\big(\langle\vec r|\Psi(t)\rangle\big) = \langle\vec r|\hat H|\Psi(t)\rangle \] \[ i\hbar\,\frac{\partial}{\partial t}\big(\Psi(r,t)\big) = \left\langle\vec r\left|\frac{\hat p^2}{2m}\right|\Psi(t)\right\rangle + \left\langle\vec r\left|V(\hat r)\right|\Psi(t)\right\rangle \]

\(\left\{\frac{d}{dt}\right.\) has to be replaced with \(\frac{\partial}{\partial t}\) as derivative is only w.r.t '\(t\)' variable.

\(\langle\vec r|\Psi(t)\rangle = \) position space wave function.

\[ \Rightarrow \quad i\hbar\,\frac{\partial}{\partial t}\Psi(r,t) = \frac{(-i\hbar)^2\nabla^2}{2m}\Psi(r,t) + V(\vec r)\,\psi(r,t) \] \[ \left\{ \begin{aligned} \Psi(r,t) &= \langle\vec r|\Psi(t)\rangle\\ V(r) &= \text{eigen value of } V(\hat r) \end{aligned} \right. \]

since we deal with hermitian operators in Q.M. (to get real eigen value)

\(V(\hat r)\) should be hermitian too.

So, \(V^{*}(\vec r)\) is simply \(V(\vec r)\)

[complex \(V(r)\) is used in case where there is absorption where probabilities are not conserved.]

Find eq\(^{\text{n}}\) becomes,

\[ \left\{ i\hbar\,\frac{\partial}{\partial t}\Psi(\vec r,t) = \frac{-\hbar^2}{2m}\nabla^2\Psi(\vec r,t) + V(\vec r)\,\Psi(\vec r,t) \right\} \]

Position space shrödinger equation

Boundary cond\(^{\text{n}}\)

1. Normalized \(\Psi(r,t)\) \[ \langle\Psi(t)|\Psi(t)\rangle = 1 = \int d^3r\ |\Psi(\vec r,t)|^2 \] (\(\Psi(r,t)\) is member of \(l_2\) space)

stationary state of the Q.Mechanical system is by def\(^{\text{n}}\) is an eigen state of total Hamiltonian of the system.

"Any stationary state is eigen state of \(\hat H\).

\(\hat H\) operator has in general infinite no of eigen states".

\[ i\hbar\,\frac{d}{dt}|\Psi(t)\rangle = H|\Psi(t)\rangle \]

lets use \(\Phi(t)\) for eigen state (of Hamiltonian) instead of \(\Psi(t)\)

\[ i\hbar\,\frac{d}{dt}|\Phi(t)\rangle = H|\Phi(t)\rangle = H_0|\Phi(t)\rangle \]

(\(H_0\) is any eigen value of \(H\).)

we know that Hamiltonian tells total energy so lets use \(E\) as eigen value of Hamiltonian

\(E\) is real eigen value. (as \(H\) is made up of physical measurables)

\[ i\hbar\,\frac{d}{dt}|\Phi(t)\rangle = E|\Phi(t)\rangle \]

\(E = \) eigen value ; \(\Phi(t) = \) eigen state. (any) of hamiltonian

\(\left\{\right.\) there can be more eigen state for one eigen value. for that eigen value should be repeated.

solution to above eq\(^{\text{n}}\) is

\[ \boxed{|\Phi(t)\rangle = e^{-\frac{iEt}{\hbar}}\,|\Phi(0)\rangle} \]

For a general state \(|\Psi(t)\rangle\)

we can write \(|\Psi(t)\rangle\) as linear combination of basis set formed from eigen state of Hamiltonian operator.

\[ |\Psi(t)\rangle = \sum_n c_n |\Phi_n(t)\rangle \]

or we can write

\[ |\Psi(t)\rangle = \sum_n c_n\,e^{-\frac{iE_n t}{\hbar}}\,|\phi_n(0)\rangle \]

or

\[ |\Psi(t)\rangle = \sum_n c_n(t)\,|\phi_n(0)\rangle \] \[ c_n(t) = c_n\,e^{-\frac{iE_n t}{\hbar}} \]

\(\left\{\right.\) Generally we use,

\[ |\Psi(t)\rangle = \sum_n c_n\,e^{-\frac{iE_n t}{\hbar}}\,|\phi_n(0)\rangle \]

\(\left.\right\}\) The abstract state of any system.

where, \(\big(|\phi_n(0)\rangle\) is a choosen time independent basis set\(\big)\)

\[ c_n = \langle\phi_n|\Psi(t)\rangle \]

where, \(\{n = \) quantum number\(\}\)

In General there may be as many quantum no. as there are d.o.f of the system.

for any particle confined in 3-D potential there will be at least 3 quantum no. There may be more if there are internal quantum no. like spin, hypercharge, strangeness for elementry particles.

\[ |\Psi(t)\rangle = \sum_n c_n\,e^{-\frac{iE_n t}{\hbar}}|\Phi_n(0)\rangle \]

is also written as

\[ \int_n c_n\,e^{-\frac{iE_n t}{\hbar}}|\phi_n(0)\rangle \]

which changes to integral sign when \(n\) is contineous.

All quantum interference phenomena happen because of the superposition of different energy states. (Beats, Interference groups, mforms. due to different frequency superposition).

\[ |\Psi(t)\rangle = \sum_n c_n e^{-\frac{iE_n t}{\hbar}}|\Phi_n(0)\rangle \]

\(\hookrightarrow\) not a fourier series. \(\{c_n|\phi_n(0)\rangle\) is aperiodic.

Here we have assumed that all eigen states of Hamiltonian are made orthogonal by Gramh-Schmidt orthonormalization.

lets look for stationary states in position basis \(\{|\vec r\rangle\}\).

\[ i\hbar\,\frac{d}{dt}|\Psi(t)\rangle = \hat H|\Psi(t)\rangle \]

taking inner product.

\[ \left\langle\vec r\left|i\hbar\frac{d}{dt}\right|\Psi(t)\right\rangle = \langle\vec r|\hat H|\Psi(t)\rangle \] \[ = \langle\vec r|E\,\Psi(t)\rangle \]

\(\{|\vec r\rangle\) is any position eigen state\(\}\).

\[ i\hbar\,\frac{\partial}{\partial t}\Psi(\vec r,t) = E\,\Psi(\vec r,t) \] \[ \Rightarrow \quad \boxed{\Psi(r,t) = e^{-\frac{iEt}{\hbar}}\,|\psi(r,0)\rangle} \]

For a stationary state \(\underline{|\psi(r,t)\rangle}\).

For any general state

\[ \Psi(r,t) = \sum_n e^{-\frac{iE_n t}{\hbar}}\,|\psi_n(r,0)\rangle \]

lets use our old notation \(|\phi_n\rangle\) for stationary states.

\[ |\phi_n(\vec r,t)\rangle = e^{-\frac{iE_n t}{\hbar}}\,|\phi_n(\vec r, 0)\rangle \]

So if work in position basis. The \(|\Psi(t)\rangle\) can be written as.

\[ \langle\vec r|\Psi(t)\rangle = \left\langle\vec r\left|\sum_n c_n e^{-\frac{iE_n t}{\hbar}}\right|\phi_n(\vec r,0)\right\rangle \] \[ = \sum_n c_n\,e^{-iE_n t/\hbar}\,\langle\vec r|\phi_n(\vec r,0)\rangle \]

where \(\langle\vec r|\Psi(t)\rangle\) is general state at time '\(t\)' in position basis.

So The shrödinger equation for stationary states \(|\phi_n(\vec r,t)\rangle\) looks like

\[ \left\{ \begin{aligned} i\hbar\,\frac{\partial}{\partial t}|\phi_n(\vec r,t)\rangle &= \frac{(-i\hbar)^2}{2m}\nabla^2\phi_n(\vec r,t) + V(\vec r)\,\phi_n(r,t)\\ &= E_n\,\phi_n(\vec r,t) \end{aligned} \right. \]

In fact we can remove '\(t\)' dependence by only considering

\[ \frac{-\hbar^2}{2m}\nabla^2\phi_n(\vec r,t) + V(\vec r)\,\phi_n(\vec r,t) = E_n\,\phi_n(\vec r,t) \]

where

\[ \phi_n(\vec r,t) = \phi_n(\vec r,0)\,e^{-\frac{iE_n t}{\hbar}} \]

using above we get,

\[ \boxed{\frac{-\hbar^2}{2m}\nabla^2\phi_n(\vec r,0) + V(\vec r)\,\phi_n(\vec r,0) = E_n\,\phi_n(\vec r,0)} \]

\(\left.\right\}\) Time Independent Shrödinger equation for stationary state.

the same equation can be written as :--

\[ \left\{\left(\frac{-\hbar^2}{2m}\nabla^2 + V(\vec r)\right)|\phi_n(\vec r)\rangle = E_n|\phi_n(\vec r)\rangle\right\} \]

eigen value equation. & Homogeneous too.

Any confinement on matter "quantizes the energy levels".

A free particle can have any energy but a confined particle in potential / box ensures that only certain packets of energy are allowed.

Q) Where is confinement in H-atom, where \(e^-\) is placed in coulomb potential?

Ans.) Hilbert noted that cond\(^{\text{n}}\) that \(\int|\psi|^2 d^3r = 1\) is enough to produce confinement / quantization

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classically :--

\[ H(x,p) = \frac{p^2}{2m} + V(x) \] \[ V(x) = \left\{ \begin{aligned} &0 && 0 < x < L\\ &\infty && \text{outside} \end{aligned} \right. \]

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old rule of Bohr's quantization (for periodic orbits) (postulate)

Area under phase space \(= \oint p\,dx = nh\)

\[ \Rightarrow \quad 2\sqrt{2mE_n}\cdot L = nh \] \[ E_n = \frac{n^2h^2}{8mL^2} \qquad \text{(semi -- classical result)} \quad \left(\oint P\,dx = nh\right) \] \[ E_n = \frac{n^2\pi^2\hbar^2}{2mL^2} \]

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using time independent shrödinger eq\(^{\text{n}}\) :--

\[ \left[\frac{-\hbar^2}{2m}\frac{d^2}{dx^2} + V(x)\right]|\phi_n(x)\rangle = E_n|\phi_n(x)\rangle \]

for \((0 < x < L)\) inside the box. \(\big(V(x) = 0\big)\)

\[ \frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\,|\phi_n(x)\rangle = E_n|\phi_n(x)\rangle \] \[ \left(\frac{d^2}{dx^2} + k_n^2\right)|\phi_n(x)\rangle = 0 \qquad \left\{k_n^2 = \frac{2mE_n}{\hbar^2}\right\} \]

Solutions is trivial

\[ |\phi_n(x)\rangle = A\cos k_n x + B\sin k_n x \]

Imposing boundary cond\(^{\text{n}}\) :--

\[ \phi_n(0) = 0 = \phi_n(L) \] \[ \Rightarrow \quad A = 0 \ , \qquad k_n L = n\pi \]

as \(B \neq 0\)

Because of

\[ \langle\phi_n(x)|\phi_n(x)\rangle = 1 \quad \text{for all space} \] \[ \int_0^{L} B^2\sin^2 k_n x\ dx = 1 \] \[ \Rightarrow \quad B^2\int_0^{L}\left(\frac{1-\cos 2k_n x}{2}\right)dx = 1 \] \[ \Rightarrow \quad B^2\left(\frac{L}{2} - \left(\frac{\sin 2k_n x}{4k_n}\right)_0^{L}\right) = 1 \] \[ B^2\left(\frac{L}{2} - 0\right) = 1 \quad \Rightarrow \quad \boxed{B = \sqrt{\tfrac{2}{L}}} \] \[ |\phi_n(x)\rangle = \sqrt{\frac{2}{L}}\,\sin\left(\frac{n\pi}{L}x\right) \]

and

\[ E_n = \frac{\hbar^2 n^2\pi^2}{2mL^2} \qquad (n = 1,2,3\cdots\infty) \]

where Bohr rule also predicts same as exact eigen values.

note :-- \(\left(n \neq 0\right.\) as \(|\phi_n(x)\rangle\) vanishes & normalization does not work\(\left.\right)\)

what is

\[ |\phi_n(x,t)\rangle = e^{-\frac{iE_n t}{\hbar}}|\phi_n(x)\rangle \qquad \& \quad e^{-\frac{iE_n t}{\hbar}}\underline{|\phi_n(0)\rangle} \]

Because of lack of symmetry, we can also see that for any Energy eigen value we have a unique eigen state, so there is no degeneracy in this problem.

\[ \phi_n(x) = \left\{ \begin{aligned} &\sqrt{\tfrac{2}{L}}\,\sin\left(\tfrac{n\pi}{L}\right)x && 0 \leq x \leq L\\ &0 && x < 0,\ x > L \end{aligned} \right. \]

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[\(\psi'(x)\) is disc has finite discontinuity at \(x = 0, L\)

So, \(\psi''(x)\) should have infinite discontinuity at \(x = 0, L\), and that discontinuity must be compensated by \(V(x)\)

\[ \left(\frac{-\hbar^2}{2m}\right)\frac{d^2\phi_n}{dx^2} + V(x)\,\phi(x) = E_n\phi(x) \]

]

\[ \int_{-\infty}^{\infty} dx\ |\phi_n(x)|^2 = 1 \] \[ [\phi_n(x)] = \frac{1}{\sqrt{\text{length}}} \qquad \text{diamensions of } |\phi_n(x)\rangle \]

we do not have following const. in \(\phi_n(x)\) due to some reasons.

const.	Reason for absence
\(c\)	no relativity
\(G\)	no gravity
\(k_B\)	assuming \(T = 0\,\)K (absolute)
\(\hbar\)

[\(\hbar\) does appear in wavefunctions of H-atom \(R(r) \propto e^{-(r/a_0)}\), \(\underline{a_0 = f(\hbar)}\)]

\(\sin(\ )\) \(\longrightarrow\) must depend on \(x\). & must be dimensionless.

\(\left(\sqrt{\tfrac{2}{L}}\right)\sin(\ )\) \(\longrightarrow\) To normalize.

So '\(\hbar\)' don't appear in this problem because of dimensional reasons.

lets check if \(|\phi_n(x)\rangle\) is momentum eigen state or not.

\[ \hat p|p\rangle = p|p\rangle \qquad (\to \text{eigen value}) \] \[ -i\hbar\,\frac{\partial}{\partial x}\phi_n(x) \neq (\ )\,\phi_n(x) \] \[ \Rightarrow \quad |\phi_n(x)\rangle \ \text{ is not a momentum eigen state.} \]

But \([H, \hat P] = 0\) as \(H = \frac{\hat p^2}{2m}\)

So we must have some common eigen states.

But we just found out that momentum eigen states are not momentum eigen states.

[\(\to\) Energy / Hamiltonian]

\(\left\{\right.\) So Only solution is "Particle is not free".

\[ \hat H = \frac{\hat p^2}{2m} + V(\hat x) \] \[ \text{so} \quad \left\{ \begin{aligned} [H, p^2] &\neq 0\\ [H, P] &\neq 0 \end{aligned} \right. \]

what would \(|p\rangle\) (momentum eigen states) looks like?

\[ \hat p\,\chi(x) = p\,\chi(x) \] \[ \Rightarrow \quad -i\hbar\,\frac{d\chi}{dx} = p\chi \] \[ \Rightarrow \quad \chi(x) \propto e^{\frac{ipx}{\hbar}} \]

so in position basis, momentum eigen states must look like this.

\[ \phi_n(x) = \sqrt{\frac{2}{L}}\left(\frac{1}{2i}\right)\left[e^{ik_n x} + e^{-ik_n x}\right] \] \[ \left(k_n = \frac{n\pi}{L}\right) \] \[ \frac{ipx}{\hbar} = \frac{i(n\pi/L)x}{1} \] \[ \boxed{p = \frac{n\pi\hbar}{L}} \qquad \& \quad p = -\frac{n\pi\hbar}{L} \]

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[Q) we know that if \(\phi(x) \in l_2\), \(\tilde\phi(p)\) must also be a member of \(l_2\). But here dirac functions are not square integrable so whats the solution?

\[ \left(\delta(k - k_0) = \int_{-\infty}^{\infty} e^{i(k-k_0)x}\,dx\right) \]

iff \((-\infty < x < \infty)\) there will be \(\tilde\phi(p) = \) delta funct at \(\pm\hbar k_n\)

But \(0 < x < L\) so \(\tilde\phi(p)\) is not delta fun of \(p\)]

lets find \(\tilde\phi(p)\)

\(\tilde\phi(p)\) is fourier transform of \(\phi(x)\).

\[ \tilde\phi(p) = \int_{-\infty}^{\infty} dx\ \phi_n(x)\,e^{-\frac{ipx}{\hbar}} \] \[ = \sqrt{\frac{2}{L}}\int dx\ \frac{e^{-\frac{ipx}{\hbar}}}{2i}\left(e^{ik_n x} - e^{-ik_n x}\right) \] \[ \tilde\phi(p) = \sqrt{\frac{2}{L}}\int_0^{L} \frac{e^{-\frac{ipx}{\hbar}}}{2i}\left(e^{ik_n x} - e^{-ik_n x}\right)dx \]

So,

\[ \underline{\tilde\phi(p) = \text{smooth function of } p,\ \in l_2} \]

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"So particle in box can have any momentum value"

\[ \langle A\rangle = \frac{\langle\Psi|A|\Psi\rangle}{\langle\Psi|\Psi\rangle} \]

since \(\Psi\) is normalized \(\Rightarrow\) \(\langle\Psi|\Psi\rangle = 1\)

\[ \langle A\rangle = \langle\Psi|A|\Psi\rangle \]

\(\langle A\rangle\) is time independent. Thats why we call it a stationary state.

\[ \langle A\rangle = \langle\Psi|A|\Psi\rangle \] \[ = \int dx' \int dx\ \langle\Psi|x\rangle\langle x|A|x'\rangle\langle x'|\Psi\rangle \] \[ \langle \hat x\rangle_n = \int dx'\int dx\ \phi_n^{*}(x)\,\langle x|\hat x|x'\rangle\,\phi_n(x) \]

(\(\to\) acts upon)

\[ \left\{ \begin{aligned} \hat x|x'\rangle &= x'|x'\rangle\\ \hat x|x\rangle &= x|x\rangle \end{aligned} \right. \] \[ = \int dx'\int dx\ \phi_n^{*}(x)\ x\ \langle x|x'\rangle\,\phi_n(x) \] \[ = \int dx'\int dx\ \phi_n^{*}(x)\ x\ \delta(x-x')\,\phi_n(x) \]

doing \(x'\) integral ; \(\big(\delta(x-x') = 1\) for \(x = x'\), \(\phi_n(x') = \phi_n(x)\) at \(x = x'\big)\)

\[ \boxed{\langle x\rangle_n = \int dx\ \phi_n^{*}(x)\ x\ \phi_n(x)} \]

\(\hookrightarrow\) calculated in eigen state \(\phi_n(x)\).

(where; \(x = \) eigen value of \(\hat x\) ; \(\langle x|\hat x = x\langle x|\) ; \(n \to n^{\text{th}}\) eigen state of Hamiltonian operator)

so,

\[ \langle x\rangle_n = \int_0^{L} dx\ \frac{2}{L}\sin^2\left(\frac{n\pi}{L}x\right)\cdot x \] \[ \langle x^2\rangle_n = \int_0^{L} dx\ \frac{2}{L}\sin^2\left(\frac{n\pi}{L}x\right)\cdot x^2 \]

compute \((\Delta x)_n\) uncertainty / std. deviation in position.

\[ (\delta x)_n = \sqrt{\langle x^2\rangle_n - \langle x\rangle_n^2} \]

similarly,

\[ \langle p\rangle_n = \int_{-\infty}^{\infty} dx\ \phi_n^{*}(x)\,(-i\hbar)\frac{d}{dx}\phi_n(x) \] \[ \langle p\rangle_n = (-i\hbar)\int_0^{L} dx\ \frac{2}{L}\sin\left(\frac{n\pi}{L}x\right)\cos\left(\frac{n\pi}{L}x\right)\left(\frac{n\pi}{L}\right) = 0 \] \[ \langle p^2\rangle_n = \int_0^{L} dx\ \frac{2}{L}\sin\left(\frac{n\pi}{L}x\right)\left(\frac{n\pi}{L}\right)^2(-i\hbar)^2\sin\left(\frac{n\pi}{L}x\right)(-1) \] \[ \langle p^2\rangle_n = \frac{2}{L}\cdot\frac{n^2\pi^2\hbar^2}{L^2}\int_0^{L}\sin^2\left(\frac{n\pi}{L}x\right)dx \] \[ = \frac{n^2\pi^2\hbar^2}{L^2} \]

also

\[ E_n = \frac{n^2\pi^2\hbar^2}{2mL^2} = \boxed{\frac{\langle p^2\rangle_n}{2m}} \]

(only iff energy is only K.E.)

\[ (\Delta p)_n = \sqrt{\langle p^2\rangle_n - \langle p\rangle^2} = \frac{n\pi\hbar}{L} \] \[ (\Delta x)_n = \sqrt{\langle x^2\rangle_n - \langle x\rangle_n^2} \] \[ \boxed{(\Delta x)_n\,(\Delta p)_n \geq \hbar} \]

Energy levels are not degenerate in this case.

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\[ \hat H = \frac{\hat p^2}{2m} \]

in this problem \(V(x) = 0\) everywhere, so

\[ [H, P] = 0 \quad \text{as } [\hat p, p] = 0 \]

[1-D simply connected space has no degeneracy, but this space is not simply connected so it may have degeneracy in 1-D. & In fact It has degeneracy of 2. (Its doubly degenerate).]

So every eigen state of \(H\) will be a momentum eigen state.

since, \(\hat p\) and \(\frac{\hat p^2}{2m}\) has \(\infty\) no of eigen states so They share complete set as, common eigen state.

Periodic Boundary condition :--

\[ \phi_n(0) = \phi_n(L) \] \[ \phi_n(x+L) = \phi_n(x) \]

(\(\phi_n(x)\) is any eigen state of \(\hat H\).)

If function value are same at \(x\), \(x+L\) so their derivative should also be equal.

\[ \phi_n'(x+L) = \phi_n'(x) \] \[ \phi_n''(x+L) = \phi_n''(x) \qquad \cdots \text{all derivative must match.} \]

"Are all \(|p\rangle\) also eigenstates of \(\hat H\)? (yes) if so, then \(|\Psi\rangle = \sum_n c_n|p\rangle\)

\[ \langle x|\hat p|p\rangle = p\,\langle x|p\rangle \] \[ -i\hbar\,\frac{\partial}{\partial x}\langle x|p\rangle = p\,\langle x|p\rangle \quad \Rightarrow \quad \langle x|p\rangle \propto e^{\frac{ipx}{\hbar}} \] \[ \frac{-\hbar^2}{2m}\frac{\partial^2\phi_n(x)}{\partial x^2} + 0 = E_n\phi_n(x) \] \[ \frac{d^2\phi_n(x)}{dx^2} + \frac{2mE_n}{\hbar^2}\phi_n(x) = 0 \] \[ \left(D^2 + k^2\right)\phi_n(x) = 0 \] \[ D = \pm ik \] \[ \phi_n(x) = A e^{ikx} + B e^{-ikx} \]

1)

\[ \phi_n(0) = \phi_n(0+L) \] \[ A + B = A e^{ikL} + B e^{-ikL} \]

2)

\[ \phi_n(x) = \phi_n(x+L) \] \[ A e^{ikx} + B e^{-ikx} = A e^{ik(x+L)} + B e^{-ik(x+L)} \] \[ A e^{ikx}\left(1 - e^{ikL}\right) + B e^{-ikx}\left(1 - e^{-ikL}\right) = 0 \] \[ \frac{A e^{ikx}}{B e^{-ikx}} = \frac{e^{-ikL} - 1}{1 - e^{ikL}} \qquad \cdots \textcircled{1} \]

3)

\[ \phi_n'(x) = \phi_n'(x+L) \]

[\(\phi_n(x) = A e^{ikx} + B e^{-ikx}\)

\(\phi_n(x) = a\cos k_n x + b\sin k_n x\)

\[ \phi_n(0) = \phi_n(L) \] \[ a = a\cos k_n L + b\sin k_n L \] \[ k_n L = 2n\pi \] \[ k_n = \frac{2n\pi}{L} = \sqrt{\frac{2mE}{\hbar^2}} \] \[ \frac{4n^2\pi^2}{L^2} = \frac{2mE_n}{\hbar^2} \] \[ E_n = \frac{2n^2\pi^2\hbar^2}{mL^2} \]

]

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\[ \phi_{n_1 n_2}(x,y) = \frac{2}{L}\sin\left(\frac{n_1\pi}{L}\cdot x\right)\sin\left(\frac{n_2\pi}{L}\cdot y\right) \] \[ E(n_1, n_2) = \frac{\hbar^2\pi^2}{2mL^2}\left(n_1^2 + n_2^2\right) \]

These energy levels are degenerate except ground state.

\(\{\)where, \(n_1, n_2 = 1, 2, 3\cdots\}\)

1. Ground state \(\Rightarrow\) \(n_1 = 1\), \(n_2 = 1\) (non-degenerate). \[ E(1,1) = \frac{\hbar^2\pi^2}{2mL^2}(2) \]
2. First excited state \((1,2)\) or \((2,1)\) degenerate (doubly degenerate) \[ E(1,2) = E(2,1) = \frac{\hbar^2\pi^2}{2mL^2}(5) \]
3. 2\(^{\text{nd}}\) excited state \((2,2)\) (non-degenerate) \[ E(2,2) = \frac{\hbar^2\pi^2}{2mL^2}(8) \]
4. 3\(^{\text{rd}}\) excited state \((1,3)\) or \((3,1)\). (doubly degenerate)

if potential were like a ractangle

\[ \phi_{n_1 n_2}(x,y) = \frac{2}{\sqrt{L_1 L_2}}\sin\left(\frac{n_1\pi}{L_1}x\right)\sin\left(\frac{n_2\pi}{L_2}x\right) \] \[ E(n_1,n_2) = \frac{\hbar^2\pi^2}{2m}\left(\frac{n_1^2}{L_1^2} + \frac{n_2^2}{L_2^2}\right) \]

Now degeneracy depends on \(L_1\) & \(L_2\) in energy levels.

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