• #Lecture 17
  • #L in Q.M
  • #Radial momentum :-
  • #comments on gauge transformation :-
  • #Digression :- Angular momentum ;-
  • #J. Schwinger method.
• #Lecture 16
  • #charged particle in a constant uniform magnetic field.
  • #Cyclotron frequency and Landau levels

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[The notebook writes these two lectures out of order: the "lec 17" heading appears first (immediately below a marginal remark saying so), and the "lec 16)" heading only comes several pages later. This page follows the notebook's physical order -- Lecture 17 first, Lecture 16 second.]

["|lec 16) is written after |lec 17)"]

The general \(\hat H\) of a charge particle in EM field induced by \(A(\vec r,t)\) & \(\phi(r,t)\) is

\[ H = \frac{(\vec P - e\vec A)^2}{2m} + e\phi \qquad \left\{\begin{aligned} A &= A(r,t)\\ \phi &= \phi(r,t)\end{aligned}\right. \]

[\(e = \) charge of particle]

Shrödinger eq\(^{\text{n}}\) in position representation.

\[ H\psi(r,t) = i\hbar\frac{\partial\psi(r,t)}{\partial t} = \frac{(-i\hbar\nabla - eA)\cdot(-i\hbar\nabla - eA)}{2m}\psi(\vec r,t) + e\phi\,\psi(\vec r,t) \]

let us work in Heisenberg picture, where rate of change of position operator can be simply

\[ \frac{dx_i}{dt} = \frac{[x_i, H]}{i\hbar} = \frac{\left[x_i\ \frac{p^2}{2m} + e\phi\right]}{i\hbar} \]

\(\left\{\begin{aligned}{} [x_i\ A_i] &= 0\\ (x_i, \phi_i) &= 0\end{aligned}\right.\)

\[ \frac{dx_i}{dt} = \frac{1}{2mi\hbar}\left([x_i\ P_\ell]P_\ell + P_\ell[x_i\ P_\ell]\right) \] \[ = \frac{[x_i\ p_\ell]P_\ell - e\underbrace{[x_i\ A_\ell]}_{0}P_\ell + P_\ell[x_i, p_\ell] - P_\ell\underbrace{[x_i, eA_i]}_{0}}{2mi\hbar} \] \[ \frac{d\hat x_i}{dt} = \frac{i\hbar P_\ell + P_\ell(i\hbar)}{2mi\hbar} = \frac{P_\ell}{m} = \frac{p_i - eA_i}{m} \]

find accelaration,

\[ \frac{d^2x_i}{dt^2} = \frac{\left[\frac{dx_i}{dt}, H\right]}{i\hbar} \] \[ = \frac{\left[\frac{p_i - eA_i}{m},\ \frac{(\vec p - e\vec A)^2}{2m} + e\phi\right]}{i\hbar} \qquad \left(P_i = p_i - eA_i\right) \] \[ \frac{d^2\vec r}{dt^2} = \frac{e}{m}\left[\vec E + \frac{1}{2}\left(\frac{d\vec r}{dt}\times\vec B - \vec B\times\frac{d\vec r}{dt}\right)\right] \]

\(\left[a = \frac{e}{m}\left(E + \underline{\vec v\times\vec B}\right)\right]\)

\(\frac{d\vec r}{dt}\times\vec B\) is not hermitian so to take care of that it is represented as \(\frac12\left(\frac{d\vec r}{dt}\times\vec B - B\times\frac{d\vec r}{dt}\right)\) which is hermitian.

\[ \vec L = \vec r\times\vec p \]

we know that \(\hat r, \hat p\) do not commute with each other so how do we say \(\vec L\) is hermitian?

reason is because non-canonical conjugate pairs do commute with each other.

\[ \vec L = \begin{vmatrix} \hat i & \hat j & \hat k\\ x & y & z\\ p_x & p_y & p_z\end{vmatrix} = \hat i(yp_z - zp_y) - \hat j(xp_z - zp_x) + \hat k(xp_y - yp_x) \] \[ L = L^\dagger = \hat i(yp_z - zp_y)^\dagger - \hat j(xp_z - zp_x)^\dagger + \hat k(xp_y - yp_x)^\dagger = L \] \[ = L,\ \left(\text{as } [r_i\,p_j] = i\hbar\,\delta_{ij}\right) \]

\([x\ p_z] = 0\) & so on

\((x\ p_y) = 0\)

\[ p_r = \vec p\cdot\hat r \ \Rightarrow \ p_r = \frac{\vec p\cdot\vec r}{|r|} = \frac{(xp_x + yp_y + zp_z)}{\sqrt{x^2+y^2+z^2}} \]

Q). is the above defn compatible with Q.M? No!

radial momentum is represented as :-

\[ p_r = \alpha\,\frac{\vec p\cdot\vec r}{|r|} + (1-\alpha)\,\frac{\vec r}{|r|}\cdot\vec p \] \[ p_r^\dagger = p_r \ \text{ at } \ \alpha = \tfrac12 \ ; \qquad p_r = \frac12\left(\frac{\vec p\cdot\vec r + \vec r\cdot\vec p}{|r|}\right) \]

\(p_r\) can't be represented as \(\vec p\cdot\hat r\) as it is not hermitian.

\[ \left(\frac{\vec p\cdot\vec r}{|r|}\right)^\dagger = \frac{\left(p_x^\dagger x^\dagger + p_y^\dagger y^\dagger + p_z^\dagger z^\dagger\right)}{r} \neq \] \[ = \frac{p_x x + p_y y + p_z z}{r} \ \neq \ \frac{xp_x + yp_y + zp_z}{r} \] \[ \text{as } \ [x\ p_x]\neq 0,\ [y, p_y]\neq 0,\ [z, p_z]\neq 0 \]

where as

\[ p_r = \frac{(\vec p\cdot\vec r + \vec r\cdot\vec p)}{2|r|} \ \text{ is hermitian.} \qquad \left(\alpha = \tfrac12\right). \]

\((L, p_r)\) are physicel measurables, so upon measurment we should get real eigen values, which is possible easily if operator is hermitian).

similerly, in laplace -- runge vector \((A)\)

we should write it as

\[ A = \frac{1}{2}\left(\vec p\times\vec L - \vec L\times\vec p\right) - \frac{mk\vec r}{|r|} \]

\(\rightarrow\) When we make a gauge transformation, No physical variables change. of \(E,\) & \(B\).

\(\rightarrow\) In the quantum version it implies that the physical quantities like probabilities (\(|\psi|^2\) & so on) should not change when we go to a different gauge. (in \(A, \phi\)).

(Q) what is the change that occurs, when we go from one gauge to another. let's ask, what happens to probability current when we switch on the E-m field.

\[ \frac{\partial\rho}{\partial t} + \nabla\cdot\vec J = 0 \]

where, \(\left\{\begin{aligned}\rho &= |\psi|^2 \quad \text{probability density}\\ J &= \text{probability current}\end{aligned}\right.\)

\[ J = \frac{\hbar}{2mi}\left[\psi^*\nabla\psi - \psi\nabla\psi^*\right] \] \[ = \frac{\hbar}{m}\,\text{Im}(\psi^*\nabla\psi) \]

Yet, there is another way of doing it, which is important in other applications of Q.M. & thats the following

let's write \(\psi(r,t)\) as.

as \(\left(|\psi|^2 = \rho\right)\)

\[ \psi(r,t) = \sqrt{\rho}\ e^{\frac{iS(r,t)}{\hbar}} \]

\(S(r,t) \longleftrightarrow\) Action or phase

so what is \(\vec J\) in this form?

\[ \nabla\psi = \nabla(\sqrt\rho)\,e^{iS/\hbar} + \sqrt\rho\,\frac{i}{\hbar}\nabla(S)\,e^{iS/\hbar} \] \[ \nabla\psi^* = \nabla(\sqrt\rho)\,e^{-iS/\hbar} - \sqrt\rho\,\frac{i}{\hbar}\nabla(s)\,e^{-iS/\hbar} \] \[ J = \frac{\hbar}{2mi}\left[\psi^*\nabla\psi - \psi\nabla\psi^*\right] \] \[ = \frac{\hbar}{2mi}\left(\sqrt\rho\,\nabla(\sqrt\rho) + \rho\,\frac{i}{\hbar}\nabla(s) - \sqrt\rho\,\nabla(\sqrt\rho) + \frac{i}{\hbar}\rho\,\nabla(s)\right) \]

[the two \(\nabla(\sqrt\rho)\) terms are struck out]

\[ = \frac{2i}{\hbar}\ \frac{\rho\,\nabla(s)\,\hbar}{2mi} \] \[ \boxed{J = \frac{\rho}{m}\nabla(s)} \]

1. what is \(\nabla(s) = ?\)) in presence of E-m field :- \[ J = \frac{\hbar}{2mi}\left[\psi^*\nabla\psi - \psi\nabla\psi^*\right] \] \[ \vec p \longrightarrow \vec p - e\vec A \] \[ -i\hbar\nabla \longrightarrow -i\hbar\nabla - e\vec A \] \[ \nabla \longrightarrow \nabla - \frac{ie\vec A}{\hbar} \] \[ J' = \frac{\hbar}{2mi}\left[\psi^*\left(\nabla - \frac{ie\vec A}{\hbar}\right)\psi - \psi\left(\nabla - \frac{ieA}{\hbar}\right)\psi^*\right] \] \[ J' = J - \frac{\hbar}{2mi}\ \frac{2ieA\,|\psi|^2}{\hbar} \] \[ J' = J - \frac{eA}{m}|\psi|^2 = \frac{\rho}{m}\left(\nabla(s) - e\vec A\right) = \frac{\rho}{m}\nabla(s)' \] \[ \nabla S' \longrightarrow \nabla S - e\vec A \] [we know that the \(|\psi|^2\) should not change upon varing the \((A,\phi)\) But Action/phase\((s)\), \(J\), changes upon transformng \((A,\phi)\to(A',\phi')\).] [If \(\rho\) plays role of charge density, \(\nabla(s)\) plays role of velocity, then \(J\) becomes current density]

so this is all happens when we have EM field, This phase fun. \(s\). its \(\nabla(s)\) is shifted by \((e\vec A)\).

\[ s \longrightarrow s - e\int \vec A\cdot d\vec l \]

so it suggest, phase of EM field is path dependent. (line integral)

let's go back to angular momentum (algebra), find its (eigen values/fun. study its some properties, then we will come back and apply it on charge particle sitting in EM field.

Q.Mechenical angular momentum turns out to be somewhat different from classical angular momentum.

For a single particle angular momentum was defined as

\[ \vec L = \vec r\times\vec p \qquad \text{---- (classical)}. \]

It turns out the same defnition is true for Q.M except \(r, p, L\) are operators.

classically we know,

\[ \{x_i\ p_j\} = \delta_{ij} \] \[ \{L_i\ L_j\} = \epsilon_{ijk}L_k \]

\(\left\{\begin{aligned} \epsilon_{ijk} \ \text{is}\ 1 \ &\text{if } i, j, k \text{ are consecutive}\\ \text{else is} -1 \ &\underline{\qquad}\ \text{odd sequenced}\\ = 0 \ &\text{if } i = j = k.\end{aligned}\right.\)

[a small cyclic (circular) arrow is drawn beside "consecutive"]

Quantum mechanicelly angular momentum is defined by these commutation relations.

\(\vec J = (J_1, J_2, J_3)\) are angular momentum operators satisfying

\[ [J_i, J_j] = i\hbar\,\epsilon_{ijk}J_k \]

So any quantum mechanical angular momentum of any system whatso ever is represented by three hermitian operators \((J_1, J_2, J_3)\) which satisfy this commutation relation.

\[ [J_i\ J_j] = i\hbar\,\epsilon_{ijk}J_k \]

You can see diamensions are matching.

The above statement is dfination and all the properties of \(J\). will come from it.

This angular momentum algebra is also called SO(3) algebra, because it has to do with rotation in 3-D. It is a lie -- algebra.

\[ [J_1, [J_2, J_3]] + \text{cyclic permutations} = 0 \qquad (\text{Jacobi identity}) \]

Given this information what can we say about angular momentum states of any system.

(remarkebly this question can be answered from this algebra and nothing more).

we need few observations:

\[ J^2 = J_1^2 + J_2^2 + J_3^2 \qquad (\text{by def}^{\text{n}}). \]

lets ask, what is \([J, J^2] = ?\)

\[ [J, J^2] = [J_1,\ J_1^2 + J_2^2 + J_3^2] = \underbrace{[J_1, J_1^2]}_{0} + [J_1, J_2]J_2 + J_2[J_1, J_2] \] \[ +\ [J_1\ J_3]J_3 + J_3[J_1, J_3] \] \[ [J_1, J^2] = i\hbar\left(J_3J_2 + J_2J_3\right) + \ -i\hbar\left(J_2J_3 + J_3J_2\right) \] \[ = i\hbar\left(J_3J_2 - J_3J_2\right) + i\hbar\left(J_2J_3 - J_2J_3\right) = 0 \]

similerly

\[ [J_2, J^2] = [J_3\ J^2] = 0 \] \[ \Rightarrow \quad [J_i\ J^2] = 0 \qquad (i = 1,2,3) \]

It also implies any linear combination of \(J_1, J_2, J_3\) will commute with \(J^2\).

\[ \boxed{[\vec J\cdot\hat n,\ J^2] = 0} \qquad \text{for any } \hat n \]

So, component of \(J\) in any direction commutes with \(J^2\). Therefore we can simultaneously find eigen-states of \(J_i\) and \(J^2\). Generally we use \(J_3\) as axis of quntisation rather than \(\vec J\cdot\hat n\). But remember any direction \((\hat n)\vec J\) can be used as axis of quantisation. No two components commute with each other.

\[ [\vec J\cdot\hat n,\ \vec J\cdot\hat n'] \neq 0 \qquad (\text{if } \hat n \neq \hat n') \]

Note :- \([\vec J\cdot\hat n, J^2] = 0\) forms maximal set of commuting operator so we can write angular momentum states depending upon only (\(j\) & \(j_n\)).

"we would like to find out what is spectrum of \(J, J_1, J_2, J_3, J^2\) or what are possible values of eigen value of \(J^2, J, J_1, J_2, J_3\) based on these commutation relations"?

we must do exactly what we did for harmonic oscillator. we have to do this purely algebrically. "(using trick of bosonisation"

In case of Harmonic oscillator we found \(a^\dagger, a\) (ladder operators), can we find similer operators here? (yes we can).

\[ J_+ = J_1 + iJ_2 \] \[ J_- = J_1 - iJ_2 = (J_+)^\dagger = (J_1 + iJ_2)^\dagger = J_1^\dagger - iJ_2^\dagger \]

what is,

\[ [J_+\ J_-] = [J_1 + iJ_2,\ J_1 - iJ_2] \] \[ = \underbrace{[J_1, J_1]}_{0} + (-i)[J_1, J_2] + i[J_2, J_1] + [J_2, J_2] \] \[ = \left(-i\,J_3 + i(-J_3)\right)(i\hbar) \] \[ = 2\hbar J_3 \] \[ \boxed{[J_+, J_-] = 2\hbar J_3} \] \[ [J_+, J_3] = [J_1 + iJ_2,\ J_3] \] \[ = [J_1, J_3] + i[J_2, J_3] \] \[ = \left(-i\hbar J_2 + i(i\hbar)J_1\right) \] \[ = -\hbar[J_1 + iJ_2] = -\hbar J_+ \] \[ \boxed{[J_+\ J_3] = -\hbar J_+} \] \[ [J_-, J_3] = [J_1 - iJ_2,\ J_3] = [J_1, J_3] - i[J_2, J_3] \] \[ = -i\hbar J_2 - i(i\hbar)J_1 \] \[ = \hbar[J_1 - iJ_2] = \hbar J_- \] \[ \boxed{[J_-, J_3] = \hbar J_-} \]

[we can Rewrite the angular momentum algebra

\[ (J_+ J_-) = 2\hbar J_3 \] \[ [J_+, J_3] = -\hbar J_+ \] \[ [J_-\ J_3] = \hbar J_- \]

advantage is that we do not have complex no "\(i\)".]

concider two different harmonic oscillator which do not talk with each other.

[\((A, B)\)]

\[ \left.\begin{aligned} A &\to a, a^\dagger\\ B &\to b, b^\dagger\end{aligned}\right\} \text{raising \& lowering operators} \]

such that

\[ [a\ a^\dagger] = [b, b^\dagger] = I \] \[ [a, b] = 0 = [a\ b^\dagger] = [a^\dagger, b^\dagger] = 0 \]

[\(a, a^\dagger\) / \(b, b^\dagger\) are chosen to be diamensionless.]

If,

\[ J_+ = (a^\dagger b)\hbar \] \[ J_- = (J_+)^\dagger = (a^\dagger b)^\dagger = (b^\dagger a)\hbar = (ab^\dagger)\hbar \] \[ [J_+, J_-] = ? \] \[ = \hbar^2\left[a^\dagger b,\ ab^\dagger\right] \] \[ = \left(a^\dagger[b, ab^\dagger] + [a^\dagger, ab^\dagger]b\right)\hbar^2 \] \[ = \hbar^2\left(a^\dagger\left(\underbrace{[b, a]}_{0}b^\dagger + a[b, b^\dagger]\right) + \left([a^\dagger\ a]b^\dagger + a\underbrace{[a^\dagger, b^\dagger]}_{0}\right)b\right) \] \[ = \hbar^2\left(a^\dagger a\,[b\,b^\dagger] + [a^\dagger a]\,b^\dagger b\right) \] \[ = \hbar^2\left(a^\dagger a\,I + (-I)\,b^\dagger b\right) \] \[ \boxed{[J_+\ J_-] = \hbar^2\left(a^\dagger a - b^\dagger b\right)} \] \[ \left\{\begin{aligned} J_1 &= \frac{J_+ + J_-}{2} = \hbar\left(\frac{a^\dagger b + ab^\dagger}{2}\right) = \text{hermitian}\\[4pt] J_2 &= \frac{J_+ - J_-}{2i} = \hbar\left(\frac{a^\dagger b - ab^\dagger}{2i}\right) = \text{hermitian}\\[4pt] J_3 &= \frac{[J_+\ J_+]}{2\hbar} = \hbar\left(\frac{a^\dagger a - b^\dagger b}{2}\right) = \text{hermitian} \end{aligned}\right. \]

This autometically implies

\[ [J_i\ J_j] = i\hbar\,\epsilon_{ijk}J_k \] \[ J^2 = J_1^2 + J_2^2 + J_3^2 \] \[ \frac{4}{\hbar^2}J^2 = \left(a^\dagger b + ab^\dagger\right)^2 - \left(a^\dagger b - ab^\dagger\right)^2 + \left(a^\dagger a - b^\dagger b\right)^2 \] \[ = a^\dagger b\,a^\dagger b + a^\dagger b\,ab^\dagger + ab^\dagger a^\dagger b + ab^\dagger ab^\dagger \] \[ -\ a^\dagger b\,a^\dagger b + a^\dagger b\,ab^\dagger + ab^\dagger a^\dagger b - ab^\dagger ab^\dagger \] \[ +\ a^\dagger a\,a^\dagger a - a^\dagger a\,b^\dagger b - b^\dagger b\,a^\dagger a + b^\dagger b\,b^\dagger b \] \[ \frac{4}{\hbar^2}J^2 = 2\left(a^\dagger b\,ab^\dagger + ab^\dagger a^\dagger b\right) + \left(a^\dagger a\right)^2 - 2a^\dagger a\,b^\dagger b + \left(b^\dagger b\right)^2 \]

lets call

\[ N_a = a^\dagger a \] \[ N_b = b^\dagger b \]

we know that \(ab = ba\), \(b^\dagger a^\dagger = a^\dagger b^\dagger\)

\[ \frac{4}{\hbar^2}J^2 = 2\left(a^\dagger a\ bb^\dagger + a\,a^\dagger\,b^\dagger b\right) + \left(N_a\right)^2 - 2N_aN_b + N_b^2 \] \[ = 2\Big(N_a(1+N_b) + (N_a+1)N_b\Big) + \left(N_a - N_b\right)^2 \] \[ = N_a^2 + N_b^2 + 2N_aN_b + 2\left(N_a + N_b\right) \] \[ = \left(N_a + N_b\right)^2 + 2\left(N_a+N_b\right) \] \[ J^2 = \frac{\hbar^2}{4}\left(N_a+N_b\right)\left(\left(N_a+N_b\right) + 2\right) \] \[ = \frac{\hbar^2\left(N_a+N_b\right)}{2}\left[\frac{\left(N_a+N_b\right)}{2} + 1\right] \]

[\(N_a, N_b\) are operators / number operators where as \(n_a, n_b\) are eigen states of these number operators]

let's call

\[ \frac{N_a + N_b}{2} = j \qquad \left(j = \left(\frac{n_a+n_b}{2}\right)\right) \quad \longrightarrow \text{(eigen values)} \] \[ \boxed{J^2 = \hbar^2\,j(j+1)} \] \[ \text{where } \left.\begin{aligned}(n_a)\ \ n_a &= 0, 1, 2\cdots\\ (n_b)\ \ n_b &= 0, 1, 2\cdots\end{aligned}\right\} \text{are eigen values of } N_a \ \& \ N_b \]

whatever be the physical system, the possible value of eigen values of \(J^2\) are given by

\[ J^2 = \hbar^2\,j(j+1) \qquad \Big| \qquad J^2 = \hbar^2\,j(j+1) \]

where "\(j\)" can only have

\[ 0, \ \tfrac12, \ 1, \ \tfrac32, \ \cdots \]

so we see half integers appears naturally ( spins have half integral values). physical interpretation is still missing.

what are eigen values of \(J_3\).

\[ J_3 = \frac{\hbar\left(a^\dagger a - b^\dagger b\right)}{2} = \frac{\hbar\left(N_a - N_b\right)}{2} \] \[ \text{eigen values of } J_3 = \frac{\hbar}{2}\left(n_a - n_b\right) \]

\(\left\{\begin{aligned}\text{given } j, \ \frac{(N_a+N_b)}{2}, \ &\text{what are eigen values of } J_3, \text{ i.e.}\\ \text{eigen values of } &\frac{(N_a - N_b)}{2}\hbar\ ?\end{aligned}\right.\)

["lec 18 is written after lec 16)"]

\[ H = \frac{(p - eA)^2}{2m} \]

we would like to find eigen values and eigen states of this hamiltonian. (in the position basis)

Since, \(\vec A = \vec A(\vec r)\)

therefore in general it does not commute with \(p\), \([A, p]\neq 0\)

However we found that \([A(r), p] = -i\hbar\,\nabla\cdot\vec A(r)\)

if \(\nabla\cdot\vec A = 0\)

\[ \Rightarrow \quad [\vec p, \vec A] = 0 \]

for constant uniform \(\vec B\).

\[ \vec A = \tfrac12\left(\vec B\times\vec r\right) \]

it is easily varified \(\nabla\times A\) gives us back \(B\).

if \(B = B\,\hat e_z\)

\[ A = \frac12\begin{vmatrix}\hat i & \hat j & \hat k\\ 0 & 0 & B\\ x & y & z\end{vmatrix} = \frac12\left(\hat i(-BY)\ -\hat j(-BX),\ 0\hat k\right) \] \[ \vec A = \left(\frac{-1}{2}\cdot Y\cdot B,\ \frac12\cdot X\cdot B,\ 0\right) \]

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0]
  \draw[->] (-3.0,0) -- (3.2,0) node[right] {$x$};
  \draw[->] (0,-2.6) -- (0,2.8) node[above] {$y$};
  \foreach \x in {-2,-1,1,2} {
    \foreach \y in {-2,-1,1,2} {
      \draw[-{Stealth}] (\x,\y) -- ++({-0.30*\y},{0.30*\x});
    }
  }
  \foreach \x in {-2,-1,1,2} {\draw[-{Stealth}] (\x,0) -- ++(0,{0.30*\x});}
  \foreach \y in {-2,-1,1,2} {\draw[-{Stealth}] (0,\y) -- ++({-0.30*\y},0);}
  \node[anchor=west] at (1.6,2.4) {The vector field $\vec A$};
  \node[anchor=west] at (1.6,1.9) {see curl $A = B\,\hat z$};
  \node[anchor=west] at (1.6,1.4) {$\nabla\times A = B\,\hat z$};
\end{tikzpicture}

so the general rule is if we want a \(\vec B\) in perticular direction you can choose the vector potential to be transverse to it, and the transversed components do not involve co-ordinate along \(\vec B\).

(Is \(\vec A\) unique ?)

(No we can always add \(A' = A + \nabla\chi\))

Is it possible to get rid of one of the components of \(\vec A\) ? to make it just \(x\) component.

\[ A' = A + \nabla\chi(r) \] \[ \chi = -\tfrac12 BXY. \] \[ \nabla\chi = \frac{\partial\chi}{\partial x}\hat i + \frac{\partial\chi}{\partial y}\hat j + \frac{\partial\chi}{\partial z}\hat k \] \[ = \left(\frac{-BY}{2}\hat i\right) + \left(\frac{-BX}{2}\hat j\right) + 0 \] \[ A' = \left(\frac{-BY}{2},\ \frac{+XB}{2},\ 0\right) + \left(\frac{-BY}{2},\ \frac{-XB}{2},\ 0\right) \] \[ A' = \left(-BY,\ 0,\ 0\right) \]

\(\left\{\begin{aligned}&\text{no physical quantity should change}\\ &|\psi|^2, H, \text{ if we work in } A' = A + \nabla\chi.\end{aligned}\right.\)

(plot \(A'\))

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\begin{tikzpicture}[scale=1.0]
  \draw[->] (-3.4,0) -- (3.6,0) node[right] {$x$};
  \draw[->] (0,-2.2) -- (0,2.4) node[above] {$y$};
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      \draw[-{Stealth}] (\x,\y) -- ++({-0.55*\y},0);
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\end{tikzpicture}

let's look at what happens to \(H\).

\[ H = \frac{(p - eA)^2}{2m} \qquad A = (-BY, 0, 0) \]

\(\left\{\begin{aligned}&\text{is } \nabla\cdot\vec A = 0\\ &\vec\nabla\cdot\vec A = \frac{\partial}{\partial x}A_x + \frac{\partial}{\partial y}A_y + \frac{\partial}{\partial z}A_z\\ &= 0 + 0 + 0 = 0\\ &[p, A] = 0\end{aligned}\right.\)

\[ p^2 = p_x^2 + p_y^2 + p_z^2 \] \[ H = \frac{(p - eA)(p - eA)}{2m} \] \[ = \frac{p^2 - 2e\,\hat A\cdot\vec p + e^2A^2}{2m} \] \[ = \frac{\left(p_x^2 + p_y^2 + p_z^2\right) - 2e\,A_x p_x + e^2A_x^2}{2m} \] \[ H = \frac{\left(p_x + eYB\right)^2}{2m} + \frac{p_y^2}{2m} + \frac{p_z^2}{2m} \qquad \longleftarrow \left(A_x = -BY\right) \]

\(\left\{\begin{aligned} p^2 &= \left(\vec p_x + \vec p_y + \vec p_z\right)\cdot\left(\vec p_x + \vec p_y + \vec p_z\right)\\ &= p_x^2 + p_y^2 + p_z^2 + 2p_x\cdot p_y + 2p_y p_z + 2p_z p_x\\ p^2 &= p_x^2 + p_y^2 + p_z^2\\ &\because \left(p_i\ p_j = \delta_{ij}\right)\end{aligned}\right.\)

Now we can solve for shrödinger eq\(^{\text{n}}\) in position basis by writing \(p_x = -i\hbar\frac{\partial}{\partial x}\) & so on but there is a easier way

\(\triangleright\) (beginning of digression)

(let's ask what happens to velocity of this particle?-)

\[ p = m\vec v + e\vec A \qquad \left(m\vec v = \vec p - e\vec A = \vec\pi\right) \] \[ \vec v = \frac{\vec p - e\vec A}{m} \]

what is

\[ [v_x\ v_y] = \frac{1}{m^2}\left[p_x - eA_x,\ p_y - eA_y\right] \] \[ = \frac{1}{m^2}\left(\underbrace{[p_x\ p_y]}_{0} - e[p_x\ A_y] - e[A_x\ p_y] + e^2\underbrace{[A_x\ A_y]}_{0}\right) \] \[ = \frac{1}{m^2}\left(e[A_y\ p_x] - e[A_x\ p_y]\right) \]

\(\left\{[f(x), p_x] = i\hbar f'(x)\right.\)

\[ [v_x\ v_y] = \frac{e}{m^2}\left(i\hbar\frac{\partial A_y}{\partial x} - i\hbar\frac{\partial A_x}{\partial y}\right) \] \[ = \frac{i\hbar e}{m^2}\left(\frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y}\right) \] \[ [v_x\ v_y] = \frac{i\hbar e}{m^2}B_z \] \[ \boxed{[v_i\ v_j] = \frac{i\hbar e}{m^2}\,\epsilon_{ijk}B_k} \qquad\qquad [\pi_i\ \pi_j] = i\hbar e\,\epsilon_{ijk}B_k \]

\(\left[\begin{vmatrix}\hat i & \hat j & \hat k\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ A_x & A_y & A_z\end{vmatrix}\right.\)

\[ [v_i\ v_j] \neq 0 \]

\(\rightarrow\) we can't measure two cartesian component of velocity simultaneusly with arbitrary precision for a charged particle in const. \(\vec B\).

\(\triangleleft\) (end of digression)

For any const \(B\). and choosing \(A(A_x, A_y, 0)\). \((A_z = 0)\)

The Hamiltonian can be written as.

\[ H = \frac{\pi_x^2 + \pi_y^2}{2m} + \frac{p_z^2}{2m} \]

\(\left\{\text{it is kind of Harmonic oscillator form of } H. \quad \frac12\left(x^2+p^2\right)\right.\)

To find eigen values of \(H\), we need to diagonalize it, so just like we did in SHO,

\[ a = x + ip \] \[ a^\dagger = x - ip \]

[\(a\) is written as \(\frac{x+ip}{\sqrt2}\) with the \(\sqrt2\) struck out]

so that \(a\,a^\dagger\) is diagonalizable.

similerly

\[ a = \frac{\pi_x + i\pi_y}{\sqrt{2eB\hbar}} \]

so,

\[ a^\dagger = \frac{\pi_x - i\pi_y}{\sqrt{2eB\hbar}} \]

\(\left\{p, A \text{ are hermitian so } \pi \text{ is also } = \pi^\dagger\right.\)

\[ [\pi_x\ \pi_y] = i\hbar e\,B_k \qquad \left(B\right) \] \[ [\pi^2] = [\hbar e B]\ ; \qquad [\pi] = \boxed{\sqrt{\hbar e B}} \]

so,

\[ a^\dagger a = \frac{1}{(2eB\hbar)}\left\{\pi_x^2 + \pi_y^2 + i\left(\pi_x\pi_y - \pi_y\pi_x\right)\right\} \] \[ \frac{1}{(2eB\hbar)}\left\{\pi_x^2 + \pi_y^2 + i^2\hbar e B\right\} \] \[ \left(a^\dagger a + \frac12\right) = \frac{\pi_x^2 + \pi_y^2}{(2eB\hbar)} \qquad \downarrow \ \text{(Landau levels)} \]

using value of \(\pi_x^2 + \pi_y^2\) in \(H\).

\[ H = \frac{(2eB\hbar)}{2m}\left(a^\dagger a + \frac12\right) + \frac{p_z^2}{2m} \] \[ H = \left(\frac{eB\hbar}{m}\right)\left(a^\dagger a + \frac12\right) + \frac{p_z^2}{2m} \]

So we now know the eigen values of \(a^\dagger a\) so we know eigen values of Half of \(H\).. We have reduced the problem into Harmonic oscilletor plus a free particle moving in \(z\)-direction.

And these levels are quantized.

\(\triangleright\)

\[ \left(\frac{eB}{m} = \text{cyclotron frequency}\right) \qquad (W_c) \] \[ qVB = \frac{mv^2}{r} \] \[ \left(\frac{1}{T} = \frac{v}{2\pi r}\right) = \frac{eB}{m} \]

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\[ L = mr^2\cdot\omega = \frac{mr^2\cdot 2\pi}{T} \] \[ \mu = iA = \frac{e}{T}\cdot\pi r^2 = \frac{er^2\omega}{2} = \frac{e}{2}\left(\frac{L}{m}\right) = \frac{eL}{2m} \] \[ \left(\frac{\mu}{L} = \frac{e}{2m} = \text{gyromagnetic ratio}\right) \]

\(\triangleleft\)

\[ H = \hbar W_c\left(a\,a^\dagger + \frac{I}{2}\right) + \frac{p_z^2}{2m} \qquad \left(W_c = \frac{eB}{m}\right) \] \[ E(n, k_z) = \hbar W_c\left(n + \frac12\right) + \frac{\hbar^2 k_z^2}{2m} \qquad [a\ a^\dagger] = 1(I) \] \[ \downarrow \ \text{(Landau levels)} \]

"So if we confine the particle in XY plane then levels would look quantized like those of simple Harmonic oscillator, but if not there is free motion in \(z\) direction it drifts in \(\hat z\) direction with any real no. \(k_z\).

These levels look non degenerate but actually are degenerate. (more the one eigen state for a single eigen value).

These landau levels have acquired enormous importance, because now it is possible to take \(e^-\) and confine them essentially to two -- diamensions. (MOSFET divices) {\(e^-\) gas in 2-D behaves very different than 3D).

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