Quantum Mechanics Lec 18-19 (V Balakrishnan) Opus 5

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[Editorial note : the explicit "Lec 18" heading in the notebook stands at the top of scan qm7-04. The three pages before it (qm7-01 .. qm7-03) carry no heading and simply finish off the Landau-level / charged-particle-in-a-magnetic-field calculation that was already running on the last page of the previous notebook (qm6-24, which breaks off mid-sentence with "So, to write down wave function explicitly, we must choose the vector potential explicitly. The wave vector potential \((\vec A(r))\)"). Those three pages are transcribed here, first, as a carry-over section so that they are not lost; the "Lec 18" material proper begins at the heading "Angular momentum states from two oscillators" below.]

Charged particle in a magnetic field -- contd.

should not present in wave functions, because final wave fun. energy eigen values should be independent of any gauge you choose.

So let's choose \(\vec A(\vec r) = (-By, 0, 0)\)

\[ H = \frac{(\bar p - e\bar A)^2}{2m} \]

[\(A = (-By, 0, 0)\)]

[\(p = (p_x, p_y, p_z)\)]

[\(p^2 = p_x^2 + p_y^2 + p_z^2\)]

\[ H = \frac{p^2 - e\bar p\cdot\bar A - e\bar A\cdot\bar p + e^2A^2}{2m} \] \[ H = \frac{p_x^2 + e^2A^2 - 2e\bar A\cdot\bar p}{2m} + \frac{p_y^2}{2m} + \frac{p_z^2}{2m} \] \[ H = \frac{(p_x + eBy)^2}{2m} + \frac{p_y^2}{2m} + \frac{p_z^2}{2m} \] \[ = \frac{p_x^2 + p_y^2 + p_z^2}{2m} + \frac{eBy}{m}p_x + \frac{e^2B^2y^2}{2m} \]

Therefore

\[ H\phi(\bar r) = E\phi(\bar r) \] \[ p_x = -i\hbar\frac{\partial}{\partial x} \] \[ p_y = -i\hbar\frac{\partial}{\partial y} \] \[ p_z = -i\hbar\frac{\partial}{\partial z} \] \[ -\frac{\hbar^2}{2m}\left(\frac{\partial^2\phi}{\partial x^2} + \frac{\partial^2\phi}{\partial y^2} + \frac{\partial^2\phi}{\partial z^2}\right) + \frac{eBy}{m}\left(-i\hbar\frac{\partial\phi}{\partial x}\right) + \frac{e^2B^2y^2}{2m}\phi(r) = E\phi(r) \] \[ \left(\frac{-\hbar^2}{2m}(\nabla^2) + \frac{ie\hbar B}{m}\,y\frac{\partial}{\partial x} + \frac{e^2B^2}{2m}y^2\right)\phi(r) = E\phi(\bar r) \]

let the trial guess be

\[ \phi(\bar r) = e^{ik_x x}\,e^{k_z z}\cdot f(y) \]

(non-trivial to guess)

\(\hookrightarrow\) as we have derivatives for eigen value.

\[ \frac{d(e^{ax})}{dx} = a\,e^{ax} \]

so we assumed it would be a fun. of exponentials.

the guess \(\phi(\bar r)\) should be a solution of above shrödinger equation.

\[ \left(\frac{\hbar^2k_z^2}{2m} + \left\{\frac{+\hbar^2}{2m}k_x^2 - \frac{\hbar^2}{2m}\frac{d^2f}{dy^2} + \frac{eB\hbar}{m}k_xy + \frac{e^2B^2}{2m}y^2\right\}\right)e^{ik_x x}\left(e^{ik_z z}\cdot f\right) = E\,e^{ik_x x}e^{ik_z z}\cdot f \] \[ \Rightarrow \quad \left(E - \frac{\hbar^2k_z^2}{2m}\right)f = \left(\frac{\hbar^2k_x^2}{2m} - \frac{\hbar^2}{2m}\frac{d^2f}{dy^2} + \frac{eB\hbar}{m}k_xy + \frac{e^2B^2}{2m}y^2\right) \] \[ -\frac{\hbar^2}{2m}\frac{d^2f}{dy^2} + \left(\frac{e^2B^2}{2m}y^2 + \frac{\hbar eB}{m}y + \frac{\hbar^2k_x^2}{2m}\right)f = \left(E - \frac{\hbar^2k_z^2}{2m}\right)f \]

[the bracket on the right is called \(\longrightarrow \varepsilon\)]

\[ \frac{-\hbar^2}{2m}\frac{d^2f}{dy'^2} + \frac{1}{2}m\omega_c^2\,y'^2f = \varepsilon f(y') \] \[ y' = (y + \cdots k_x) \]

\(\downarrow\)

eq\(^{\text{n}}\) of simple harmonic oscillator in coordinate \('y'\).

\[ \phi(\bar r) = A_n\,e^{ik_x x}\,e^{ik_z z}\cdot e^{-\frac{y'^2m\omega}{2\hbar}}\cdot H_n\left(y\sqrt{\frac{m\omega}{2\hbar}}\right) \]

\(\hookrightarrow\) sol\(^{\text{n}}\) from SHO.

\[ E(k_x,k_z) = \hbar\omega_c\left(n+\frac{1}{2}\right) + \frac{k_z^2\hbar^2}{2m} \] \[ H_n\left(y\sqrt{\frac{m\omega}{2\hbar}}\right) = \text{hermite polynomials} \] \[ H_n(x) = e^{-x^2}\left(\frac{-d}{dx}\right)^n e^{x^2} \]

where is the centre of circle?

centre can appear anywhere in \(xy\) plane, so this infinite degeneracy appear through \('k_x'\) which does not appear in energy at all.

every level is infinitely degenerate, in the sense that for every different value of \(k_x\) \(\exists\) a different eigen state for the same energy level.

as long as \(L >> \left(\sqrt{\frac{2\hbar}{m\omega}} = \lambda \ \text{say}\right)\).

(size of box) \(\hookrightarrow\) radius of cyclotron orbit

Then it is a good approximation and degeneracy \(\propto \left(\frac{L}{\lambda}\right)^2\)

when box \(\to\infty\)

degeneracy becomes infinite degeneracy.

If box is very small we have to put potential because of box. in \(H\).

[Lec 18)

Recall we have discovered eigen values of \(J^2\) and \(J\) are \(\hbar^2j(j+1)\) and \(m\hbar\) respectively.

[\(J^2 \doteq \hbar^2j(j+1)\) \(\downarrow\) eigen values (and not the operator)]

\[ J^2 \doteq \hbar^2 j(j+1) \qquad j = 0,\ \tfrac{1}{2},\ 1,\ \tfrac{3}{2},\ \cdots \] \[ J_3 \doteq \hbar m \qquad m = -j,\ -j-1,\ ---\ 0,\ j-2,\ j-1,\ j\ = \{(2j+1)\ \text{values}\} \]

where

\[ j = \frac{(n_a+n_b)}{2}, \qquad m = \frac{n_a-n_b}{2} \qquad \Rightarrow \quad \begin{cases} n_a = j+m\\ n_b = j-m\end{cases} \] \[ m = \frac{0-4}{2},\ \frac{1-3}{2},\ \frac{2-2}{2},\ \frac{3-1}{2},\ \frac{4-0}{2} \] \[ m = -2,\ -1,\ 0,\ 1,\ 2 \]

[for given \(j = 2\) ; \(\frac{N_a+N_b}{2} = 2\) ; \(N_a+N_b = 4\). \(N_a+N_b = 4\) can be obtained from \((0,4)\) \((4,0)\) \((1,3)\) \((3,1)\) \((2,2)\) which produce \(m = -j, -j+1, \cdots 0, j-1, j\)]

Now we have two oscillators, so the eigen state depends on both \(n_a\) & \(n_b\).

\[ |n_a, n_b\rangle = \text{Abstract state vector corresponding to } (\hat n_a) \text{ and } (\hat n_b) \text{ operators.} \]

The states we are interested in are angular momentum states labelled by \(j, m\).

\[ |j,m\rangle \quad \text{or} \quad |n_a, n_b\rangle \quad \text{or} \quad |n_a = j+m,\ n_b = j-m\rangle \]

such that

\[ J^2|j,m\rangle = \hbar^2 j(j+1)|j,m\rangle \]

and

\[ J_3|j,m\rangle = \hbar\, m\,|j,m\rangle \] \[ |j,m\rangle = |n_a = j+m,\ n_b = j-m\rangle \]

\(\{\)Abstract state in basis formed from \(|j\rangle\otimes|m\rangle\}\) ; \(\{\)(abstract state in) basis formed from \(|n_a\rangle\otimes|n_b\rangle\}\)

\[ |n_a, n_b\rangle = \frac{(a^\dagger)^{n_a}}{\sqrt{n_a!}}\frac{(b^\dagger)^{n_b}}{\sqrt{n_b!}}|n_a=0,\ n_b=0\rangle \]

\(\{\)Recall that \(|n\rangle = \frac{(a^\dagger)^n}{\sqrt{n!}}|0\rangle\)

\(\triangleright\) "what do we mean by two ket vectors \(|n_a,n_b\rangle\) ?"

\(|n_a,n_b\rangle\) is direct product of these two hilbert spaces.

The idea is following :-

\(\mathcal{H}_a : \{|n_a\rangle\}\)

\(\mathcal{H}_b : \{|n_b\rangle\}\)

\(\exists\) for oscillators \(a\), a hilbert space \(\mathcal{H}_a\) which has states/vectors labelled by \(\{|n_a\rangle\}\).

similarly for oscillator \(b\), \(\exists\) a hilbert space \(\mathcal{H}_b\) defined by states \(\{|n_b\rangle\}\)

operator/Oscillator \(\qquad\) space \(\qquad\) basis set

\(a^\dagger,\ a\) \(\xrightarrow{\text{acts upon}}\) \(\mathcal{H}_a\) \(\qquad\) \(\{|n_a\rangle\}\)

\(b^\dagger,\ b\) \(\xrightarrow{\text{acts upon}}\) \(\mathcal{H}_b\) \(\qquad\) \(\{|n_b\rangle\}\)

(direct sum of Hilbert space)

\[ \mathcal{H}_a \oplus \mathcal{H}_b = \mathcal{H} \]

\(\hookrightarrow\) This space is span by following states/basis:

\[ \{|n_a\rangle \otimes |n_b\rangle\} \]

or \(\{|n_a, n_b\rangle\}\) short hand notation

So,

\[ |n_a, n_b\rangle = \frac{(a^\dagger)^{n_a}}{\sqrt{n_a!}}|n_a=0\rangle \otimes \frac{(b^\dagger)^{n_b}}{\sqrt{n_b!}}|n_b=0\rangle \]

\(\hookrightarrow\) direct product / Tensor product / cartesian product

It is Just like plane is direct product of \(R\otimes R = R^2\);

\[ x \in R \] \[ y \in R \] \[ (x,y) \in R^2 = R\otimes R \]

Note ; If \(\mathcal{H}_a\) has 10 states, \(\mathcal{H}_b\) has 15 states then \(\mathcal{H}\) \(\{|n_a\rangle\otimes|n_b\rangle\}\) has \(10\times15 = 150\) states.

let us find \(|j,m\rangle\) ?

\[ |n_a,n_b\rangle = |j,m\rangle = \frac{(a^\dagger)^{j+m}(b^\dagger)^{j-m}}{\sqrt{(j+m)!\,(j-m)!}}\,|n_a=0,\ n_b=0\rangle \] \[ \left\{|j,m\rangle = \frac{(a^\dagger)^{j+m}(b^\dagger)^{j-m}}{\sqrt{(j+m)!\,(j-m)!}}\left|\frac{0+0}{2},\ \frac{0-0}{2}\right\rangle\right. \]

If \(|0,0\rangle\) is normalized to unity, it is gaurantee that \(|j,m\rangle\) is normalized to unity because of \(\sqrt{(j+m)!\,(j-m)!}\)

So, in space of angular momentum states \(|j,m\rangle\) The orthonormality condition becomes

\[ \langle j,m|j',m'\rangle = \delta_{jj'}\,\delta_{mm'}. \]

for a system whose total angular quantum no \((j)\) is specified, then the angular momentum states are \((2j+1)\) in total. So the direct product space will be \((2j+1)\) dimensional.

\(\hookrightarrow\) so diamension of operators (angular momentum) will be \((2j+1)\times(2j+1)\).

So we can represent them in form of finite-diamensional matrices

once it is finite diamensional matrix, we know how to find basis/natural basis.

let's assume \(|j,m\rangle\) is angular momentum state in which \(j\) is specified.

what is \(J_+|j,m\rangle = ?\) \(\longrightarrow\) state in \(|j\rangle\) & \(|m\rangle\) basis

\[ J_+ = J_1 + iJ_2 = (a^\dagger b)\hbar \]

\(\{I_+ = (a^\dagger b)\hbar\) in harmonic oscillator basis \(\{|n_a\rangle\}\ \{|n_b\rangle\}\)

\[ J_+|j,m\rangle = \hbar(a^\dagger b)\left|\frac{n_a+n_b}{2},\ \frac{n_a-n_b}{2}\right\rangle \]

\(|j,m\rangle\) & \(|n_a,n_b\rangle\) Both represents angular momentum states, let's first find \(J_+|n_a,n_b\rangle = ?\) in harmonic oscillator basis spanned by \(\{|n_a\rangle\otimes|n_b\rangle\}\)

first let us work in \(|n_a,n_b\rangle\) basis.

\[ J_+|j,m\rangle = \hbar(a^\dagger b)|n_a,n_b\rangle = a^\dagger b\,|j+m,\ j-m\rangle \] \[ J_+|n_a,n_b\rangle = \hbar\,a^\dagger|j+m\rangle \otimes b|j-m\rangle \] \[ = \hbar\sqrt{j+m+1}\,|j+m+1\rangle \otimes \sqrt{j-m}\,|j-m-1\rangle \]

\[ J_+|n_a,n_b\rangle = \hbar\sqrt{(j+m+1)(j-m)}\,\big|\underbrace{j+m+1}_{n_a},\ \underbrace{j-m-1}_{n_b}\big\rangle \]

In \(|j,m\rangle\) basis

\[ J_+|j,m\rangle = \hbar\sqrt{(j+m+1)(j-m)}\left|\frac{(j+m+1)+(j-m-1)}{2},\ \frac{j+m+1-j+m+1}{2}\right\rangle \] \[ J_+|j,m\rangle = \hbar\sqrt{(j+m+1)(j-m)}\,|j,\ m+1\rangle \]

\(\left\{\text{for } m = j \ \Rightarrow\ J_+|j,j\rangle = 0 \ ;\ J_-|j,-j\rangle = 0\right.\)

similarly,

\[ J_-|j,m\rangle = \hbar\sqrt{(j+m)(j-m+1)}\,|j,\ m-1\rangle \]

For a given \(j\), there are \(2j+1\) eigen states of angular momentum.

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[\(\checkmark\) states \(\otimes\) (not energy)]

So \(J_+\) operators increase the eigen value of \(J_3\) by 1.

\(J_\pm\) ; for example \(j = 1\)

\[ m = -1,\ 0,\ 1 \]

total angular momentum states are 3.

But since, there are no states above \(|1,1\rangle\)

\[ J_+|1,1\rangle = 0 \qquad \text{or} \qquad \boxed{J_+|j,j\rangle = 0} \]

similarly there are no states below \(m = -j\)

\[ \boxed{J_-|j,-j\rangle = 0} \]

Therefore in this space of \(\{|j,m\rangle\}\) we can represent \(J_+\) operator as,

\[ \boxed{(J_\pm)^{2j+1} = 0} \]

\(\hookrightarrow\) nil potent matrix. \(M^n = 0\).

\(J_+\) & \(J_-\) are not hermitian, but are hermitian conjugate of each other.

(watch from 25:00 to 45:00)

\(\triangleright\) We know that \(J_1\ J_2\ J_3\) are operators which can be expressed as matrices (for fixed \(j\), \((2j+1)\times(2j+1)\) diamensional matrices), but here we call \(J_1, J_2, J_3\) act as if they are vectors. (calling collection of 3 operators as component of a vector \(J\)). (Though \('J'\) is also an operator).

How that is justified.?

"\(J_1, J_2, J_3\) transform like component of a vector under physical rotations of coordinate system."

If we have a transformation under which a system is invarient, then physical quantities don't change under this transformation.

It implies \(\langle\psi|A|\psi\rangle\) must remain invarient under a transformation. Suppose transformation acts on states themselves

so, \(|\Psi\rangle \longrightarrow U|\psi\rangle\) ; \(U =\) unitary operator

\[ \langle\psi|A|\psi\rangle \longrightarrow \langle\psi|U^\dagger A U|\psi\rangle \]

we can see use heisenberg picture suggesting

\[ \left\{A \longrightarrow U^\dagger A U\right. \]

[\('U'\) depends on object it acts upon it.]

Since \('U'\) is unitary \(U^\dagger = U^{-1}\)

\[ \boxed{A \longrightarrow U^{-1}AU} \qquad \Rightarrow \quad \text{rotation(transformation)}. \]

Just like a vector under rotation changes transforms from \(v \to v'\) under rotation about \(z\) axis.

\[ v_1' = v_1\cos\alpha + v_2\sin\alpha \] \[ v_2' = -v_1\sin\alpha + v_2\cos\alpha \] \[ v_3' = v_3 \]

\(\left\{v_1, v_2, v_3\right.\) can be matrices also\(\}\) so operators component can be transform like a vector.

Here \(v'\) is a vector which can be written as linear combinations of \(v_1\) and \(v_2\).

Note :-

\[ |\psi(t)\rangle = e^{\frac{-iHt}{\hbar}}|\psi(0)\rangle \]

\(\leftarrow\) state vectors transform like -this and operators transform like \(\longrightarrow \left(e^{iHt/\hbar}A(0)e^{-iHt/\hbar}\right)\) under rotation

Similarly

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

\[ \left\{e^{\lambda A}Be^{-\lambda A} = B + \lambda[A,B] + \frac{\lambda^2}{2!}[A,[A,B]] + \frac{\lambda^3}{3!}[A,[A,[A,B]]]\right\} + \cdots \]

\[ 1)\quad e^{i\gamma J_3}J_1e^{-i\gamma J_3} = J_1' = J_1 + \lambda\,i\hbar J_2 + \frac{\lambda^2}{2!}(-J_1\,i\hbar) + \frac{\lambda^3}{3!}(-i\hbar J_2) \] \[ = J_1\cos\gamma + J_2\sin\gamma \] \[ 2)\quad e^{i\gamma J_3}J_2e^{-i\gamma J_3} = J_2' = -J_1\sin\gamma + J_2\cos\gamma \] \[ 3)\quad e^{i\gamma J_3}J_3e^{-i\gamma J_3} = J_3 = J_3' = J_3 \]

linear combination of \(J_1\) & \(J_2\)

The abstract representation of rotation of operators about an axis \(\hat n\) through an angle \('\theta'\) is

\[ R(\theta,\hat n) = e^{i\vec J\cdot\hat n\theta} \]

unitary operator ; \(\vec J =\) hermitian operator

So \(J_1, J_2, J_3\) transforms like a component of a vector.

Representation of \(R(\theta,\hat n)\) depends on the state it acts upon.

Note :- orbital angular momentum \(\vec L\) can only take integer eigen values \((0,1,2,\cdots)\) ; this result arrived from the requirement of wave function to be single-valued.

Note :- the representation of physical transformation (translation, parity, rotation etc of coordinate system) would be represented by an abstract operator ; the actual representation of this unitary operator would depend upon the object it acts upon.

[\(R(\theta,\hat n) = e^{i\vec J\cdot\hat n\theta}\) ; we will do several representations of rotation group, one of them for angular momentum operator acting on physical wavefunctions \(\psi(R,\theta,\phi)\)]

How does angular momentum operators look like for \(j = \frac{1}{2}\) ;

[The experiments suggested that the value of spin (intrinsic) angular momentum corresponds to \(j = \frac{1}{2}\) ; \(\{|\frac{1}{2},m\rangle\}\) = set of eigen states of spin angular mom. op\(^{\text{r}}\)]

we are looking at case where

\[ J^2 \doteq \hbar^2\,\tfrac{1}{2}\left(\tfrac{1}{2}+1\right) \]

are eigen values of \(J^2\) operator.

It turns out that particles like \(e^-\), \(p\), \(n\) have a property called intrinsic angular momentum, even in the rest frame of the particle there is an angular momentum.

For point particles like \(e^-\), you still have associated an intrinsic angular momentum, even in the frame where linear momentum of \(e^-\) is zero. It is like charge / mass of \(e^-\). This property is intrinsic property.

and experiments have shown us that \(j = \frac{1}{2}\) for \(e^-\) spin.

\[ j = \tfrac{1}{2} \quad \Rightarrow \quad (2j+1 = 2) \]

So every operator can be represented with \((2j+1)\times(2j+1)\) i.e. \(2\times2\) matrices.

\[ J^2 \doteq \hbar^2\,\tfrac{1}{2}\left(\tfrac{1}{2}+1\right) \] \[ J_3 \doteq \pm\frac{\hbar}{2} \qquad m = -\tfrac{1}{2},\ \tfrac{1}{2} \]

"we shouldn't picture spin as spinning \(e^-\), or proton, first of all \(e^-\) is elementary particle, it does not have structure. \(P^+\) do have structure, but however the origin of spin is in relativistic Q.M. It's not a non relativistic concept nor and not a classical concept at all"

It turns out, every elementary particle have intrinsic property called spin. (just like rest mass). so \((\vec L \neq \vec r\times\vec p)\)

This property is dictated by requirement of lorentz invarience. (Invarience of laws of physics under lorentz transformations). No mechanical model is possible to represent spin angular momentum.

Note :- The photons also have spin, but since, photon has zero rest mass, the idea of spin (the origin and meaning of spin) for zero rest mass particles is slightly different from what it is for non-zero rest mass particles.

spin quantum no: \((s)\) or \((j)\)

\[ s = \tfrac{1}{2} \quad \text{for } e^-,\ p,\ n,\ \nu_e \] \[ = 1 \quad \text{for } \gamma,\ W^\pm,\ Z^0 \] \[ = 0 \quad \text{for } \pi^0,\ \pi^\pm,\ K^0 \]

[these are elementary particles]

So these quantum particles have intrinsic angular momentum even in rest frame.

since \(J^2\) commutes with \(J_1, J_2, J_3\)

\[ J^2 = \hbar^2\left(\tfrac{3}{4}\right)\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix} \]

\(\rightarrow\) (schur's lemma theorem) \(\hookrightarrow\) unit matrix

[even when \(\vec p = 0\) so it can not be of form \(\vec r\times\vec p\) but this operator transforms like an vector / angular momentum]

States :-

\[ \{|j,m\rangle\} = \left\{\left|\tfrac{1}{2},m\right\rangle\right\} \] \[ = \left\{\left|\tfrac{1}{2},\tfrac{1}{2}\right\rangle \ \&\ \left|\tfrac{1}{2},-\tfrac{1}{2}\right\rangle\right\} \] \[ \text{or} \quad \{|\uparrow\rangle,\ |\downarrow\rangle\} \]

\(|\uparrow\rangle\) spin up ; \(|\downarrow\rangle\) spin down

\[ J_3 \doteq \pm\hbar/2 \quad (\text{eigen values of } J_3) \] \[ J_3|\uparrow\rangle = \frac{\hbar}{2}|\uparrow\rangle \qquad \text{or} \qquad J_3\left|\tfrac{1}{2},\tfrac{1}{2}\right\rangle = \frac{\hbar}{2}\left|\tfrac{1}{2},\tfrac{1}{2}\right\rangle \]

\(\left\{|1\rangle = |\uparrow\rangle\ ;\ |2\rangle = |2\rangle\right.\) [?]

\(\hookrightarrow\) eigen value

\[ J_3|\downarrow\rangle = -\frac{\hbar}{2}|\downarrow\rangle \] \[ \text{or} \quad J_3\left|\tfrac{1}{2},-\tfrac{1}{2}\right\rangle = -\frac{\hbar}{2}\left|\tfrac{1}{2},-\tfrac{1}{2}\right\rangle \]

\(\left\{\langle\uparrow|\uparrow\rangle = 1 = \langle\downarrow|\downarrow\rangle\ ;\ \langle\uparrow|\downarrow\rangle = 0\right.\)

\[ \boxed{\hat J_3|j,m\rangle = m\hbar\,|j,m\rangle} \]

\(\longrightarrow\) meaning of eigen states

\[ |\uparrow\rangle = \begin{pmatrix}1\\0\end{pmatrix} \qquad |\downarrow\rangle = \begin{pmatrix}0\\1\end{pmatrix} \]

representation in natural basis.

\[ J_{\ell m} = \langle \ell|J_3|m\rangle \] \[ \begin{cases}\langle 1|J_3|1\rangle & \langle 1|J_3|2\rangle\\ \langle 2|J_3|1\rangle & \langle 2|J_3|2\rangle\end{cases} \]

so \(J_3\) should be : \(\nearrow \sigma_3\) (pauli matrix)

\[ J_3 = \frac{\hbar}{2}\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix} \]

[\(|\uparrow\rangle\) & \(|\downarrow\rangle\) states implies magnetic moment which either points along the direction of ext. \(\vec B\) or opposite to ext \(\vec B\).]

such that

\[ J_3\begin{pmatrix}1\\0\end{pmatrix} = \frac{\hbar}{2}\begin{pmatrix}1\\0\end{pmatrix} \] \[ J_3\begin{pmatrix}0\\1\end{pmatrix} = -\frac{\hbar}{2}\begin{pmatrix}0\\1\end{pmatrix} \]

(If we fix \('j'\) and work in \('2j+1'\) space)

Any operator in a subspace \((2j+1)\) diamensionality which commutes with all the operators in subspace, must be some multiple of unit matrix.

\[ [J^2,\ \vec J\cdot\hat n] = 0 \] \[ J^2 = (\text{no.})\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix} \]

\(J_1, J_2, J_3\) forms lie algebra of rotation group ; & \(J^2\) commutes with all \(J_1, J_2, J_3\)

Any operator in lie algebra which commutes with its generators of Lie algebra is called Casimir operator. So, \(J^2\) is casimir operator.

no. of casimir operators is called the rank of the algebra.

In this case it is one. (which is \(J^2\) only). there is no other operator which commutes with its generators \((J_1, J_2, J_3)\).

[\(J^2 = J_1^2 + J_2^2 + J_3^2\) is only one casimir operator]

(Q) *Can be \(J_1\) and \(J_2\) are also diagonal matrices ?* (just like \(J_3\))

No, if they were diagonal matrix they could have commute with each other. so \(J_1\) & \(J_2\) can't be diagonal.

\[ [J_1,J_2] \neq 0 \neq [J_2,J_3] \neq [J_1,J_3] \neq 0 \]

of course we could have chose different basis such that any one of \(J_1, J_2, J_3\) is diagonal matrix and rest is not.

[Lec 19)

The spin operator for an \(e^-\) \((\vec S)\) has three components \(S_x, S_y, S_z\)

is represented by

\[ \vec S_i = \frac{\hbar}{2}\vec\sigma_i \]

with spin quantum no \((s = \tfrac{1}{2})\)

which obey angular momentum algebra

for \(e^-\), \(s = \tfrac{1}{2}\) \(\Rightarrow\) \(j = \tfrac{1}{2}\)

\(j\) or \(s\)

\[ \left\{\begin{aligned}\left|\tfrac{1}{2},\tfrac{1}{2}\right\rangle &= |\uparrow\rangle = \begin{pmatrix}1\\0\end{pmatrix}\\[4pt] \left|\tfrac{1}{2},-\tfrac{1}{2}\right\rangle &= |\downarrow\rangle = \begin{pmatrix}0\\1\end{pmatrix}\end{aligned}\right. \] \[ \sigma_1 = \begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}, \quad \sigma_2 = \begin{pmatrix}0 & -i\\ i & 0\end{pmatrix}, \quad \sigma_3 = \begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix} \qquad \sigma_i^2 = I \]

[Notice only one operator is diagonal matrix]

\(\downarrow\) (representation in natural basis)

\[ \langle\uparrow|\uparrow\rangle = \langle\downarrow|\downarrow\rangle = 1 \] \[ \langle\uparrow|\downarrow\rangle = \langle\downarrow|\uparrow\rangle = 0 \]

\(\}\) orthonormality.

"\(\vec S\) is simply \(J\) for \(j = \tfrac{1}{2}\) (\(e^-\), \(p\), \(n\), \(\nu_e\))

\[ \vec J_{j=1/2} = \vec S. \text{"} \]

"If we took case of a stone \(m = 1\) kg performing circular motion with radius \(r = 1\) m, has angular momentum

\[ L = I\omega = mr^2\omega = \omega \]

but this \('L'\) should be \(\sim \hbar\,l\)

\(\hookrightarrow\) some angular momentum quantum no.

\[ \hbar\,l \sim 1 \]

(for stone)

\[ (l \sim 10^{34}) \quad \text{so as } l \text{ is very enormous } (l = j). \]

The discreteness of angular momentum completely washed out. (as values of \(m\) varies from \(-j\) to \(+j\) in steps of 1.) That is why we see states as contineous rather then discrete. \((-10^{+34},\ -10^{34}+1,\ ---\ ,\ 10^{34}-1,\ 10^{34})\).

(Q) So question arises how to detect such small \(s = 1/2\)

"How do we detect / find the spin angular momentum". ?

The spin angular momentum also implies magnetic moment for the \(e^-\), \((\mu_e)\), and that \(\mu_e\) (magnetic moment) would couple to applied magnetic fields, and then we can menupulate the spin of \(e^-\) using the magnetic field.

\[ \vec\mu_e = \frac{g_e\,e}{2m_e}\,\vec S \]

[remember \(\vec S\) is simply the \(J\) for \(j = 1/2\) / \(s = 1/2\)]

\[ = \frac{-(2)|e|}{2m_e}\cdot\frac{\hbar}{2}\vec\sigma \] \[ = -\left(\frac{|e|\hbar}{2m_e}\right)\sigma \]

\(\longrightarrow\) Bohr magneton \((\mu_B)\)

\[ \mu_B = \frac{|e|\hbar}{2m_e} \]

is natural unit of magnetic moment for charge \(|e|\) and mass \(m_e\).

\[ \boxed{\vec\mu_e = -\mu_B\,\vec\sigma} \]

If we switch on \(\vec B\). There will be extra term in \(\hat H\) of charge particle

\(\left(\hat H = \frac{(p-eA)^2}{2m} + e\phi + \hat H_{\text{magnetic}}\right)\)

\[ H_{\text{magnetic}} = -\vec\mu_e\cdot\vec B \]

\(\hookrightarrow\) P.E of magnetic moment in applied \(\vec B\).

\[ H_{\text{magnetic}} = +\mu_B\,\vec\sigma\cdot\vec B \]

(our expectations :-)

Classically) If we have a magnetic diapole moment placed in an external magnetic field at non zero angle \(\theta\). then the diapole moment vector precesses around the \(\vec B_{\text{ext}}\).

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  \draw[->] (0,0) -- (0,2.4) node[above] {$B$};
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["precession frequency is called larmor frequency. If the \(\vec B\) is sufficiently non uniform / strong then we know that diapole moment also experience force along with torque. And such \(\vec B_{\text{ext}}\) if strong can seperate out magnetic diapole moment states. It is what done in Stern Gerlach exp. which is root of Quantum computing & quantum information"]

\[ \left(\frac{dL}{dt} = \tau\ ;\ \frac{d\mu}{dt} = \kappa\tau\right) \]

(classical magnetic moment)

\[ \frac{d\mu_e}{dt} = \kappa(\tau) = \kappa(\mu_e\times B) \]

\(\hookrightarrow\) constant

\(L \propto \mu_e\) so,

\[ \frac{d\mu_e}{dt} = (\kappa)\,\mu_e\times B \]

\(\hookrightarrow\) provides larmor frequency

From Ernfest theorem, the expectation value of any operator obey classical equation. So

\[ \frac{d\langle\mu\rangle}{dt} = (\ )\cdot\langle\mu_e\rangle\times B \]

\(\}\) should behave quantum mechanically

let us ask what happens to spin states in const \(\vec B\). as time passes. :-

Since \(|\uparrow\rangle\) and \(|\downarrow\rangle\) form basis, we can write \(|\Psi(t)\rangle\) as linear combination of those basis. (\(|\Psi(t)\rangle\) = angular momentum state).

\[ |\Psi(t)\rangle = a(t)|\uparrow\rangle + b(t)|\downarrow\rangle \]

\(\left\{\langle\uparrow|\downarrow\rangle = 0\ ;\ \langle\uparrow|\uparrow\rangle = 1\ ;\ \langle\downarrow|\downarrow\rangle = 1\right.\)

\[ |a(t)|^2 + |b(t)|^2 = 1 \qquad \text{for all time} \]

let us look at simplest system.

at \(t = 0\) we only have \(|\uparrow\rangle\) state (or prepare \(|\uparrow\rangle\) states)

means,

\[ \begin{cases} a(0) = 1\\ b(0) = 0\end{cases} \] \[ |\psi(0)\rangle = |\uparrow\rangle \]

and what happens if we switch on the hamiltonian.

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

\(\left(\sigma\cdot B = \sigma_1 B_x + \sigma_2 B_2 + \sigma_3 B_z\right)\)

\[ H_{\text{magnetic}} = \mu_B\,\sigma_3 B \]

\((\vec\sigma\cdot\vec B = \sigma_3\cdot B)\).

\[ |\psi(t)\rangle = e^{\frac{-iHt}{\hbar}}|\psi(0)\rangle \] \[ = e^{\frac{-iB\mu_B\sigma_3 t}{\hbar}}|\psi(0)\rangle \] \[ = e^{\frac{-iB\mu_B\sigma_3 t}{\hbar}}|\uparrow\rangle \] \[ = e^{\frac{-iB\mu_B t}{\hbar}}|\uparrow\rangle \]

[\(e^{-i\alpha\sigma_3}\begin{pmatrix}1\\0\end{pmatrix} = e^{-i\alpha}\begin{pmatrix}1\\0\end{pmatrix}\)]

[\(\sigma_3|\uparrow\rangle = \begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix} = \begin{pmatrix}1\\0\end{pmatrix}\)]

[\(J_3|S_3\rangle = \frac{\hbar}{2}\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}\) (sign)]

[so nothing happens upon applying ext. \(\vec B\). \(H|\Psi(0)\rangle = E|\Psi(0)\rangle\)]

Even if we start from arbitrary state \(|\downarrow\rangle\)

\[ |\psi(t)\rangle = e^{\frac{+iB\mu_B t}{\hbar}}|\downarrow\rangle \]

[\(e^{-i\alpha\sigma_3}\begin{pmatrix}0\\1\end{pmatrix} = e^{i\alpha}\begin{pmatrix}0\\1\end{pmatrix}\) ; \(\left(\alpha = \frac{B\mu_B t}{\hbar}\right)\)]

[\(|\downarrow\rangle\) remains in same state with eigen value \(\frac{\hbar}{2}\left(\frac{iB\mu_B t}{\hbar}\right)\) [?]]

Interesting things occur if we start with arbitrary wave function

\[ |\psi(0)\rangle = a(0)|\uparrow\rangle + b(0)|\downarrow\rangle \]

for \(B = B\,\hat e_x\) let \(B = b_x\hat e_x\) (in \(x\) direction only)

\[ |\psi(t)\rangle = e^{\frac{-iHt}{\hbar}}|\psi(0)\rangle \] \[ H = \mu_B\,\vec\sigma\cdot\vec B = \mu_B(\sigma_1)\cdot B = +(B)\frac{|e|\hbar}{2m}\cdot\sigma_1 \] \[ |\psi(t)\rangle = e^{-\frac{i}{\hbar}\frac{\mu_B B|e|\hbar}{2m}\sigma_1 t}|\psi(0)\rangle \] \[ = e^{-\frac{i}{2}\frac{|e|B}{m}\sigma_1 t}|\psi(0)\rangle \]

\(\hookrightarrow \omega_c\) (cyclotron frequency)

\[ = e^{-\frac{i\omega_c t}{2}\sigma_1}|\psi(0)\rangle \]

let, if \(|\psi(0)\rangle = |\uparrow\rangle\) \(\left(\begin{aligned}b(0) &= 0\\ a(0) &= 1\end{aligned}\right)\)

So,

\[ |\psi(t)\rangle = e^{-\frac{i\omega_c t}{2}\sigma_1}|\uparrow\rangle \]

here, \(e^{-\frac{i\omega_c t}{2}\sigma_1}\) is not a phase factor as \(|\uparrow\rangle\) is not eigen state of \(\sigma_1\)

also let's write \(e^{\frac{-i\omega_c t}{2}\sigma_1}\) in matrix form.

[\(e^{\alpha x} = 1 + \frac{(\alpha x)}{1!} + \frac{(\alpha x)^2}{2!} + \cdots\)]

\[ e^{-i\vec a\cdot\vec\sigma} = (\cos a)\,I - i\left(\frac{\vec a\cdot\vec\sigma}{|a|}\right)\sin(a) \]

\(\left\{a = \frac{\omega_c t}{2}\ ;\ \frac{\vec a\cdot\vec\sigma}{|a|} = \sigma_1\right.\)

\(\hookrightarrow\) generalization of \(e^{i\theta} = \cos\theta + i\sin\theta\)

\[ |\Psi(t)\rangle = \left[\left(\cos\frac{\omega_c t}{2}\right)I - i\sigma_1\left(\sin\frac{\omega_c t}{2}\right)\right]|\uparrow\rangle \] \[ I = \begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}, \qquad \sigma_1 = \begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix} \] \[ = \begin{pmatrix}\cos\frac{\omega_c t}{2} & -i\sin\frac{\omega_c t}{2}\\[4pt] -i\sin\frac{\omega_c t}{2} & \cos\frac{\omega_c t}{2}\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix} \] \[ = \begin{pmatrix}\cos\frac{\omega_c t}{2}\\[4pt] -i\sin\frac{\omega_c t}{2}\end{pmatrix} = \cos\frac{\omega_c t}{2}|\uparrow\rangle + \left(-i\sin\frac{\omega_c t}{2}\right)|\downarrow\rangle \] \[ \text{or} \quad \left(\cos\frac{\omega_c t}{2}\begin{pmatrix}1\\0\end{pmatrix} + \left(-i\sin\frac{\omega_c t}{2}\right)\begin{pmatrix}0\\1\end{pmatrix}\right) \] \[ \begin{cases} a(t) = \cos\dfrac{\omega_c t}{2}\\[8pt] b(t) = -i\sin\dfrac{\omega_c t}{2}\end{cases} \qquad \left(|a(t)|^2 + |b(t)|^2 = 1\right) \]

also, we can see that \(|\Psi(t)\rangle\) remains normalized.

Q.) Now, what is significance of these coefficients ?

\(|a(t)|^2\) = probability that at time \('t'\) the state is \(|\uparrow\rangle\)

\(|b(t)|^2\) = Probability that the state is down \(|\downarrow\rangle\).

Such that \(|a|^2 + |b|^2 = 1\)

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\(\left\{|b(t)|^2 = 1\ ;\ \frac{\omega_c t}{2} = n\pi\right.\)

So, quantum mechanically, this \(H\) \((B = B\hat e_x)\) plays the role of spin flip operator. It flips the spin from up to down and down to up repeatedly.

\[ |\uparrow\rangle \longrightarrow |\downarrow\rangle \] \[ |\downarrow\rangle \longrightarrow |\uparrow\rangle \]

so we can't say unlike classically, to that the system is in \(|\uparrow\rangle\) state or \(|\downarrow\rangle\) state. It exist in superposition of states. except those instances where \(a(0) = 0\) or \(b(0) = 0\).

for only \(\left(t = \frac{2n\pi}{\omega_c}\right)\) system is in \(|\downarrow\rangle\) state, at for \(t = \frac{2\pi(n+\frac{1}{2})}{\omega_c}\) system is in \(|\uparrow\rangle\) state).

suppose we have eigen state of \(\sigma_1\) \(|\rightarrow\rangle\)

\[ |\rightarrow\rangle = a|\uparrow\rangle + b|\downarrow\rangle \]

(thus any spin-state in this hilbert space can be written as superposition of \(|\uparrow\rangle\) and \(|\downarrow\rangle\) states.)

For the case \(B = B\,e_z\)

\[ H = \mu_B\,\vec\sigma\cdot\vec B = \mu_B\,\sigma_3 B \qquad \Rightarrow \qquad \begin{aligned}H|\uparrow\rangle &= E_1|\uparrow\rangle\\ H|\downarrow\rangle &= E_2|\downarrow\rangle\end{aligned} \]

the eigen states

since \(\sigma_3\) has two eigen values \(+1\), \(-1\).

\[ E_1 = \mu_B B \qquad E_2 = -\mu_B B \]

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"any state can be written as superposition, \(\exists\) infinite no. of superpositions possible. (\(a, b \in\) any number).

we start with two level system but we can have \(\infty\)-no of possibilities, this is called qubit."

so it is superposition which leads to all the interesting phenomena.

Suppose we have such 2 \(e^-\) each with two level systems. we look at spin states of 2 electrons together.

[we are here not interested in coulamb interaction, other d.of, not interested in P.E, not interested in K.E due to motion]

we will look only at spin-states of system.

what is hilbert space looks like

\[ \mathcal{H}_1 \qquad \{|\uparrow\rangle_1,\ |\downarrow\rangle_1\} \] \[ \mathcal{H}_2 \qquad \{|\uparrow\rangle_2,\ |\downarrow\rangle_2\} \]

what are possible states of full / combined system of \(e^-\)s.

There are four possible states to start with.

\[ |\uparrow\uparrow\rangle,\ |\uparrow\downarrow\rangle,\ |\downarrow\uparrow\rangle,\ |\downarrow\downarrow\rangle. \]

\(\triangleright\)) How to add angular momenta in Q.M ?

let \(J_1\) and \(J_2\) be two total angular momentum operators for \(e^-_1\) and \(e^-_2\)

components of \(J_1\) commutes with components of \(J_2\)

also, \([J_1, J_2] = 0\). i.e. \([J_{1i}, J_{2j}] = 0\)

[for \(e_1\): \(J_1 \to J_1^2, J_{11}, J_{12}, J_{13}\) ; \(e_2\): \(J_2 \to J_2^2, J_{21}, J_{22}, J_{23}\)]

\[ \vec J_1 + \vec J_2 = \vec J \]

\(\longrightarrow\) Total angular momentum.

So let us ask

(Q) what is/are possible eigen values of \(J/J^2\) given the eigen values of \(J_1/J_1^2\) or \(J_2/J_2^2\).

Theory of add\(^{\text{n}}\) of angular momenta :-

\[ J^2_{\text{e.value}} = j(j+1)\hbar^2 \] \[ J_{1\,\text{e.v.}} = j_1(j_1+1)\hbar^2 \] \[ J_{2\,\text{e.v}} = j_2(j_2+1)\hbar^2 \]

Then the rule is, for total quantum no \('j'\) is :-

\[ j = |j_1-j_2| \ \text{to}\ (j_1+j_2) \]

\(\hookrightarrow\) quantum no of \(\vec J\).

[\(J\) is an operator so it does not add up like vectors, operators have their own peculiarity.]

But, any component adds up linearly

for \(\left(\vec J_1 + \vec J_2 = \vec J\right)\)

(example :-)

\[ J_{13} = J_1^z = m_1\hbar \] \[ J_{23} = J_2^z = m_2\hbar \]

[\(m_1 = -j_1,\ -j_1+1,\ ---\ j_1-1,\ j_1\) ; \(m_2 = -j_2,\ -j_2+1,\ ---\ j_2-1,\ j_2\)]

\[ J^z = (m_1+m_2)\hbar \qquad m_1+m_2 = -j,\ -j+1\ --,\ j-1,\ j \]

\(\rightarrow\) *what if we add spins of two \(e^-\) :-*

what happens if we solve for \(j_1 = \frac{1}{2}\) & \(j_2 = \frac{1}{2}\) (\(j = \frac{1}{2}\) for both \(e^-\))

so \(J\) is represented as \(\vec S\).

\[ \vec S = \vec S_1 + \vec S_2 = \vec S \]

possible values for total spin quantum no. \(S = ?\))

\(\because\) \(S_1 = \frac{1}{2}\), \(S_2 = \frac{1}{2}\)

\[ s = |j_1-j_2| \ \text{to}\ j_1+j_2 = 0 \ \text{to}\ 1 \ \text{in steps of } 1 \] \[ (s = 0 \ \text{or}\ 1) \]

\('s'\) is total spin quantum no.

[\(s_1 = \frac{1}{2}\) ; \(m_1 = -j\) to \(+j\) ; \(s_2 = \frac{1}{2}\) ; \(m_2 = -j\) to \(j\) ; \(m_2 = -\frac{1}{2}, \frac{1}{2}\)]

[so \('s'\) runs from \(|s_1-s_2|\) to \((s_1+s_2)\) \(\Rightarrow\) 0 to 1 ; \((|\frac{1}{2}-\frac{1}{2}|\) to \(|\frac{1}{2}+\frac{1}{2}|)\)]

\(\Rightarrow\) So, if we take two fermions we put them together, we now create a boson (integral spin quantum no \((s)\) 0, 1, 2 ...).

"This is how super conductivity occurs"

Q). what are the possible states?

\[ |\uparrow\rangle_1 \otimes |\uparrow\rangle_2 = |\uparrow\uparrow\rangle \]

\(\longrightarrow\) representation.

[where \(|\uparrow\rangle_1 = |\frac{1}{2},\frac{1}{2}\rangle_1 = \begin{pmatrix}1\\0\end{pmatrix}\) ; \(|\downarrow\rangle_1 = |\frac{1}{2},-\frac{1}{2}\rangle_1 = \begin{pmatrix}0\\1\end{pmatrix}\)]

So we have 4 - states.

\[ |\uparrow\uparrow\rangle,\ |\uparrow\downarrow\rangle,\ |\downarrow\uparrow\rangle,\ |\downarrow\downarrow\rangle. \]

[rough work in a box on this page :

\[ 2\sin^2\theta \Rightarrow 1-\cos2\theta \] \[ 1 - \cos(\theta+\theta) \] \[ 1 - [\cos^2\theta - \sin^2\theta] = 1 - (1-\sin^2\theta - \sin^2\theta) = 1-1+2\sin^2\theta \]

]

\(\Rightarrow\) So they form a basis in full Hilbert space.

So every spin-state (of combined system of 2 \(e^-\)) can be written as superposition of these above mentioned four states.

\[ S = 0 \quad \text{or} \quad 1 \] \[ (m_1+m_2) = S^z = 0 \quad \text{or} \quad -1,\ 0,\ 1 \]

\[ S = 0, \qquad S^z = 0 \quad \} \ \text{singlet state} \]

(\(-s\) to \(+s\) in steps of 1)

\[ S = 1, \qquad S^z = \begin{matrix}-1\\ 0\\ 1\end{matrix} \quad \} \ \text{triplet state} \]

(\(m_s = -s\) to \(+s\) in steps of 1)

(let us write states) :-

\[ |s, s_z\rangle \quad \text{or} \quad |j,m\rangle \]

spin singlet state

\[ |0,0\rangle = \big(|\uparrow\downarrow\rangle - |\downarrow\uparrow\rangle\big)/\sqrt2 \]

\(\longrightarrow\) anti-symmetric under exchange of 2 \(e^-\).

spin triplet states are symmetric under exchange

\[ |1,1\rangle = |\uparrow\uparrow\rangle \qquad \text{symmetric} \] \[ |1,-1\rangle = |\downarrow\downarrow\rangle \qquad \text{symmetric} \] \[ |1,0\rangle = \big(|\uparrow\downarrow\rangle + |\downarrow\uparrow\rangle\big)/\sqrt2 \quad (\text{normalized state}) \]

\(\}\) symmetric under the exchange of 2 electrons.

\[ \langle s_1, s_1^z|s_2, s_2^z\rangle = \delta_{s_1s_2}\,\delta_{s_1^z s_2^z} \]

\(\}\) orthogonal states.

If we have two fermions which are absolutly identical then no matter what states these fermions are in, total state must be anti-symmetric under exchange.

But this is only spin part of wave function. There is also a special part, if \(e^-\) are moving around constitutes \(\psi(r,t)\) then Total wave function \(\Psi = \psi(r,t) * \psi(\text{spin})\), and that \(|\Psi\rangle\) should be anti-symmetric.

So, in singlet state special part of wavefunction must be symmetric and in triplet states special part of state must be anti-symmetric.

\(|1,0\rangle\) should be symmetric under exchange so, it would be superposition of \(|\uparrow\downarrow\rangle\) and \(|\downarrow\uparrow\rangle\) state.

all states should be orthogonal with each other so

\[ |0,0\rangle = \big(|\uparrow\downarrow\rangle - |\downarrow\uparrow\rangle\big)/\sqrt2 \]

after exchange

\[ |0,0\rangle = \big(|\downarrow\uparrow\rangle - |\uparrow\downarrow\rangle\big)\sqrt2 \]

anti-symmetric under exchange

overview | << prev

Contents

Lecture 18

Charged particle in a magnetic field -- contd.

Angular momentum states from two oscillators

Direct product of Hilbert spaces

For a given 'j'

Components of 'J' transform like a vector under rotation :-

Rotation transformation of states and operators :- (Rotation representation)

Transformation of operators :-

j = 1/2 (very important case)

schur's lemma theorem :-

Lecture 19

Spin operator for an electron

How do we detect / find the spin angular momentum ?

Quantum mechanically :-

Spin states in a constant magnetic field

Entanglement

Theory of addition of angular momenta :-

pauli exclusion principle :-