• #Lecture 10
  • #Ground state of H. (1S) (spherically symmetric).
  • #Particle in Finite well (potential well) :
  • #Attractive delta -- fun. potential :-
  • #WHAT IF we have two such delta -- potential wells

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[The notebook does not carry a separate "Lec 9" heading -- the material for it runs on inside the pages transcribed on the Lec 7-8 page. This page begins at the "lec 10" heading in the scans.]

\(\rightarrow\) "symmetry implies degeneracy".

\(\rightarrow\) If a problem is integrable in classical mechanics does not imply that same problem will be integrable in quantum mechanically.

ex.

\[ H(x,p) = \frac{p^2}{2m} + \frac{1}{2}m\omega^2x^2 + \lambda x^4 \]

classically Integrable (1 C.O.M requred for 1.D.of \(= H\))

But quantum mechanically we can't find exect eigen values of \(H\). we can find it with arbitary accuracy using some perturbation method.

Nature of symmetry in Q.M has very -- very deep profound implications as opposed to classical mechanics.

ex In H- atom, we assumed that \(e^-\) orbit around nucleous.

\[ L \neq 0 \]

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according to bohr theory

\[ \boxed{L \neq 0} \ \in \ L = mvr = n\hbar \qquad (n = 1,2,3) \]

\(n = \) angular quantum no.

\[ \frac{mv^2}{r} = \frac{Ze^2}{r^2} \] \[ E_n \propto \frac{-1}{n^2} \qquad (n = 1,2,3\cdots) \] \[ n = 1,\ l = 0,\ m = 0 \]

eigen fun. can be written as

\[ \left\{\phi_{100}(r,\theta,\phi) \propto e^{-r/a_0}\right. \] \[ l = 0 \quad \Rightarrow \quad L = 0 \]

So, How is that quantum theory suggests that angular momentum of \(e^-\) in ground state is zero but classical mechanics / intution says its not zero.

The \(e^-\) does not orbit with radius \(a_0\). we can find \(e^-\) anywhere b/w \(0\) to \(\infty\) \((0 < r < \infty)\), and probability of finding it b/w \(r\), \(r+dr\) is

\[ P(r)\,dr = |\phi_{100}(r)|^2\ dV \] \[ = e^{-\frac{2r}{a_0}}\,(4\pi r^2)\,dr \]

\(\left\{\phi_{100}(r,\theta,\phi) = \phi(r)\right.\) (as no dependence on \(\theta, \phi\))

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  \draw[thick] (0,0) .. controls (0.7,0.2) and (0.9,2.3) .. (1.3,2.3)
      .. controls (1.8,2.3) and (2.2,0.5) .. (3.2,0.2)
      .. controls (3.9,0.08) and (4.3,0.04) .. (4.9,0.03);
  \draw[dashed] (1.3,0) -- (1.3,2.3);
  \node[below] at (1.3,-0.05) {$a_0$};
  \foreach \x in {0.75,0.85,0.95,1.05} {\draw (\x,0) -- (\x,{2.3*(\x-0.0)/1.3});}
\end{tikzpicture}

\(\{P(r)\) is maximum at \(r = a_0\}\)

still It does not answer How \(L = 0\).

\(\{\)classically, This could only happen if \(e^-\) passes through center\(\}\)

How it is that the \(e^-\) predominently at \(r = a_0\) has zero angular momentum.

Classically different solutions which differ in direction of \(\vec L\) are related by rotation transformations. we can go from one solution (rotation is in \(xy\) plane) to another solution (rotation is in \(yz\) plane) by rotation of coordinate axis.

So the group of transformations (rotation group) under which the Hamiltonian is invarient. Each element of that group of rotations takes you from one possible solution to another possible solution.

And once you fix coordinate system and specify the initial conditions, the orbit is fixed.

Now Q.M says superposition is valid. So any solution is superposition of all solutions. It says if two solutions are related by a symmetry transformation the general solution is superposition of two solutions.

Now its easy to see that if take the orbit and put it in all possible planes and add all \(\vec L\). we get \(\vec L = 0\).

Thats the reason in Q.M we can still sustain a zero \(\vec L\) solution; even though the \(e^-\) has overwhelmingly probability to be at non-zero distance from origin. This is very profound & deep aspect of Q.M.

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      .. controls (2.2,0.05) and (2.8,0.03) .. (3.2,0.02);
  \node[below left] at (0,-0.03) {$0$};
  \node[below right] at (1.7,-0.03) {$L$};
  \node[left] at (-0.1,-2.2) {$-V_0$};
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what are possible values of energy states?

\[ \phi_n'' + k_n^2\phi_n - V(x)\,\phi_n = 0 \]

\(\hookrightarrow\) finite discountinity. compensated by finite discontinuity of \(\phi_n''(x)\).

There is a finite probability that a particle tunnel.

Now,

lets make the problem simpler by assuming that width of the well goes to zero & simultaneously the \(V(x) \to \infty\) such that product is finite. we would like to see bound state (i.e. potential can retain the particle).

lets assume \(V(x)\) is a negative delta function.

\[ \frac{-\hbar^2}{2m}\phi''(x) + V(x)\,\phi(x) = E\,\phi(x) \]

\(\left\{\phi(x) = \right.\) eigen state of \(H\). ; \(E = \) eigen values of \(H\)

lets put \(\delta\)-fun at \(x = 0\), so \(V(x) = -\lambda\,\delta(x)\)

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  \draw[->] (-2.2,0) -- (2.2,0) node[right] {$x$};
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  \node[above right] at (0.05,0.05) {$0$};
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\[ [\lambda] = ML^3T^{-2} \]

because, \([\delta(x)] = L^{-1}\)

\(E\) must depend on \(\lambda, \hbar, m\)

\[ E \propto \lambda^{\alpha}\,\hbar^{\beta}\,m^{\gamma} \] \[ [ML^2T^{-2}] \propto [ML^3T^{-2}]^{\alpha}\,[ML^2T^{-1}]^{\beta}\,m^{\gamma} \] \[ [M^1L^2T^{-2}] = k\ M^{\alpha+\beta+\gamma}\,L^{3\alpha+2\beta}\,T^{-\beta-2\alpha} \] \[ \begin{aligned} \alpha + \beta + \gamma &= 1\\ 2\beta + 3\alpha &= 2\\ -2\alpha - \beta &= -2 \end{aligned} \qquad \Rightarrow \qquad \begin{aligned} \beta &= -2\\ \gamma &= +1\\ \alpha &= 2 \end{aligned} \] \[ \therefore \quad E \propto \frac{m\lambda^2}{\hbar^2} \]

For \(x > 0\)) \(\Rightarrow\) \(V(x) = 0\)

\[ \left(\frac{-\hbar^2}{2m}\phi''(x) = E\,\phi(x)\right) \quad \Rightarrow \quad \phi''(x) = -\frac{2mE}{\hbar^2}\phi(x) \]

\(\left\{E < 0\right.\) for bound states so better to write

\[ \phi''(x) = +\frac{2m|E|}{\hbar^2}\phi(x) \] \[ \phi''(x) = \frac{2m|E|}{\hbar^2}\phi(x) \] \[ \phi''(x) - k^2\phi(x) = 0 \qquad \left(k = \sqrt{\frac{2m|E|}{\hbar^2}}\right) \] \[ (D^2 - k^2)\phi(x) = 0 \qquad (D = \pm k) \qquad (D = d/dx) \] \[ \phi(x) = A e^{kx} + B e^{-kx} \] \[ \lim_{x\to\infty}\|\phi(x)\| \longrightarrow \text{finite} \quad \text{so} \quad A = 0 \] \[ \phi(x) = B e^{-kx} \]

for \(x < 0\))

\[ \phi(x) = C e^{kx} + D e^{-kx} \]

normalizability \(\Rightarrow\) \(D = 0\) \(\left[\lim_{x\to-\infty}\|\phi(x)\| \to \text{finite}\right.\)

\[ \phi(x) = C e^{kx} \]

Since \(\phi(x)\) is contineous.

\[ \phi(x)\Big|_{x\to0^-} = \phi(x)\Big|_{x=0} = \phi(x)\Big|_{x\to0^+} \] \[ \Rightarrow \quad \boxed{C = B} \]

So,

\[ \phi(x) = B e^{kx} \]

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  \node[anchor=west] at (0.9,1.95) {only one state.};
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\[ \phi'(x(0^-)) \neq \phi'(x(0^+)) \] \[ \Rightarrow \quad \phi''(x(0)) \ \text{ has infinite discontinuity.} \]

for \(x = 0\))

\[ \phi''(x) + \frac{2m\lambda}{\hbar^2}\delta(x)\,\phi(x) = k^2\phi(x) \qquad \left\{k^2 = \frac{-2mE}{\hbar^2}\right\} \]

Integrating from \(x = -\epsilon\) to \(x = +\epsilon\) & let \(\epsilon \to 0\)

\[ \int_{-\epsilon}^{\epsilon} dx\left\{\phi''(x) + \frac{2m\lambda}{\hbar^2}\delta(x)\,\phi(x)\right\} = \int_{-\epsilon}^{\epsilon} k^2\phi(x)\,dx \] \[ \left(\frac{d\phi(x)}{dx}\right)_{0^-}^{0^+} + \frac{2m\lambda}{\hbar^2}\phi(0) = k^2\left(\int_0^{\epsilon} B e^{-kx}dx + \int_{-\epsilon}^{0} B e^{kx}dx\right) = 0 \] \[ \left(\frac{d\phi}{dx}\bigg|_{x\to0^+} - \frac{d\phi}{dx}\bigg|_{x\to0^-}\right) + \frac{2m\lambda}{\hbar^2}B = 0 \qquad \big(\phi(0) = B\big) \] \[ \left\{ \begin{aligned} x > 0, \ \phi(x) &= B e^{-kx}, && \phi'(x) = -kB e^{-kx}\\ x < 0, \ \phi(x) &= B e^{kx}, && \phi'(x) = +kB e^{+kx} \end{aligned} \right. \] \[ -2kB + \frac{2m\lambda}{\hbar^2}B = 0 \] \[ B\left(2k - \frac{2m\lambda}{\hbar^2}\right) = 0 \] \[ B \neq 0, \qquad k = \frac{m\lambda}{\hbar^2} \ ; \qquad k^2 = \frac{2m|E|}{\hbar^2} = \frac{m^2\lambda^2}{\hbar^4} \] \[ \Rightarrow \quad |E| = \frac{m\lambda^2}{2\hbar^2} \]

So, for bound state \(E < 0\)

Note :--

\[ \boxed{E = \frac{-m\lambda^2}{2\hbar^2}} \]

(only one bound state) as no quantum no exists.

So a single point of \(\infty\)-discontinuity in potential supports one B. state

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  \node[above right] at (0.05,0.05) {$0$};
  \node[anchor=north] at (-2.3,-0.6) {$Ae^{kx}$};
  \node[anchor=north] at (0,-0.6) {$Be^{-kx} + Ce^{kx}$};
  \node[anchor=north] at (2.3,-0.6) {$Be^{-kx}$};
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\[ V(x) = -\lambda\left[\delta(x-a) + \delta(x+a)\right] \]

\(\{5\) eq\(^{\text{n}}\) unknown \((A, B, C, D, k)\}\)

\(\phi(-a)\) is contineous & differentiable \(\Rightarrow\) 2 equations

\(\phi(+a)\) \(\Rightarrow\) 2 equations ; (normalization)

For two states which are independent

\[ H \sim \begin{pmatrix} E & 0\\ 0 & E\end{pmatrix} \qquad \text{in some vector space.} \]

now, if we switch on the coupling b/w the two

\[ H \sim \begin{pmatrix} E & \varepsilon\\ \varepsilon & E\end{pmatrix} \]

eigen values \(\Rightarrow\) \(E_{1,2} = E \pm \varepsilon\)

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  \draw[->] (0,-1.9) -- (0,2.2) node[above] {$\phi(x)$};
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\end{tikzpicture}

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