Extra Credit 6

Last time

1-D Wave equation: \(\frac{\partial^2 u}{\partial x^2} = \frac{1}{v^2}\frac{\partial^2 u}{\partial t^2} \)

* \(u(x,t)\): displacements as function of \(x,t\)

* 2^nd-order: solution is sum of 2 linearly independent functions

* general solution by separation of variables

* boundary conditions give specific physical system

* “normal modes” — octaves, nodes, Fourier series, “quantization”

* The pluck: superposition of normal modes, time-evolving wavepacket

Problem Set #2: time evolution of plucked system

* More complicated for separation of 2-D rectangular drum. Two separation constants.

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Today: Begin Quantum Mechanics

The 1-D Schrödinger equation is very similar to the 1-D wave equation. It is a postulate. Cannot be derived, but it is motivated in Chapter 3 of McQuarrie. You can only determine whether it fails to reproduce experimental observations. This is one of the weirdnesses of Quantum Mechanics.

We are always trying to break things (story about the Exploratorium in San Francisco)

1. Operators: Tells us to do something to the function its right

Examples: Âf = g , operator denoted by  (“^” hat)

* take derivative \(\begin{cases}
\frac{d}{dx}f(x) = f'(x) \\
\frac{d}{dx}(af(x)+bg(x)) = \underbrace{af'(x)+bg'(x)}_{\text{linear operator}} \end{cases}\)

* integrate \(\int dx (af(x) + bg(x)) = \underbrace{a \int dxf + b \int dxg}_{\text{linear operator}} \)

* take square root \(\sqrt{(af(x) + bg(x))} = \underbrace{ {[af(x)+bg(x)]}^\frac{1}{2}} _{\text{NOT linear operator}} \)

We are interested in linear operators in Quantum Mechanics. (part of McQuarrie’s postulate #2)

2. Eigenvalue Equations

Âf(x) = af(x)

a is an eigenvalue of the operator  .
f(x) is a specific eigenfunction that “belongs” to the eigenvalue a

more explicit notation Â\(f_n(x) = a_n f_n(x) \)

Operator An Eigenfunction Its eigenvalue
\(\hat{A} = \frac{d}{dx}\) \(e^{ax}\) a
\(\hat{B} = \frac{d^2}{dx^2}\) \(sin bx + cos bx\) \(-b^2\)
\(\hat{C} = x\frac{d}{dx}\) \(ax^2\) \(n\)

3. Important Operators in Quantum Mechanics (part of McQuarrie's postulate #2)

For every physical quantity there is a linear operator

coordinate \(\hat{x} = x\)

momentum \(\hat{p_x} = -i \hbar \frac{\partial}{\partial x}\) (at first glance, this seems surprising. Why?)

kinetic energy \(\hat{T} = \frac{\hat{p^2}}{2m} = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} \)

potential energy \(\hat{V}(x) = V(x)\)

energy \(\hat{H} = \hat{T} + \hat{V} = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + V(x)\) (the “Hamiltonian”)

Note that these choices for \(\hat{x}\) and \(\hat{p}\) are dimensionally correct, but their “truthiness” is based on whether they give the expected results.

4. There is a very important fundamental property that lies behind the uncertainty principle: non-commutation of two operators. \(\hat{x}\hat{p} \neq \hat{p}\hat{x}\)

To find out what this difference between \(\hat{x}\hat{p}\) and \(\hat{p}\hat{x}\) is, apply the commutator, \([ \hat{x}, \hat{p}] \equiv \hat{x}\hat{p} − \hat{p}\hat{x} \), to an arbitrary function.

\( \hat{x}\hat{p}f(x) = x(i \hbar)\frac{df}{dx} = i \hbar x \frac{df}{dx}\)

\(\hat{p}\hat{x}f(x) = (-i \hbar) \frac{d}{dx}(xf) = (-i \hbar) \left[ f + x\frac{df}{dx} \right] \)

\([\hat{x},\hat{p}] \equiv \hat{x}\hat{p}-\hat{p}\hat{x} = i\hbar \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, a \,\, non-zero \,\, "commutator" \)

We will eventually see that this non-commutation is the reason we cannot sharply specify both x and px.

5. Wavefunctions (McQuarrie’s postulate #1)

\(\Psi (x)\): state of the system – contains everything that can be known. Strangely, \(\Psi (x)\) itself can never be directly observed. The central quantity of quantum mechanics is not observable. This should bother you!

* \(\Psi (x)\) is a “probability amplitude” – similar to the amplitude of a wave (can be positive or negative)
* \(\Psi (x)\) can exhibit interference

* probability of finding particle between x, x + dx is \(\Psi^* (x) \Psi(x)dx\) (\(\Psi^*\) is the complex conjugate of \(\Psi\))

6. Average value of observable \(\hat{A}\) in state \(\Psi\)? Expectation value. (part of McQuarrie’s postulate #4)

\(\langle A\rangle = \frac{\int \Psi^* \hat{A}\Psi dx}{\int \Psi^* \Psi dx} \)

Note that the denominator is needed when the wavefunction is not normalized to one.

7. Schrödinger Equation

\( \hat{H} \Psi_n = E_n \Psi_n\) \(\Psi_n\) is an eigenfunction of \(\hat{H}\) that belongs to the specific energy eigenvalue, \(E_n\). (part of McQuarrie’s postulate #5)

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Let’s look at two of the simplest quantum mechanical problems. They are also very important because they appear repeatedly.

1. Free particle: \(V(x) = V_0\) (constant potential)

\[\hat{H} = -\frac{\hbar}{2m} \frac{d^2}{dx^2} + V_0\]

\[\hat{H}\Psi = E \Psi, \,\,\, move \, V_0 \,\,\, to \,\,\, RHS\]

\[-\frac{\hbar}{2m} \frac{d^2}{dx^2} \Psi = (E-V_0) \Psi\]

\[\frac{d^2}{dx^2}\Psi = \frac{-2m(E-V_0}{\hbar^2}\Psi\]

Note that if \(E < V_0\), then on the RHS we need \(\Psi\) multiplied by a negative number. Therefore \(\Psi\) must contain complex exponentials. This is the physically reasonable situation.

But if \(E < V_0\) (how is such a thing possible?), then on the RHS we need \(\Psi\) multiplied by a positive number. \(\Psi\) must contain real exponentials.

\( \begin{align} \left. \begin {array}{cc}
e^{+kx} \,\,\, diverges\,\,\, to \,\,\, \infty \,\,\, as \,\,\, x \longrightarrow+\infty \\
e^{-kx} \,\,\, diverges \,\,\, to \,\,\, \infty \,\,\, as \,\,\, x \longrightarrow -\infty
\end{array} \right\} \end{align} \) unphysical [but useful for |x| finite (tunneling)]

So, when \(E < V_0\) , we find \(\Psi (x)\) by trying \(\Psi = ae^{+ikx} + be^{-ikx}\) (two linearly independent terms)

\[\frac{d^2\Psi}{dx^2}=-k^2(ae^{ikx} + be^{-ikx})\]

\[-\frac{2m(E-V_0}{\hbar^2}=-k^2\]

Solve for E,
\[E_k = \frac{{(\hbar k)}^2}{2m} + V_0\]

You show that \(\begin{cases}
* \Psi = ae^{ikx} \,\,\, is\,\,\, eigenfunction \,\,\, of \,\,\, \hat{p} \\
* with \,\,\, eigenvalue\,\,\, \hbar k \\
* and\,\,\, \langle p \rangle = \hbar k. \end{cases}\)

No quantization of E because k can have any real value.

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Non-Lecture

What is the average value of momentum for \(\Psi = ae^{ikx} + be^{-ikx}\) ?

\[\langle p \rangle = \frac{\int_{-\infty}^{\infty}dx \Psi^*\hat{p} \Psi }{\int_{-\infty}^{\infty}dx \Psi^* \Psi}\swarrow{\text{normalization integral}} \]

\[=\frac{\int_{-\infty}^{\infty} dx (a^*e^{-ikx}+b^*e^{ikx})(-i\hbar)\frac{d}{dx}(ae^{ikx}+be^{-ikx})}{\int_{-\infty}^{\infty} dx (a^*e^{-ikx}+b^*e^{ikx})(ae^{ikx}+be^{-ikx})}\]

\[=\frac{-i\hbar \int_{-\infty}^{\infty} dx (a^*e^{-ikx}+b^*e^{ikx})(ik)(ae^{ikx}+be^{-ikx})}{\int_{-\infty}^{\infty} dx (|a|^2 + |b|^2 + a*be^{-2ikx}+ab*e^{2ikx})}\]

\[=\frac{\hbar k \int_{-\infty}^{\infty} dx (|a|^2-|b|^2 + ab*e^{2ikx} - a*be^{-2ikx})}{\int_{-\infty}^{\infty} dx (|a|^2 + |b|^2 + ab*e^{2ikx}+a*be^{-2ikx})}\]

Integrals from –∞ to +∞ over oscillatory functions like \(e^{\pm i2kx}\) are always equal to zero. Why?

\[\langle p \rangle = \hbar k \frac{|a|^2-|b|^2}{|a|^2+|b|^2}\]

\[ if \,\,\, a = 0 \,\,\,\,\,\,\,\,\, \langle p \rangle = -\hbar k\]

\[if \,\,\, b = 0 \,\,\,\,\,\,\,\,\, \langle p \rangle = +\hbar k\]

\[\frac{|a|^2}{|a|^2+|b|^2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \frac{is\,\,\, fraction \,\,\,of\,\,\, the \,\,\,observations}{in\,\,\, \Psi \,\,\, which\,\,\, have \,\,\,p>0} \]

\[\frac{|b|^2}{|a|^2+|b|^2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \frac{is\,\,\, fraction \,\,\,of\,\,\, the \,\,\,observations}{in\,\,\, \Psi \,\,\, which\,\,\, have \,\,\,p<0} \]

End of Non-Lecture

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Free particle: it is possible to specify momentum sharply, but if we do that we will find that the particle must be delocalized over all space.

For a free particle, \(\Psi^*(x)\Psi(x)dx\) is delocalized over all space. If we have chosen only one value of \(|k|\), \(\Psi^* \Psi\) can be oscillatory, but it must be positive everywhere. Oscillations occur when \(e^{ikx}\) is added to \(e^{-ikx}\) .

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Non-Lecture

\[\Psi = ae^{ikx} + be^{-ikx}\]

\[\Psi^* \Psi = |a|^2 + |b|^2 + 2Re[ab*e^{2ikx}], \,\,\, but \,\,\, if \,\,\, a,b \,\,\, are \,\,\, real \]

\[\Psi^* \Psi = \underbrace{a^2 + b^2}_{\text{constant}} + \underbrace{2ab cos2kx}_{\text{oscillatory}} \]

Note that \(\Psi^* \Psi \geq 0\) everywhere. For \(x \)where \(cos 2bx\) has its maximum negative value, \(cos 2kx = –1\), then \(\Psi^* \Psi = (a-b)^2\) . Thus \(\Psi^* \Psi \geq 0\) for all \(x\) because \((a–b) 2 \geq 0\) if \(a\),\(b\) are real.

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Sometimes it is difficult to understand the quantum mechanical free particle wavefunction (because it is not normalized to 1 over a finite region of space). The particle in a box is the problem that we can most easily understand completely. This is where we begin to become comfortable with some of the mysteries of Quantum Mechanics.

* insight into electronic absorption spectra of conjugated molecules.
* derivation of the ideal gas law in 5.62!
* very easy integrals

Particle in a box, of length a, with infinitely high walls.

“infinite box”

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

\( \begin{align} \left. \begin {array}{cc}
V(x) = 0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, 0 \leq x \leq a \\
V(x) = \infty \,\,\,\,\,\,\,\,\,\, x<0,x>a
\end{array} \right\} \end{align} \) very convenient because \(\int_{-\infty}^{\infty} dx \Psi^* V(x) \Psi = 0\).

(convince yourself of this!)

\(\Psi (x)\) must be continuous everywhere.

\(\Psi (x) = 0\) everywhere outside of box (otherwise \(\int_{-\infty}^{\infty} \Psi^* V \Psi = \infty \)

\(\Psi(0) = \Psi (a) = 0\) at edges of box.

Inside box, this looks like the free particle, which we have already solved.

\(\hat{H} \Psi = E \Psi\) Schrodinger Equation

\( -\frac{\hbar^2}{2m} \frac{d^2}{dx^2} \Psi = E \Psi \) (\(V(x) = 0\) inside the box)

\(\frac{d^2}{dx^2} \Psi = -\frac{2m}{\hbar^2}E\Psi = -k^2 \Psi\)

\(k^2 \equiv \frac{2mE}{\hbar^2}\)

\(\Psi (x) = A sinkx + Bcoskx\) satisfies Schrodinger Equation (it is the general solution)

Apply boundary conditions:

\(\Psi (0) = B = 0\) therefore \(B = 0\)

\(\Psi (a) = Asinka = 0\) therefore \(Asinka = 0\) (quantization!)

\(ka = n \pi \,\,\,\,\,\,\,\,\, k = \frac{n \pi}{a} \,\,\,\,\,\,\,\,\, n \) is an integer

\(\Psi = Asin \frac{n \pi}{a}x\)

\(A^2 \int_{0}^{a} dx sin^2 \frac{n \pi}{a} x = A^2 \frac{a}{2} =1 \)

\(A = \left(\frac{2}{a} \right)^\frac{1}{2}\)

\( \boxed{\Psi_n = \left(\frac{2}{a} \right)^\frac{1}{2} sin\frac{n \pi}{a}x}\) is the complete set of eigenfunctions for a particle in a box. Now find the energies for each value of n.

\[\hat{H} \Psi_n = -\frac{\hbar^2}{2m} \frac{d^2}{dx^2} \left(\frac{2}{a} \right)^\frac{1}{2} sin\frac{n \pi}{a}x\]

\[= + \frac{\hbar^2}{2m}\left(\frac{n \pi}{a} \right)^2 \Psi_n \]

\[= \frac{h^2 n^2}{8ma^2} \Psi_n\]

\[E_n = n^2 \underbrace{\frac{h^2}{8ma^2}}_{\text{\(E_1\)}} = n^2 E_1 \,\,\,\,\,\,\, n = 1,2,3... \,\,\,\,\,\, \text{(never forget this!)} \]

\(n = 0\) means the box is empty
what would a negative value of n mean?

\(n–1\) nodes, nodes are equally spaced. All lobes between nodes have the same shape.

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Summary: Some fundamental mathematical aspects of Quantum Mechanics.

Initial solutions of two-simplest Quantum Mechanical problems.

* Free Particle

* Particle in an infinite 1-D box

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Next Lecture:

1. * more about the particle in 1-D box

* Zero-point energy (this is unexpected)

* \(\Delta x \Delta p\) vs. \(n\) (\(n = 1\) gives minimum uncertainty)

2. particle in 3-D box

* separation of variables

* degeneracy