Measurement: Eigenvalues and Expectation Values

Chem 3240 · Lecture 3.6

Davit Potoyan

How the postulates connect

• Today: the two middle boxes, superposition and measurement
• One copy gives one reading; the probabilities p_n show up over many copies

A vector: build it, then project

\langle u_1 \vert v\rangle = c_1\,\underbrace{\langle u_1 \vert u_1\rangle}_{1} + c_2\,\underbrace{\langle u_1 \vert u_2\rangle}_{0} = c_1

Example: components of a vector

Write \lvert v\rangle = (3, 4) in the tilted axes \lvert u_1\rangle = \tfrac{1}{\sqrt{2}}(1, 1) and \lvert u_2\rangle = \tfrac{1}{\sqrt{2}}(1, -1).

Each coefficient is a projection:

c_1 = \langle u_1 \vert v\rangle = \tfrac{3 + 4}{\sqrt{2}} = \tfrac{7}{\sqrt{2}}, \qquad c_2 = \langle u_2 \vert v\rangle = \tfrac{3 - 4}{\sqrt{2}} = -\tfrac{1}{\sqrt{2}}

\lvert v\rangle = c_1\lvert u_1\rangle + c_2\lvert u_2\rangle, \qquad c_1^2 + c_2^2 = \tfrac{49 + 1}{2} = 25 = \langle v \vert v\rangle

• Perpendicular unit axes: coefficients are projections, and their squares add up to the length squared

A function: the same two steps

• Build: the box states \lvert\psi_n\rangle are the axes, orthonormal, \langle \psi_m \vert \psi_n\rangle = \delta_{mn}

\lvert\psi\rangle = c_1\lvert\psi_1\rangle + c_2\lvert\psi_2\rangle + c_3\lvert\psi_3\rangle + \cdots

• Project: the bra \langle \psi_m\rvert picks out one coefficient, exactly as for the vector

\langle \psi_m \vert \psi\rangle = \sum_n c_n\,\underbrace{\langle \psi_m \vert \psi_n\rangle}_{\delta_{mn}} = c_m

• The only new step: a bracket of two functions is an integral, \langle \psi_m \vert \psi\rangle = \int \psi_m^*\,\psi\,dx

Build a function from box states

• Each box state is perpendicular to all the others (right): each c_n is found on its own
• A narrow state needs many: short wavelengths, high n

One structure, two settings

	a vector in a plane	a function
axes	\lvert u_1\rangle, \lvert u_2\rangle	eigenfunctions \lvert\phi_1\rangle, \lvert\phi_2\rangle, \dots
orthonormal	\langle u_j \vert u_k\rangle = \delta_{jk}	\langle \phi_m \vert \phi_n\rangle = \delta_{mn}
build	\lvert v\rangle = \sum_k c_k\lvert u_k\rangle	\lvert\psi\rangle = \sum_n c_n\lvert\phi_n\rangle
project	c_k = \langle u_k \vert v\rangle	c_n = \langle \phi_n \vert \psi\rangle
the bracket	multiply matching components, add	multiply point by point, integrate

• Lecture 3.5’s grid made this literal: the integral is the dot product of the sampled values, times \Delta x

Eigenfunctions are coordinates

\lvert\psi\rangle = \sum_n c_n\,\lvert\phi_n\rangle, \qquad c_n = \langle \phi_n \vert \psi \rangle = \int \phi_n^*\,\psi\,dx

• The eigenfunctions \phi_n of an observable are the axes; c_n says how far \psi reaches along axis n
• Each c_n is a projection, found exactly as for the vector: multiply, then add up

Three one-line derivations

All three use only orthonormality, \langle \phi_m \vert \phi_n \rangle = \delta_{mn}:

\begin{aligned} \text{coefficient:}\quad & \langle \phi_m \vert \psi \rangle = \sum_n c_n\,\langle \phi_m \vert \phi_n \rangle = c_m \\[6pt] \text{normalization:}\quad & \langle \psi \vert \psi \rangle = \sum_m\sum_n c_m^*\,c_n\,\delta_{mn} = \sum_n \lvert c_n\rvert^2 = 1 \\[6pt] \text{average:}\quad & \langle \psi \vert \hat{A} \vert \psi \rangle = \sum_m\sum_n c_m^*\,c_n\,a_n\,\delta_{mn} = \sum_n \lvert c_n\rvert^2\,a_n \end{aligned}

• Same as the vector: squared components add up to one, the length of a normalized state

Example: how much ground state is in a parabola?

Expand \psi = \sqrt{30/L^5}\;x(L-x) in box states \psi_n = \sqrt{2/L}\,\sin(n\pi x/L).

Two integrations by parts:

c_n = \int_0^L \psi_n\,\psi\,dx = \frac{8\sqrt{15}}{(n\pi)^3}\ \ (n \text{ odd}), \qquad c_n = 0\ \ (n \text{ even: odd about } L/2)

\lvert c_1\rvert^2 = \frac{960}{\pi^6} = 0.9986, \qquad \lvert c_3\rvert^2 = 0.0014, \qquad \lvert c_5\rvert^2 = 0.0001

• A smooth, spread-out state is almost all ground state

A measurement returns an eigenvalue

\text{reading } a_n \text{ with probability}\quad p_n = \lvert c_n\rvert^2 = \lvert\langle \phi_n \vert \psi \rangle\rvert^2

Live: measure it yourself

Averages and spreads

\langle A \rangle = \sum_n p_n\,a_n = \langle \psi \vert \hat{A} \vert \psi \rangle

\sigma_A^2 = \sum_n p_n\,\big(a_n - \langle A \rangle\big)^2 = \langle A^2 \rangle - \langle A \rangle^2

• \langle A\rangle: the average of many readings, usually not a possible reading itself
• \sigma_A = 0 only if all the probability sits on one eigenvalue: only eigenstates are sharp

Example: readings, average, spread

\psi = \tfrac{1}{2}\psi_1 + \tfrac{1}{2}\psi_2 + \tfrac{1}{\sqrt{2}}\psi_3 in a box. What can an energy measurement give, how often, and what are \langle E\rangle and \sigma_E?

E_1,\ 4E_1,\ 9E_1 \quad\text{with}\quad p = \tfrac{1}{4},\ \tfrac{1}{4},\ \tfrac{1}{2}

\langle E \rangle = \tfrac{1}{4}(1) + \tfrac{1}{4}(4) + \tfrac{1}{2}(9) = 5.75\,E_1

\langle E^2 \rangle = \tfrac{1}{4}(1) + \tfrac{1}{4}(16) + \tfrac{1}{2}(81) = 44.75\,E_1^2

\sigma_E = \sqrt{44.75 - 5.75^2}\;E_1 = 3.42\,E_1

After the measurement

• A reading of a_n leaves the state \phi_n (collapse): measure again, get a_n every time
• The probabilities live in many identically prepared copies, not in one particle

Spikes and plane waves

operator	eigenfunctions	sharp	spread over
\hat{x}	spikes \delta(x - x_0)	position x_0	every momentum
\hat{p}	plane waves e^{ipx/\hbar}	momentum p	every position

• [\hat{x},\hat{p}] = i\hbar: no shared eigenfunctions. Each eigenfunction of one is a sum of many eigenfunctions of the other
• On a grid the spikes are the columns with a single 1: the eigenvectors of the diagonal X

Each is a sum of the other

\psi(x) = \underbrace{\sum_k c_k\,e^{ip_kx/\hbar}}_{\text{many momenta}} \qquad\qquad \psi(x) = \underbrace{\sum_j \psi(x_j)\,\big[\text{spike at } x_j\big]}_{\text{many positions}}

From the sum to the Fourier transform

• Momentum takes any value: the sum over k becomes an integral, and c_k becomes a function \phi(p)

\psi(x) = \int_{-\infty}^{\infty} \phi(p)\,\frac{e^{ipx/\hbar}}{\sqrt{2\pi\hbar}}\,dp, \qquad \phi(p) = \langle p \vert \psi \rangle = \int_{-\infty}^{\infty} \frac{e^{-ipx/\hbar}}{\sqrt{2\pi\hbar}}\,\psi(x)\,dx

• \phi(p) is a projection, like c_n = \langle \psi_n \vert \psi \rangle: the momentum wavefunction
• \lvert\phi(p)\rvert^2\,dp: the probability of a momentum reading between p and p + dp

Example: a Gaussian in momentum

Which momenta make up \psi(x) \propto e^{-x^2/4\sigma_x^2}\,e^{ip_0x/\hbar}?

Put it into \phi(p) and complete the square in the exponent:

\phi(p) \propto e^{-(p - p_0)^2/4\sigma_p^2}, \qquad \sigma_p = \frac{\hbar}{2\sigma_x}

A Gaussian again, centered at p_0: narrow in x means wide in p, and \sigma_x\sigma_p = \hbar/2.

Live: squeeze it, widen it

• Squeeze: a spike, from every momentum. Widen: one plane wave, over every position

Sharp energy or one-way momentum

• Box state: sharp E, but half moving right, half left: spread p
• Packet: one hump in p, but spread E: no box state is sharp in both

The uncertainty principle

\sigma_A\,\sigma_B \geq \tfrac{1}{2}\big\lvert\langle[\hat{A},\hat{B}]\rangle\big\rvert \qquad\Longrightarrow\qquad \sigma_x\,\sigma_p \geq \hbar/2

Example: \sigma_x\sigma_p for the box ground state

For \psi_1 = \sqrt{2/L}\,\sin(\pi x/L), find \sigma_x, \sigma_p and their product.

\langle x\rangle = \tfrac{L}{2}, \quad \langle x^2\rangle = L^2\left(\tfrac{1}{3} - \tfrac{1}{2\pi^2}\right) \;\Rightarrow\; \sigma_x = 0.181\,L

\langle p\rangle = 0, \quad \langle p^2\rangle = 2mE_1 = \left(\tfrac{\pi\hbar}{L}\right)^2 \;\Rightarrow\; \sigma_p = \tfrac{\pi\hbar}{L}

\sigma_x\,\sigma_p = 0.181\,\pi\,\hbar = 0.568\,\hbar \;\geq\; 0.5\,\hbar

• Independent of L: squeeze the box and \sigma_p grows, so confinement costs energy (\propto 1/L^2)

Takeaway

Expand the state in the eigenfunctions of the observable: the coefficients are projections, c_n = \langle \phi_n \vert \psi \rangle. A measurement returns one eigenvalue a_n, with probability \lvert c_n\rvert^2, and leaves the system in \phi_n. Averages are weighted sums, \langle A\rangle = \sum_n \lvert c_n\rvert^2 a_n, sharp only in eigenstates; and observables that do not commute cannot be sharp together: a spike in position is a sum of every plane wave, and \sigma_x\sigma_p \geq \hbar/2.