QM II: Postulates of Quantum Mechanics

9.1.Statement of Postulates

A particle is described by a normalised wave function $\psi(x)$. Recall that the space of square-normalisable wave functions forms a complex vector space with Hermitian inner product
\begin{equation} \langle \psi_1,\psi_2\rangle = \int^\infty_{-\infty} {\rm d}x \, \overline{\psi_1(x)}\psi_2(x) \, . \end{equation}
In this notation, a normalised wave function obeys $\langle \psi,\psi\rangle = \displaystyle\int^\infty_{-\infty}|\psi(x)|^2 = 1$.

There are five postulates in quantum mechanics: 1) wave function postulate, 2) observables postulate, 3) measurement outcome postulate, 4) collapse postulate, 5) evolution postulate.
Measurable quantities or observables are represented by Hermitian operators $A(x,p)$ constructed from polynomial or real analytic functions of the elementary position and momentum operators,
\begin{align} \hat x & = x \\ \hat p & = - i \hbar \frac{\partial}{\partial x} \, . \end{align}
For example, the Hamiltonian operator is $\hat H = \displaystyle\frac{\hat p^2}{2m} + V(x) $.
The possible outcomes of a measurement of $A$ are its eigenvalues $a$. How we assign probabilities to these outcomes depends on whether the spectrum of eigenvalues is discrete or continuous:
- Discrete: $\{a_j\}$. Choose a basis of eigenfunctions $\psi_j(x)$ that obey $\langle \phi_i , \phi_j\rangle = \delta_{ij}$. The probability to find the eigenvalue $a_j$ is $P_j : = | \langle \phi_j , \psi \rangle |^2$.
- Continuous: $a\in\mathbb{R}$. Choose a basis of eigenfunctions $\psi_a(x)$ that obey $\langle \phi_a , \phi_{a'}\rangle = \delta(a-a')$. The probability density for measurements of the observable is $P(a) : = | \langle \phi_a , \psi \rangle |^2$.
If a measurement of $A$ yields the result $a_j$ (or $a$ for a continuous spectrum), the wave function immediately after the measurement is $\phi_j(x)$ (or $\phi_a(x)$ for a continuous spectrum).
As long as no measurements are made, the wave function evolves in time according to the Schrödinger equation,
\begin{equation} i\hbar\frac{\partial\psi(x,t)}{\partial t} = \hat{H} \psi(x,t)\,. \end{equation}

Of these, the 4th postulate is perhaps the least understood and the most mysterious. It has led to various interpretations of quantum mechanics. For actual computations, it often does not matter how this postulate is interpreted, and we will not go into these matters here.

9.2.Discussion of Postulates

Let us emphasise some important observations / consistency checks:

The outcomes of measurements are real numbers. This is consistent with the fact that observables are represented by Hermitian operators, whose eigenvalues are real numbers.
To compute probabilities it is convenient to expand the wave function in an orthonormal basis of eigenfunctions of $A$
- Discrete: If we expand the wave function
  \begin{equation} \psi(x) =\sum_j c_j \phi_j(x)\, , \end{equation}
  where $c_j = \langle \phi_j , \psi \rangle$. The probability to measure $a_j$ is therefore the the modulus squared of the coefficient,
  \begin{align} P_j & := | \langle \phi_j , \psi \rangle|^2 = |c_j|^2 %& = | \sum_j c_j \langle \psi_i,\psi_j\rangle |^2 \\ \, . \end{align}
  Consistency requires that the total probability is $1$. This follows from the normalisation of the wave function,
  \begin{equation} \langle \psi , \psi \rangle = \sum_j|c_j|^2 = \sum_j P_j = 1\, . \end{equation}
- Continuous: We expand the wave function
  \begin{equation} \psi(x) = \displaystyle\int^\infty_{-\infty} {\rm d}a \, c(a) \phi_a(x)\, , \end{equation}
  where $c(a) = \langle \phi_a , \psi\rangle$. The probability distribution is therefore the modulus squared of the coefficient function,
  \begin{equation} P(a) := | \langle \phi_a , \psi \rangle|^2 = |c(a)|^2 \, . \end{equation}
  Consistency requires the total probability is $1$. This follows from the normalisation of the wave function,
  \begin{equation} \langle \psi , \psi \rangle = \int^\infty_{-\infty} {\rm d}a \, |c(a)|^2 = \int^\infty_{-\infty} {\rm d}a \, P(a) = 1 \, . \end{equation}
A measurement of $A$ that yields an eigenvalue $a_j$ / $a$ causes the wave function to immediately collapse to the corresponding eigenfunction,
\begin{equation} \psi(x) \to \psi_j(x) \, /\, \psi_a(x) \end{equation}
in the discrete / continuous case. In the continuous case, you should be concerned that the wave function immediately after the measurement is a not square-normalisable,
\begin{equation} \langle \phi_a,\phi_a \rangle = \infty\, . \end{equation}
This assumes an ``idealised" measurement of a continuous parameter where the outcome is determined with infinite accuracy. A real-world measurement will always have some experimental uncertainty $\epsilon$. The wave function immediately after the measurement is then a square-normalisable wave function
\begin{equation} \int {\rm d}a' \, \delta_\epsilon(a-a') \phi_{a'}(x) \, , \end{equation}
where $\delta_\epsilon(a-a')$ is sharply peaked around $a' =a$ with width $\epsilon$. The idealised measurement corresponds to the limit $\epsilon \to 0$ where $\delta_\epsilon(a-a') \to \delta(a-a')$. This is a convenient and useful mathematical ideal.

9.3.Expectation Values

The definition $\langle A \rangle = \langle \psi, A \psi\rangle$ is backed up by a computation of the expectation value using a decomposition of the wave function in terms of eigenfunctions of $\hat{A}$.

The expectation value of a Hermitian operator is defined by

\begin{equation} \langle A \rangle = \langle \psi, A \psi \rangle = \int^\infty_{-\infty} {\rm d}x \, \overline{\psi(x)} \, A \cdot \psi(x) \, . \end{equation}

This expectation value can be computed by acting with the differential operator $A$ and evaluating the integral. However, it is often more convenient to first expand the wave function in an orthonormal basis of eigenfunctions of $A$. This leads to the expected formulae using the probabilities $P_j$ or probability distribution $P(a)$.

Discrete: The expectation value is
\begin{align} \langle A\rangle & =\sum_{ij} \bar c_i c_j \langle \phi_i , A \cdot \phi_j \rangle \\ & = \sum_{ij} a_j \bar c_i c_j \langle \phi_i , \phi_j \rangle \\ & = \sum_j a_j |c_j|^2 \\ & = \sum_j a_j P_j \, . \end{align}
Continuous: The expectation value is
\begin{align} \langle A\rangle & = \int^\infty_{-\infty} {\rm d}a \, {\rm d}a' \, \overline{c(a)} \, c(a') \langle \phi_a , A \cdot \phi_{a'} \rangle \\ & = \int^\infty_{-\infty} {\rm d}a \, {\rm d}a' \, a' \, \overline{c(a)} \, c(a') \langle \phi_a , \phi_{a'} \rangle \\ & = \int^\infty_{-\infty} {\rm d}a \, {\rm d}a' \, a' \, \overline{c(a)} \, c(a') \delta(a-a')\\ & = \int^\infty_{-\infty} {\rm d}a \, a |c(a)|^2 \\ & =\int^\infty_{-\infty} {\rm d}a\, a P(a) \, . \end{align}

A similar result holds for the expectation value of any real analytic function $f(A)$.

These expressions make it manifest that we are computing an average, because we are summing/integrating the possible outcomes of the measurement – the eigenvalues – with the probabilities to find the system in the state corresponding to each eigenvalue.

9.4.Position Revisited

Eigenfunctions of the position operator are Dirac delta functions. If you insist on square-normalisable wave functions, use narrow-width Gaussians instead.

The position operator $\hat x$ multiplies a wave function by $x$. The spectrum of the position operator is continuous with an orthonormal basis of eigenfunctions

\begin{equation} \phi_{x'}(x) = \delta(x-x') \end{equation}

labelled by $x'\in \mathbb{R}$.

They form eigenfunctions because

\begin{equation} \hat x \cdot \phi_{x'}(x) = x\delta(x-x') = x'\delta(x-x') = x' \phi_{x'}(x) \end{equation}

and are orthonormal,

\begin{align} \langle \phi_{x_1} , \phi_{x_2} \rangle & = \int^\infty_{-\infty} {\rm d}x \, \phi_{x_1}(x) \phi_{x_2}(x) \\\nonumber & = \int^\infty_{-\infty} {\rm d}x \, \delta (x-x_1)\delta(x-x_2) \\ & = \delta(x_1-x_2)\, . \end{align}

As mentioned above, these wave functions are not square-normalisable: $\langle \phi_{x'}, \phi_{x'}\rangle=\infty$. Nevertheless, we regard $\phi_{x'}(x)$ as the wave function immediately after an ‘`idealised" measurement that measures the position $x\rq{}$ with infinite accuracy.

Figure 9.1: An eigenstate of the position operator, with eigenvalue $x'$.

We can expand any wave function in this basis,

\begin{align} \psi(x) & = \int^\infty_{-\infty} {\rm d}x' c(x') \phi_{x'}(x) \\ \nonumber & = \int^\infty_{-\infty} {\rm d}x' c(x') \delta(x-x') \\ & = c(x) \, , \end{align}

so the coefficient function is the wave function itself. The probability density for position measurements according to the postulates are

\begin{align} P(x) = | \langle \phi_{x} , \psi \rangle |^2 = \left| \, \int^\infty_{-\infty} {\rm d}x' \delta(x-x') \psi(x') \, \right|^2 = |\psi(x)|^2 \, . \end{align}

This recovers our original probabilistic interpretation of the wave function!

9.5.Momentum Revisited

Eigenfunctions of the momentum operator are plane waves, which are not unit-normalisable on the real line either.

The momentum operator is $\hat p = - i \hbar \partial_x$. In the last lecture we showed that the spectrum of the momentum operator is continuous with an orthonormal basis of eigenfunctions labelled by $p\in \mathbb{R}$,

\begin{equation} \phi_{p}(x) = \frac{1}{\sqrt{2\pi \hbar}} e^{ipx/\hbar} \, . \end{equation}

In particular,

\begin{equation} \langle \phi_p , \phi_{p'} \rangle = \frac{1}{2\pi \hbar}\int^\infty_{-\infty} {\rm d}p \, e^{i(p'-p)x/\hbar} = \delta(p-p') \, . \end{equation}

As above, these wave functions are not square-normalisable, but arise from an ``idealised" measurement that determines the momentum with infinite accuracy.

We can expand any wave function in the momentum basis

\begin{align} \psi(x) & = \int^\infty_{-\infty} {\rm d}p \, c(p) \phi_p(x) \\ & = \frac{1}{\sqrt{2\pi \hbar}} \int^\infty_{-\infty} {\rm d}p \, c(p) e^{ipx/\hbar} \, . \end{align}

The coefficients can be computed using the inner product and orthonormality,

\begin{align} c(p) & = \langle \phi_p , \psi\rangle \\ & = \int^\infty_{-\infty} {\rm d}x \, \overline{\phi_p(x)} \psi(x) \\ & = \frac{1}{\sqrt{2\pi \hbar}} \int^\infty_{-\infty} {\rm d}x \, \psi(x)e^{-ipx/\hbar} \, . \end{align}

The momentum probability density is

\begin{align} \widetilde P(p) = | c(p)|^2 \, . \end{align}

9.6.Problems

Momentum space representation
Recall that classically there is a symmetry between position and momentum. In fact, we could also have formulated the postulates of quantum mechanics in terms of a so-called momentum wave function $\widetilde\psi(p)$ with momentum operator $\hat p = p$ and position operator $\hat x = i \hbar \partial_p$. Can you work out the commutator $[\hat{x},\hat{p}]$ in this representation? We will return to this formulation in a later chapter.

Solution:
▶
In this representation, we get
\begin{equation} \begin{aligned} \left[ \hat x , \hat p \right] \, \widetilde\psi(p) & = \hat x (\hat p \, \psi(x) ) - \hat p (\hat x \, \psi(x) ) \\ & = i\hbar \frac{\partial}{\partial p}\left( p \psi(p) \right) - p \left( i\hbar \frac{\partial}{\partial p} \psi(p) \right) & = i \hbar \, \psi(p) \, . \end{aligned} \end{equation}
Which of course gives the same result as we derived before using the position representation.
2
Wave function collapse
A quantum mechanical system is, at $t=0$, prepared in a state described by the wave function
\begin{equation} \psi(t=0, x) = C\left( \frac{1}{\sqrt{2}} \psi_{E=1}(x) + e^{i\alpha} \psi_{E=2}(x)\right)\,, \end{equation}
where the wave functions on the right-hand side are orthonormal energy eigenfunctions. Both $C$ and $\alpha$ are real constants.
1. Determine the constant $C$.
2. An energy measurement is made. What are the possible outcomes, and what are the probabilities of those outcomes?
3. Subsequently, the position of the particle is measured. What do you know about the wave function of the system immediately after this measurement?
Solution:
▶
1. Because $\langle \psi_1 |\psi_2\rangle=0$ and each of the wave functions is normalised, we have
  \begin{equation} \langle \psi | \psi\rangle = C^2 \left( \frac{1}{2} |\psi_1|^2 + |\psi_2|^2 \right) = C^2 \frac{3}{2}\,,\quad\Rightarrow\quad C = \sqrt{2/3}\,. \end{equation}
2. Either $E=1$ with probability $1/3$, or $E=2$ with probability $2/3$.
3. The position operator does not commute with the energy operator, and in particular the energy eigenstates are not position eigenstates. Rather, the wave function will now be an eigenfunction of the position operator, which is a Dirac delta function at the position at which the particle was measured to be.

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1.	Why Quantum Mechanics?
2.	The Double Slit Experiment
3.	Wave function and Probabilities
4.	Momentum and Planck's constant
5.	Schrödinger's Equation
6.	The Hilbert Space
7.	Hermitian Operators
8.	The Spectrum of a Hermitian Operator
9.	Postulates of Quantum Mechanics
9.1.	Statement of Postulates
9.2.	Discussion of Postulates
9.3.	Expectation Values
9.4.	Position Revisited
9.5.	Momentum Revisited
9.6.	Problems
10.	Commutators and Uncertainty Principle
11.	Energy Revisited
12.	Stationary states
13.	Case Study: The Free Particle
14.	Two-particle systems
15.	Simple Harmonic Oscillator
16.	The Continuity Equation
17.	Scattering Problems
18.	Tunnelling
19.	Momentum-space Wave function
20.	Ehrenfest's Theorem
21.	Bibliography