Lectures on Quantum Mechanics 3: The Statistical Interpretation

God's dice God's dice

Probability and Measurement

There are four properties of wavefunctions that we must assume for QM to work. The wavefunction Ψ\Psi must:

1) Be square integrable throughout all space (particles exist and the probability of finding the particle is 100% in the entire space) ΨΨdx=1\int _{ -\infty }^{ \infty }{ { \Psi }^{ * }\Psi } dx\quad =1

where Ψ{ \Psi }^{ * } is the complex conjugate of Ψ\Psi

  • Arbitrary wavefunctions can be normalized such that the above equation holds

2) Particles can only exist if the wavefunction decays to zero faster than 1x\frac{1}{\sqrt{|x|}} as x±x \rightarrow \pm \infty

3) The wavefunction stays normalized as it evolves in time in order for the Schrödinger equation to be continually solved ddtΨΨdx=0 \frac { d }{ dt } \int _{ -\infty }^{ \infty }{ { \Psi }^{ * }\Psi } dx=0

4) The average is defined to be the expectation value

If you have encountered statistics before, the above concepts may seem vaguely familiar. To those who haven’t, the good news is that you must now learn to be well acquainted with statistics because QM is all about the behaviour of wavefunctions and its corresponding probabilities. In the orthodox interpretation, the measurable quantity (say the position of a particle) is spread all over space, where in some places there is a higher probability of the particle being there. We can calculate this probability by evaluating the integral of the modulus of the wavefunction ΨΨ{\Psi}^{*}\Psi.

before before after after

But where is the physics behind this? In the quantum description in the world, a particle can exist everywhere in space at varying degrees of probability as prescribed by its wavefunction. Once we make a measurement of the particle, the particle will lock down on a location by collapsing the wavefunction, producing a sharp peak where the probability becomes 100%. Now if we immediately measure the particle again, we would discover that the sharp peak stay where it was last measured. However, if we were to stop measuring the particle and give it some time, the sharp peak will gradually spread out.

Expectation values and Observables

Since QM is a statistical theory of physics, it follows that everything we observe follows statistical rules. Let us begin with two important quantities: position xx and momentum pp. In classical mechanics x,px,p are just numbers, so this is still true in QM but with a twist. In QM, we take the expectation of an operator acting to the probability of the wavefunction.

<x^>=Ψx^Ψdx\left<\hat{x}\right> = \int_{-\infty}^{\infty}{\Psi}^{*}\hat{x}{\Psi}dx

<p^>=Ψp^Ψdx\left<\hat{p}\right> = \int_{-\infty}^{\infty}{\Psi}^{*}\hat{p}{\Psi}dx

Since the momentum operator is a differential operator of the form p^=ix\hat{p} =-i\hbar\frac{\partial}{\partial x} then

<p^>=Ψ(iΨx)dx\left<\hat{p}\right> = \int_{-\infty}^{\infty}{\Psi}^{*}\left(-i\hbar\frac{\partial \Psi}{\partial x} \right) dx

We will accept these expressions for now and study them in more detail later. But first, let us understand why we care about the expectation values of physical quantities. Consider the following scenarios:

1) In one experiment you make note of nn measurements and divide sum of the measurements by nn (average)

2) You take the average of repeated measurements on nn (if best infinite) identically conditioned experiments (expectation value)

From a scientific point of view, which scenario sounds more experimentally sound? Hopefully you chose 2), because that is how science is done. Every experimenter should be able to repeat an experiment and agree that their results corroborate with physical principles. You should think of the expectation value as a stronger average.

Variance and Uncertainty

When we measure the position or momentum of a particle with some wavefunction Ψ\Psi , we will notice a problem. Although we know the expectation value of said variables, the wavefunction still has some spread Δx=x<x>\Delta x = x - \left<x\right> or Δp=p<p>\Delta p = p - \left<p\right>. Now it is experimentally meaningless to talk about the spread because we will produce useless data if most of the numerical measurement equals the expectation value. Thus we use the statistical measure called variance (σ2{\sigma}^{2}) to circumvent this problem. Variance is calculated as the expectation of the square of the spread, for example

σx2=<(Δx)2>=<x2><x>2.{{\sigma}_{x}}^{2} = \left<{(\Delta x)}^{2}\right> = \left<{x}^{2}\right> - {\left<x\right>}^{2}.

We take the square root of the variance to calculate standard deviation, which measures the variation of measurable quantities. The standard deviation is convenient to QM because we can now discuss the uncertainty of, let's say, the position or momentum of a particle. We will later derive the Heisenberg Uncertainty Principle (HUP) using operators and Dirac notation, so for now we shall take a sneak peak of the HUP:

σxσp2.{\sigma}_{x}{\sigma}_{p} \ge \frac{\hbar}{2}.

Visit my set Lectures on Quantum Mechanics for more notes.


  1. Normalizing the Wavefunction

  2. Normalizing the Complex Plane Wave

  3. Standard Deviation of Position and Momentum

  4. Show that the energy eigenvalues of a wavefunction must be real (as opposed to complex numbers) in order to be normalized for all time tt. Hint: Let En=E0+iΛ{E}_{n} = {E}_{0} + i \Lambda for real numbers E0,Λ {E}_{0}, \Lambda.

  5. Infinite Square Well

  6. Conservation of Probability Let the probability density be defined ρ=Ψ(x,t)Ψ(x,t)\rho = {\Psi}^{*}(x,t) \Psi(x,t). In QM the probability current is calculated by J(x,t)=2mi(ΨΨxΨΨx).J(x,t)=\frac{\hbar}{2mi}\left({\Psi}^{*}\frac{\partial \Psi}{\partial x}-\Psi \frac{\partial {\Psi}^{*}}{\partial x}\right).

Use the Schrödinger equation to show that ρt+Jx=0.\frac{\partial \rho}{\partial t}+\frac{\partial J}{\partial x}=0.

In fact, the conservation of probability extends to three dimensions ρt+J=0.\frac{\partial \rho}{\partial t}+\nabla \cdot J=0.

Note by Steven Zheng
6 years, 6 months ago

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ain't wave mechanics probably mathematical mechanics...there is a ton of maths operators and stuff

incredible mind - 6 years, 6 months ago

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What do you mean?

Steven Zheng - 6 years, 6 months ago

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@Steven Zheng hey could u please upload a few questions on expectation of Hamiltonian operator.The previous question was very nice.

pawan dogra - 6 years ago

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Then they will be pretty much the same question. The thing I find hard about posting university level physics problems is that they are largely derivation based, not numerical.

Steven Zheng - 6 years ago

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Thanks man appreciate it

pawan dogra - 6 years ago

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