College Math Teaching

July 19, 2011

Quantum Mechanics and Undergraduate Mathematics IV: measuring an observable (example)

Ok, we have to relate the observables to the state of the system. We know that the only possible “values” of the observable are the eigenvalues of the operator and the relation of the operator to the state vector provides the density function. But what does this measurement do to the state? That is, immediately after a measurement is taken, what is the state?

True, the system undergoes a "time evolution" but once an observable is measured, an immediate (termed "successive") measurement will yield the same value; a "repeated" measurement (one made giving the system to undergo a time evolution) might give a different value.

So we get:

Postulate 4 A measurement of an observable generally (?) causes a drastic, uncontrollable alteration in the state vector of the system; immediately after the measurement it will coincide with the eigenvector corresponding to the eigenvalue obtained in the measurement.

Note: we assume that our observable operators have distinct eigenvalues; that is, no two distinct eigenvectors have the same eigenvalue.

That is, if we measure an observable with operator A and obtain measurement a_i then the new system eigenvector is \alpha_i regardless of what \psi was prior to measurement. Of course, this eigenvector can (and usually will) evolve with time.

Roughly speaking, here is what is going on:
Say the system is in state \psi . We measure and observable with operator A . We can only obtain one of the eigenvalues \alpha_k as a measurement. Recall: remember all of those “orbitals” from chemistry class? Those were the energy levels of the electrons and the orbital level was a permissible energy state that we could obtain by a measurement.

Now if we get \alpha_k as a measurement, the new state vector is \alpha_k . One might say that we started with a probability density function (given the state and the observable), we made a measurement, and now, for a brief instant anyway, our density function “collapsed” to the density function P(A = a_k)  = 1 .

This situation (brief) coincides with our classical intuition of an observable “having a value”.

Example (based on our calculation in the previous post):

For the purposes of this example, we’ll set our Hilbert space to the the square integrable piecewise smooth functions on [-\pi, \pi] and let our “state vector” \psi(x) =\left\{ \begin{array}{c}1/\sqrt{\pi}, 0 < x \leq \pi \\ 0,-\pi \leq x \leq 0  \end{array}\right.

Now suppose our observable corresponds to the eigenfunctions mentioned in this post, and we measure “-4” for our observable. This is the eigenvalue for (1/\sqrt{\pi})sin(2x) so our new state vector is (1/\sqrt{\pi})sin(2x) .

So what happens if a different observable is measured IMMEDIATELY (e. g., no chance for a time evolution to take place).

Example We’ll still use the space of square integrable functions over [-\pi, \pi]
One might recall the Legendre polynomials which are eigenfucntions of the following operator:
d/dt((1-t^2) dP_n/dt) = -(n)(n+1) P_n(t) . These polynomials obey the orthogonality relation \int^{1}_{-1} P_m(t)P_n(t)dt = 2/(2n+1) \delta_{m,n} hence \int^{1}_{-1} P_m(t)P_m(t)dt = 2/(2m+1) .
The first few of these are P_0 = 1, P_1  =t, P_2 = (1/2)(3t^2-1), P_3 = (1/2)(5t^3 - 3t), ..

We can adjust these polynomials by the change of variable t =x/\pi and multiply each polynomial P_m by the factor sqrt{2/(\pi (2m+1) } to obtain an orthonormal eigenbasis. Of course, one has to adjust the operator by the chain rule.

So for this example, let P_n denote the adjusted Legendre polynomial with eigenvalue -n(n+1) .

Now back to our original state vector which was changed to state function (1/\sqrt{\pi})sin(2x) .

Now suppose eigenvalue -6 = -2(3) is observed as an observable with the Lengendre operator; this corresponds to eigenvector \sqrt{(2/5)(1/\pi)}(1/2)(3(x/\pi)^2 -1) which is now the new state vector.

Now if we were to do an immediate measurement of the first observable, we’d have to a Fourier like expansion of our new state vector; hence the probability density function for the observables changes from the initial measurement. Bottom line: the order in which the observations are taken matters….in general.

The case in which the order wouldn’t matter: if the second observable had the state vector (from the first measurement) as an element of its eigenbasis.

We will state this as a general principle in our next post.



  1. […] So, what am I doing, blogging wise? Mostly writing about mathematical topics: here, here, here and here. […]

    Pingback by 19 July 2011: Swim progress « blueollie — July 19, 2011 @ 7:07 pm

  2. […] builds on our previous example. We start with a state and we will make three successive observations of observables which have […]

    Pingback by Quantum Mechanics and Undergraduate Mathematics V: compatible observables « College Math Teaching — July 25, 2011 @ 2:27 pm

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