Eigenspaces of observables

  • I
  • Thread starter Erland
  • Start date
  • #1
Erland
Science Advisor
738
136
I'm trying to grasp the subtleties of quantum mechanics, and that's not easy :).

Let H be a two dimensional complex Hilbert space of quantum states. An observable is a self adjoint operator A on H. Let ##(|\phi_1\rangle,|\phi_2\rangle)## be an orthonormal eigenbasis to A, with corresponding (real) eigenvalues ##a_1## and ##a_2##.

Now, let us make a measurement of A upon a state ##|\psi\rangle=b_1|\phi_1\rangle+b_2|\phi_2\rangle##, where ##|b_1|^2+|b_2|^2=1##.

If ##a_1\neq a_2##, then, with probability ##b_i##, the result of the measurement is ##a_i## and the state ##|\psi\rangle## is "collapsed" into the state ##|\phi_i\rangle## (##i=1,\,2##).

But if ##a_1=a_2##, then the state does not collapse, but remains unchanged after the measurement: ##|\psi\rangle=b_1|\phi_1\rangle+b_2|\phi_2\rangle##.

This is how it must be, if the collapse in both cases means a normalized orthogonal projection onto an eigenspace - the difference is that we have two eigenspaces if ##a_1\neq a_2##, but only one if ##a_1=a_2##.

The above is my understanding after reading p. 2-3 here: http://www.theory.caltech.edu/people/preskill/ph229/notes/chap2.pdf

But if this is correct, there is a strange discontinuity in the dependence of the eigenvalues. Assume that ##b_1=b_2=1/\sqrt 2##, and ##a_1=a_2=1##. Then the state after the measurement is the same as before the measurement: ##|\psi\rangle=(1/\sqrt 2) \phi_1+(1/\sqrt 2)\phi_2##.
But if we change the eigenvalues very slightly so that ##a_1=0.999## and ##a_2=1.001##, then, after the measurement, the new state will be ##\phi_i## with probability ##1/2## ##(i=1,\,2)##.
This is not at all close to the state in the previous case, despite the very small differences in the eigenvalues.

Is there really such a discontunity, and if so, why? Or did I misunderstand something?
 

Answers and Replies

  • #2
1,283
345
But if a1=a2a1=a2a_1=a_2, then the state does not collapse, but remains unchanged after the measurement:
If the observable is degenerate measuring it doesn't change your knowledge of the system. If an arbitrarily small difference is detectable then your information about the system changes. Of course any real measurement would not be so unconstrained. The word collapse invokes an image of measurement I don't feel is particularly accurate. Another thing to consider is that any measurement system must interact (through a Hamiltonian term) with the system. It would be quite odd to have ##a_1\ne a_2## and have the very same energy. Usually, it takes some work to separate systems.
 
  • #3
A. Neumaier
Science Advisor
Insights Author
7,491
3,387
Is there really such a discontunity, and if so, why? Or did I misunderstand something?
Excellent observation.

You need to remember that the introductory postulates are valid for highly idealized experiments only. The typical illustrative experiments assume that the eigenvalues are half integers. then your problem doesn't arise.

As soon as eigenvalues can move continuously, your argument shows that the postulates can become physically meaningless in borderline situations. This proves that the idealization assumed is no longer applicable and the simple collapse model no longer applies.
 
  • Like
Likes bhobba
  • #4
But if we change the eigenvalues very slightly so that ##a_1=0.999## and ##a_2=1.001##, then, after the measurement, the new state will be ##\phi_i## with probability ##1/2## ##(i=1,\,2)##.
Can we relate this to something like a Stern-Gerlach with a very weak magnet, where the observable is the final position on the screen?
 
  • #5
Simon Phoenix
Science Advisor
Gold Member
291
224
If your operator A has degenerate eigenvalues, then for this space isn't it just proportional to the identity operator?
 
  • #6
1,283
345
Can we relate this to something like a Stern-Gerlach with a very weak magnet, where the observable is the final position on the screen?
Yes, then all replies given are still valid and ##A## is the screen position. Screens are not unlimited precision. In this version of the experiment particles at the two positions are so close they can no longer be resolved. Recall the width of the particle beams are quite finite and will overlap in the limit you suggest.
 
  • #7
Erland
Science Advisor
738
136
If your operator A has degenerate eigenvalues, then for this space isn't it just proportional to the identity operator?
As I understand it, yes. In the given degenerate case, A is the identity operator, hence it represents no measure at all.
 
  • #8
Erland
Science Advisor
738
136
Thank you all for your replies!

But isn't the "collapse model" the heart of the Copenhagen interpretation? I thought that quantum mechanics according to the Copenhagen interpretation was supposed to be a final theory, not reducible to anything more basic. If what Colby and Neumaier write is true, then this view must be wrong, since the collapse model is not really appliciable in my example. Thus, it seems that Einstein was right when he said that QM is not the final theory.

So, is there any more final theory? Perhaps the many worlds interpretation, which seems to have gained a lot of support during the latest decades...?
 
  • #9
A. Neumaier
Science Advisor
Insights Author
7,491
3,387
Quantum mechanics is much more than the Copenhagen interpretation; the latter is only day 1 in a long education in QM. The simple collapse model is a caricature of most real life measurements. More flexible and realistic models are POVM based.

Teaching is generally done first by example, and only much later includes the full complexities of a subject. (Not only in QM but everywhere.)
 

Related Threads on Eigenspaces of observables

  • Last Post
5
Replies
102
Views
13K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
0
Views
971
  • Last Post
Replies
20
Views
2K
  • Last Post
Replies
15
Views
8K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
1
Views
970
  • Last Post
Replies
20
Views
3K
Top