Uncertainty principle

  • I
  • Thread starter dRic2
  • Start date
  • #1
dRic2
Gold Member
813
197
I'm having trouble trying to remember this:

suppose I have an operator ##A## and an eigenfunction ##\psi = c_1 \psi_1 + c_2 \psi_2## for ##A## so that

$$ A\psi = a\psi$$ The expected value is

##\langle A \rangle = \langle \psi|A|\psi \rangle = a##

After a measurement ##a## can be ##a_1## or ##a_2## and the probability of a particular outcome is ##|c_i|^2##.
Let's say the outcome of my measurement is ##a_1##. Now I'm certain about this, right? The uncertainty of ##a_1## is now ##0## because I measured it and so I forced the particle to be in the state with eigenvalue ##a_1##.

Now suppose I take the operator ##B## incompatible with ##A## and I try to evaluate:

$$\langle B \rangle = \langle \psi_1|B|\psi_1 \rangle = b$$

(because I know that after my measurement the particle is in the state ##\psi_1##).

This last operation is totally meaningless because, since ##a_1## is certain, the uncertainty of ##b## should be ##\infty## so, even if I can calculate an "expected value" for ##B## it has no physical significance because its uncertainty is ##\infty## , right ?

Is it correct?

Thanks
Ric
 

Answers and Replies

  • #2
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
16,116
8,141
There are too many misconceptions there to address them all. Let's take the central issue.

If a system is in the state ##\psi = c_1\psi_1 + c_2\psi_2##, where ##\psi_1, \psi_2## are eigenstates of operator/observable ##A##, corresponding to eigenvalues ##a_1, a_2##, then:

##\langle A \rangle = \langle \psi | A | \psi \rangle = |c_1|^2 a_1 + |c_2|^2 a_2 = a##

And the uncertainty in ##A##, can be calculated as:

##\sigma_A^2 = \langle (A - a)^2 \rangle ##

If observable ##A## is measured, and the result is ##a_1##, then the system is in the state ##\psi_1##. The expected value of ##A## is now ##a_1## and there is no uncertainty in ##A##: for the state ##\psi_1##, we have ##\sigma_A^2 = 0##

Now, ##\psi_1## will be a superposition of the eigenstates of operator/observable ##B##. For example:

##\psi_1 = d_1 \phi_1 + d_2 \phi_2 + d_3 \phi_3##

Where ##\phi_1## etc. are eigenstates of observable ##B##.

A measurement of ##B## (of the system in state ##\psi_1##) will return the eigenvalues ##b_1, b_2, b_3## with probabilities ##|d_1|^2## etc. And the uncertainty in ##B## can be calculated from this.

On a second point:

You may be thinking of the uncertainty in position and momentum. But the eigenvalues of those observables form a continuous spectrum and are not covered by the standard linear algebra (for a discrete spectrum). The states are not physically realisable and, if you assume the system is in an eigenstate of position, then the uncertainty in momentum is undefined (infinite, if you like).
 
  • #3
dRic2
Gold Member
813
197
There are too many misconceptions there to address them all. Let's take the central issue.

If a system is in the state ψ=c1ψ1+c2ψ2ψ=c1ψ1+c2ψ2\psi = c_1\psi_1 + c_2\psi_2, where ψ1,ψ2ψ1,ψ2\psi_1, \psi_2 are eigenstates of operator/observable AAA, corresponding to eigenvalues a1,a2a1,a2a_1, a_2, then:

⟨A⟩=⟨ψ|A|ψ⟩=|c1|2a1+|c2|2a2=a⟨A⟩=⟨ψ|A|ψ⟩=|c1|2a1+|c2|2a2=a\langle A \rangle = \langle \psi | A | \psi \rangle = |c_1|^2 a_1 + |c_2|^2 a_2 = a

And the uncertainty in AAA, can be calculated as:

σ2A=⟨(A−a)2⟩σA2=⟨(A−a)2⟩\sigma_A^2 = \langle (A - a)^2 \rangle

If observable AAA is measured, and the result is a1a1a_1, then the system is in the state ψ1ψ1\psi_1. The expected value of AAA is now a1a1a_1 and there is no uncertainty in AAA: for the state ψ1ψ1\psi_1, we have σ2A=0σA2=0\sigma_A^2 = 0
This is clear to me. It may not seem because I think I phrased it very badly


But I don't understand this:

Now, ##\psi_1## will be a superposition of the eigenstates of operator/observable ##B##. For example:
If ##[A, B] ≠ 0## how can ##\psi_1## be an eigenfunction for ##B##?
 
  • #4
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
16,116
8,141
If ##[A, B] ≠ 0## how can ##\psi_1## be an eigenfunction for ##B##?
It isn't an eigenfunction of ##B##.
 
  • #5
dRic2
Gold Member
813
197
Please correct me if I'm wrong: If ##\psi_1## is a linear combination of ##ϕ_1, ϕ_2, ... ϕ_n ## and ##ϕ_i## are eigenfunctions of ##B##, ##\psi_1## should be an eigenfunction of ##B##.
 
  • #6
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
16,116
8,141
Please correct me if I'm wrong: If ##\psi_1## is a linear combination of ##ϕ_1, ϕ_2, ... ϕ_n ## and ##ϕ_i## are eigenfunctions of ##B##, ##\psi_1## should be an eigenfunction of ##B##.
That is wrong. A well-behaved operator will have a spectrum of eigenfunctions that span the space. So, every function can be expressed as a linear combination of eigenfunctions of that operator. In QM it is generally assumed that each observable operator under consideration has this property: a complete set of orthogonal eigenfunctions (that span the space).

In the case of a finite dimensional vector space, this can be proved for "normal" operators. This is the finite dimensional spectral theorem.

PS in my example:

##B\psi_1 = d_1B\phi_1 + d_2B\phi_2 + d_3 B\phi_2 = b_1 d_1 \phi_1 + b_2 d_2 \phi_2 + b_3 d_3 \phi_3##

And ##\psi_1## is not an eigenfunction of ##B## unless ##b_1 = b_2 = b_3##.
 
  • Like
Likes dRic2
  • #7
Nugatory
Mentor
13,217
6,067
Please correct me if I'm wrong: If ##\psi_1## is a linear combination of ##ϕ_1, ϕ_2, ... ϕ_n ## and ##ϕ_i## are eigenfunctions of ##B##, ##\psi_1## should be an eigenfunction of ##B##.
No, a sum of different eigenstates is not in general an eigenstate. Suppose that ##\phi_1## and ##\phi_2## are eigenstates of ##B##, with eigenvalues ##\beta_1## and ##\beta_2## respectively. Now, ##B(\phi_1+\phi_2)=\beta_1\phi_1+\beta_2\phi_2##, and that cannot be a multiple of ##\phi_1+\phi_2## except in the special case ##\beta_1=\beta_2##.
 
  • Like
Likes dRic2
  • #8
dRic2
Gold Member
813
197
Thank you, very much. I will study again this topic more carefully
 
  • #9
dRic2
Gold Member
813
197
You may be thinking of the uncertainty in position and momentum.
I was actually thinking about ##L_x##, ##L_y## and ##L_z## and their uncertainties in H atom
 

Related Threads on Uncertainty principle

  • Last Post
3
Replies
70
Views
4K
  • Last Post
Replies
1
Views
1K
Replies
8
Views
6K
Replies
2
Views
722
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
24
Views
3K
  • Last Post
Replies
8
Views
5K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
0
Views
997
Top