Question about Uncertainty Principle

  • #1
504
35
In the Infinite Square Well problem, an energy eigenstate is in an equal superposition of two momentum eigenstates with eigenvalues that are opposite in sign(like standing waves that are formed by two wavefunctions corresponding to "opposite momentums").
So, for every energy eigenstate, we always have two possible outcomes for the momentum in a momentum measurement.
So, the question is, why does the uncertainty in the momentum change(increase) for different energy eigenstates if each eigenstate has exactly two outcomes?
Thanks in advance!
 

Answers and Replies

  • #2
George Jones
Staff Emeritus
Science Advisor
Gold Member
7,539
1,354
What is an expression for momentum uncertainty in terms of expectation values?
 
  • #3
vanhees71
Science Advisor
Insights Author
Gold Member
2021 Award
20,109
10,850
Since in the infinite-wall example you cannot define a momentum operator and thus no momentum uncertainty the question is misleading to begin with. It's an important exercise to figure out that for this geometry no self-adjoint operator that generates translations exists! Look for the topic in this forum. I've once posted a discussion on it.
 
  • #4
504
35
What is an expression for momentum uncertainty in terms of expectation values?
<p^2>-<p>^2. So, two opposite in sign momentums give <p>=0 but not <p^2>=0. But again, i can't quite get the intuition behind being more uncertain about the momentum in eigenstates of higher energy. Mathematically, i understand it in the way i explained it. But the intuition is unclear for me. We always have two possible values of momentum.
 
  • #5
504
35
Since in the infinite-wall example you cannot define a momentum operator and thus no momentum uncertainty the question is misleading to begin with. It's an important exercise to figure out that for this geometry no self-adjoint operator that generates translations exists! Look for the topic in this forum. I've once posted a discussion on it.
What do you mean by you can't define a momentum operator? Can't we act on a wavefunction with the momentum operator in just the same way as in any other problem? We can also find the expectation value of the momentum. I am feeling that i am getting something fundamental wrong and that's a bad thing! Help, SOS! :D
 
  • #6
1,116
72
The higher the energy the greater the momentum. Since you don't know which direction the particle is travelling in there is more uncertainty due to the bigger gap between +p and -p.
 
  • #7
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
21,548
12,863
<p^2>-<p>^2. So, two opposite in sign momentums give <p>=0 but not <p^2>=0. But again, i can't quite get the intuition behind being more uncertain about the momentum in eigenstates of higher energy. Mathematically, i understand it in the way i explained it. But the intuition is unclear for me. We always have two possible values of momentum.

Suppose you had a job and you knew that next week you would be paid either $99 or $101. Well, the average is $100 and there is some uncertainty.

Suppose you knew that next week you would be paid either $50 or $150? The average is still $100 and you still have only two possible values.

Would you say that the "uncertainty" (technically, it's the "variance" we should be talking about) is more, less or the same in the second case?

On a more general point, you are going to run into all sorts of problems in physics if you start with a priori "intuitive" ideas about what certain words mean. For example, the term "expected" value in stats and QM is not really a good word. It really should be the "mean" value. The "expected value" for the roll of a die is 3.5, which is, of course, an impossible outcome.

Words (and your a priori expectations of what you think they should mean) will lead you astray, where the maths and the mathematical definition of those words will not.
 
  • #8
vanhees71
Science Advisor
Insights Author
Gold Member
2021 Award
20,109
10,850
What do you mean by you can't define a momentum operator? Can't we act on a wavefunction with the momentum operator in just the same way as in any other problem? We can also find the expectation value of the momentum. I am feeling that i am getting something fundamental wrong and that's a bad thing! Help, SOS! :D

For a short answer, see

https://www.physicsforums.com/threads/particle-in-a-box-in-momentum-basis.694158/#post-4398736

For the mathematical details, see

http://arxiv.org/abs/quant-ph/0103153v1
 
  • #9
504
35
Wow, lot's of great stuff there! As a beginner on QM, i find these to be very enlightening and subtle!
But, i have one more question. In order for an observable to be attributed to an operator, the operator must be Hermitian or self-adjoint and why? Also, what is their difference? Because there is a lot of confusion among people.
Also, how to detect if an operator is not a good one(not self-adjoint or not Hermitian) in a problem(or in general)?
 
  • #10
George Jones
Staff Emeritus
Science Advisor
Gold Member
7,539
1,354
What is an expression for momentum uncertainty in terms of expectation values?

For the mathematical details, see

http://arxiv.org/abs/quant-ph/0103153v1

Mea culpa! I was more careful in

I think this is a bit subtle. What is the momentum operator for the square well? See

http://arxiv.org/abs/quant-ph/0103153.

what is their difference? Because there is a lot of confusion among people.?

I don't think that there is a completely standard usage of the term "Hermitian"; see the discussion at

https://www.physicsforums.com/threa...bles-are-hermitian.728302/page-2#post-4605707
 

Related Threads on Question about Uncertainty Principle

  • Last Post
Replies
22
Views
3K
  • Last Post
Replies
5
Views
945
Replies
22
Views
4K
  • Last Post
Replies
5
Views
846
Replies
6
Views
861
Replies
1
Views
2K
Replies
1
Views
939
  • Last Post
Replies
4
Views
2K
Replies
21
Views
4K
  • Last Post
Replies
2
Views
727
Top