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Intuitive Understanding for Commutation

by sokrates
Tags: commutation, intuitive
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sokrates
#1
Sep22-09, 05:02 PM
P: 483
I am trying to get a grip on the commutation properties of operators.
Different authors get to those differently: some start from translator operators, some relate those to Poisson brackets, etc...

My objective is to get a good intuitive feeling of what commuting and not commuting observables imply about the physical system? What are good physical examples to justify this?
What other ways have people used to link ,for instance, uncertainty relation and commutativity?

Insightful, less mathematical answers (IF AT ALL POSSIBLE) are greatly appreciated as I already have plenty of access to the mathematical foundation.

Thanks,

sokrates
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morphemera
#2
Sep22-09, 08:39 PM
P: 24
To my understanding, commutator measures the degree of the 'commutativeness' of two operations. For commutator with value 0, the associated operations are commutate. If the magnitude of commutator is small, then the operation are considered commute slightly.

For example, a commutator with Planck constant is small in our daily live scale, so it is essentially zero, or the operations can be treated as commute. As another example, considering walking on Earth surface, the operation are walking toward north and walking toward east. These two operations are not commute, but as far as two operators are small, then the different between the destination of the walk is also small.
RedX
#3
Sep22-09, 10:55 PM
P: 969
Imagine at a certain time you take as your state the eigenstate of one of the operators in the commutator. Then the eigenvalue can serve as a label for the state. Then view the other operator as a transformation. The commutator tells you how much the transformation destroys the label of your state at that moment in time.

sokrates
#4
Sep22-09, 11:15 PM
P: 483
Intuitive Understanding for Commutation

Similar to what happens after sequential Stern Gerlach measurements with non-commuting operators?

I think I get it. But still, I can't see why it sits so deep in the formal theory. Why is it so important?
quantumkiko
#5
Sep23-09, 12:50 AM
P: 29
Quote Quote by sokrates View Post

My objective is to get a good intuitive feeling of what commuting and not commuting observables imply about the physical system? What are good physical examples to justify this?
Let me try.

Operators in quantum mechanics are also called observables. The name stems from the fact that operators represent what can be observed or measured in a system. Position, momentum and energy are the observables we are introduced to in quantum mechanics. We can measure/observe them.

Remember that to observe a system, we have to somehow "shine light" on it. And we can only make measurements if we have a means to observe the system, which in our case by using light. This act of "shining light" on a system mathematically means making an operator operate on a mathematical representation of the system, the state. We have to disturb the state. We cannot make an observation unless we disturb the state.

Now what does commutation have to do with this? For example, we first want to measure the position of a particle at some instant of time. So we operate a position operator on the particle's state, etc. We'll get the position measurement needed. Now suppose we want to make another position measurement right after the previous one. We again operate a position operator on the state, etc. And this will, of course, give us the same position measurement we had from the first one. Two simultaneous measurements of the same observable are no different from each other. This is what [x, x] = 0 means.

Let's now make a momentum measurement of the particle we were left with, right after we've measured its position. Operating a momentum operator on its state, and proceeding with the mathematics, we'll get our momentum measurement. Simple. Now here's the fuzzy part. Suppose we go back to measure the position of the particle right after our momentum measurement. The position of the particle will no longer be equal to the position measurement we had before. This is what [x, p] not equal to zero means. The observables x and p do not commute. The outcome depends on the order of measurement, xp or px. We measure one observable at the expense of destroying the certainty of the other.

If we want to make measurements at different instances of time. Then we first have to make the system evolve in time. We do this by applying the Hamiltonian operator H first. Intuitively, the commutation relations of H with x and p determines how the position and momentum measurements change in time.
Fredrik
#6
Sep23-09, 04:36 AM
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Quote Quote by sokrates View Post
But still, I can't see why it sits so deep in the formal theory. Why is it so important?
Because without it, there's no uncertainty principle.
sokrates
#7
Sep23-09, 09:33 AM
P: 483
^^ Not something I get either. It is a circular argument.
meopemuk
#8
Sep23-09, 12:39 PM
P: 1,746
Quote Quote by sokrates View Post
But still, I can't see why it sits so deep in the formal theory. Why is it so important?

The true reason for commutators is related to the fundamental role played by (space-time) symmetries in quantum theory. The principle of relativity demands that the group of inertial transformations of observers (either the Galilei group or the Poincare group) should be represented by unitary operators in the Hilbert space of any physical system. The Hermitian generators of this group representation are identified with operators of basic observables (momentum, energy, angular momentum,..). These generators form a Lie algebra, and the Lie bracket is represented by commutator in the Hilbert space.

So, commutators between various observables are uniquely determined by the symmetry group. Unfortunately, this symmetry-based view on quantum mechanics is not well represented in textbooks. The best references known to me are

L. E. Ballentine, "Quantum Mechanics: A Modern Development", (World Scientific, Singapore,1998)

S. Weinberg, "The quantum theory of fields", vol. 1 (University Press, Cambridge, 1995); chapters 2-3.
RedX
#9
Sep23-09, 01:35 PM
P: 969
As meopemuk mentioned, commutators are important in Lie groups where they specify how elements multiply. This is useful not only for quantum mechanics but also classical mechanics. Say you want to rotate about the x-axis 37 degrees, and then the y-axis by 46 degrees. Then you can ask the question through what angle and what axis are these two operations equivalent to a single operation? So you need to know how two group elements multiply. But knowing the commutation relations of the generators specifies the group and tells you how the group multiplies (e.g., you can use the Baker-Campbell-Hausdorff formula which involves commutators). So one important use of commutators is to specify non-Abelian transformations.

Just to be a little clearer on what I said earlier, assume T|t>=t|t>. Then consider the transformation Q.

TQ|t>=QT|t>+X|t>
=t Q|t>+X|t>

where X is the commutator.

So you are finding T acting on the new state, Q|t>, and it too has eigenvalue t, up to the X|t> term at the end. So the commutator X takes the state Q|t> out of the subspace labeled by eigenvalue t.
sokrates
#10
Sep23-09, 02:25 PM
P: 483
Many thanks to everyone for the insights...!


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