# Question about Commutators and Uncertainty

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• physwiz222
physwiz222
TL;DR Summary
What does Position and Momentum not commuting have to do with particles not being allowed to simultaneously have defined position and momenta. I want a conceptual explanation.
In Quantum Mechanics a particle cannot have both a defined position and momentum due to the uncertainty principle we all know that. A reason for this is because the Position and Momentum Operators dont commute but it can be demonstrated with Fourier Transforms.

I know how to mathematically derive the generalized uncertainty principle and what uncertainty principle physically means. What I want to know is what is the connection between Position and Momentum not commuting the order mattering and a particle not being able to have simultaneously defined x and p. Like how are these 2 things connected physically.

I want to emphasize I am not looking for a mathematical derivation I know how the Generalized Uncertainty Principle is derived for any pair of observables and how Uncertainty related to Fourier Transforms. What I want is a conceptual explanation for why the fact that x and p operators dont commute means the particle cant have both simultaneously defined. I know the Math and how its derived I want a conceptual reason.

physwiz222 said:
What I want to know is what is the connection between Position and Momentum not commuting the order mattering and a particle not being able to have simultaneously defined x and p
Since the position and momentum operators do not commute, they cannot have any common eigenstates. That means it is impossible to find a state that has both a definite position and a definite momentum.

Vanadium 50, vanhees71, topsquark and 1 other person
It may surprise you that this feature is not unique to quantum mechanics. There is a classical equivalent related to a fairly deep theorem in symplectic geometry (the language of classical mechanics) called the non-squeezing theorem. Roughly, it tells you that no symplectic transformation (think canonical transformation, including normal time evolution in classical mechanics) can take a ball of radius R>r and send it through a hole of radius r in a plane defined by a pair of conjugate variable (i.e., x and p_x). Anyway, that's just a fun fact for you. Take it or leave it.

Last edited:
apostolosdt, topsquark and vanhees71
This is just the content of the standard Robertson-Heisenberg uncertainty relation, i.e., for any state the standard deviations of two observables obey
$$\Delta A \Delta B \geq \frac{1}{2} |\langle \mathrm{i} [\hat{A},\hat{B}] \rangle|.$$
It follows from the positive definiteness of the scalar product in Hilbert space and Born's rule about the probabilities for the outcome of measurement, i.e., from very basic principles of QT.

For a position-vector component and the momentum component in the same direction you have
$$[\hat{x},\hat{p}]=\mathrm{i} \hbar \hat{1},$$
and thus, independent from the state
$$\Delta x \Delta p \geq \hbar/2.$$

physwiz222 said:
What I want is a conceptual explanation for why the fact that x and p operators dont commute means the particle cant have both simultaneously defined. I know the Math and how its derived I want a conceptual reason.
Operators are abstract mathematical objects. It's impossible to say anything about them without mathematics.

Perhaps you could find a conceptual explanation for position and momentum being incompatible observables, free from the mathematical basis of QM?

vanhees71 said:
This is just the content of the standard Robertson-Heisenberg uncertainty relation, i.e., for any state the standard deviations of two observables obey
$$\Delta A \Delta B \geq \frac{1}{2} |\langle \mathrm{i} [\hat{A},\hat{B}] \rangle|.$$
It follows from the positive definiteness of the scalar product in Hilbert space and Born's rule about the probabilities for the outcome of measurement, i.e., from very basic principles of QT.

For a position-vector component and the momentum component in the same direction you have
$$[\hat{x},\hat{p}]=\mathrm{i} \hbar \hat{1},$$
and thus, independent from the state
$$\Delta x \Delta p \geq \hbar/2.$$
i already said I know how its derived mathematically. I know all this. I asked this for a conceptual reason for what position and momentum not commuting has to do with there being no common eigenstates.

PeroK said:
Operators are abstract mathematical objects. It's impossible to say anything about them without mathematics.

Perhaps you could find a conceptual explanation for position and momentum being incompatible observables, free from the mathematical basis of QM?
What I want is what does x and p not commuting order mattering have to do with there being no common eigenstates. Like I want a conceptual explanation as to what does non commuting operators have to do with there not being common eigenstates.

physwiz222 said:
What I want is what does x and p not commuting order mattering have to do with there being no common eigenstates. Like I want a conceptual explanation as to what does non commuting operators have to do with there not being common eigenstates.
Non commuting operators can have common eigenstates. They can't, however, share a complete set of common eigenstates.

physwiz222 said:
Like I want a conceptual explanation as to what does non commuting operators have to do with there not being common eigenstates.
There may not be a satisfying answer to that question. It’s like asking for a conceptual rather than a mathematical reason why ##\sin x=x-\frac{x^3}{6}+\frac{x^5}{120}…..##
It’s easy enough to derive the result, but aside from the mathematical derivation no obvious reason why that particular result and not some other makes sense conceptually.

physwiz222 said:
I asked this for a conceptual reason for what position and momentum not commuting has to do with there being no common eigenstates.
physwiz222 said:
What I want is what does x and p not commuting order mattering have to do with there being no common eigenstates.
Because that's how non-commuting operators work. As @Haborix commented in post #3, this property is not even unique to QM; it's true in any framework where you have non-commuting operators.

PeroK said:
Non commuting operators can have common eigenstates.
Yes, but this is not the case for position and momentum, which are the particular operators the OP asked about.

Nugatory said:
There may not be a satisfying answer to that question. It’s like asking for a conceptual rather than a mathematical reason why ##\sin x=x-\frac{x^3}{6}+\frac{x^5}{120}…..##
It’s easy enough to derive the result, but aside from the mathematical derivation no obvious reason why that particular result and not some other makes sense conceptually.
But you can see by graphing this series on a graphing calculator and seeing how including more terms the approximation of sin(x) gets better

physwiz222 said:
I know how its derived mathematically. I know all this. I asked this for a conceptual reason
What is the conceptual reason that 2+2 = 4?

Mathematics is nothing more than reasoning via a chain of concepts. If you "understand the mathematics but not the concepts" you don't understand the mathematics.

PeterDonis said:
Yes, but this is not the case for position and momentum, which are the particular operators the OP asked about.
Yes, but it complicates any conceptual explanation if the general case is more subtle - as indeed it is in QM.

Position and momentum are in some ways a special case. Not least because the HUP applies to any state. The general uncertainty principle in general gives an inequality that depends on the state vector.

I would say the nearest to a conceptual explanation would be why momentum is represented by the differential operator.

physwiz222 said:
i already said I know how its derived mathematically. I know all this. I asked this for a conceptual reason for what position and momentum not commuting has to do with there being no common eigenstates.
If you know the derivation, this should be clear.

It's also not saying that there are no common eigenstates at all, it's only saying that there are no complete sets of common eigenstates.

E.g., for angular momentum the state ##J=0,M=0## is an eigenstate for all three angular-momentum components, which are not commuting.

vanhees71 said:
If you know the derivation, this should be clear.

It's also not saying that there are no common eigenstates at all, it's only saying that there are no complete sets of common eigenstates.

E.g., for angular momentum the state ##J=0,M=0## is an eigenstate for all three angular-momentum components, which are not commuting.
I was asking specifically about x and p which have no common eigenstates. I should have been more precise.

vanhees71
At the risk of repeating others, the story I would tell is this. Taking the particular operators x and p, their commutation relation tells me that any particular eigenstate of one of the operators is necessarily transformed by the other operator to a new ray in Hilbert space (i.e., doesn't just change the magnitude). Knowing that, I can now deduce that were I to prepare a physical system in an eigenstate of one of the operators, then that state must be a sum of multiple eigenstates of the other operator. So, when I go to take a measurement of momentum for a state prepared in an eigenstate of position, I will necessarily get a distribution over the multiple values of the momentum. In other words, the commutation tells me that I can never get around measuring a distribution of values for one of the operators. Now, this rushes over a ton of subtleties, most glaringly that eigenstates of position and momentum are not in the physical Hilbert space of states. Your quest to "physicalize" every intermediate step of quantum mechanics will probably be unfruitful but I understand the urge.

Haborix said:
At the risk of repeating others, the story I would tell is this. Taking the particular operators x and p, their commutation relation tells me that any particular eigenstate of one of the operators is necessarily transformed by the other operator to a new ray in Hilbert space (i.e., doesn't just change the magnitude). Knowing that, I can now deduce that were I to prepare a physical system in an eigenstate of one of the operators, then that state must be a sum of multiple eigenstates of the other operator. So, when I go to take a measurement of momentum for a state prepared in an eigenstate of position, I will necessarily get a distribution over the multiple values of the momentum. In other words, the commutation tells me that I can never get around measuring a distribution of values for one of the operators. Now, this rushes over a ton of subtleties, most glaringly that eigenstates of position and momentum are not in the physical Hilbert space of states. Your quest to "physicalize" every intermediate step of quantum mechanics will probably be unfruitful but I understand the urge.
Makes sense thanks for the explanation

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