# Understanding the Set of Commuting Observables

Hi,

I've had a question ever since my quantum classes that's pretty simple I guess, but still seems to elude me. So here it is:

One text I used for quantum (Liboff's "Introductory Quantum Mechanics") says that in classical mechanics, there is a "vector of the state" of a system, that contains all the information in a mechanical system. This vector he claims contains the positions and velocities of all particles of a system (this is all just chapter 1). Then he goes on to say that in quantum mechanics, the vector of the state of the system cannot contain all these quantities simultaneously--this is the nature of the Heisenberg uncertainty principle.

But wait a second--why do the positions and velocities of a system yield all the information in classical mechanics? Is Liboff talking about instantaneous positions and velocities, or is he talking about the trajectories (world line functions)

$$\vec x(t)$$ and $$\frac{d \vec x(t)}{dt}$$?

Because, basically, if you have determined the trajectory function

$$\vec x(t)$$ (a map $$\Re^3 \rightarrow \Re$$),

which you can do in mechanics, then you can also find the velocity $$\vec v(t)$$ at all points, the acceleration $$\vec a(t)$$...hell, I could find the 20th derivative of position if I wanted to! So what's with saying the system is "totally specified" by giving positions and velocities? I would argue that all you need is the position function $$\vec x(t)$$.

So then that's where my quantum mechanics questions come in. What would the Heisenberg uncertainty principle have to say about, for instance, the uncertainty in position and acceleration ? What about jerk? What about the seventh derivative of position? If the seventh derivative of position were the "observable" I wanted to measure in a quantum system, how would I do it? What would the operator associated with this observable be? This is one of the main reasons quantum has always confused me. In classical mechanics, you really can't seperate position, velocity, acceleration, etc. They're all just derivatives of the same fundamental world line (trajectory) function $$x(t)$$.

If anyone has any insight, it would be greatly appreciated. Thanks a lot!

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In quantum mechanics, particles do not have trajectories like those found in classical mechanics. And so, the connection between the velocity and the position of the particle is as simple as a single derivative.

The closest you can ever get to trajectories in quantum mechanics are probability functions, which only tells you the average position of the particle as a function of time, $$\overline{x}(t)$$. The astute student should then realize that taking the derivative of $$\overline{x}(t)$$ will not yield $$\overline{v}(t)$$.

Finding $$\overline{v}(t)$$ involves the change of representation of the "vector of the state" from the position representation to the momentum representation.

Galileo
Homework Helper
I think he's talking about instantaneous position and velocity (or momentum), because, as you said, if you know the position of the particle at all times, you can calculate anything you want. Basically you already know everything about the system.
But knowing the instantaneous position of a particle doesn't allow you to know it's momentum.
However, if you know both the position and momentum of a particle in a certain instant, you can calculate the subsequent motion of the particle. This is simply because Newtons 2nd law F=ma is a second order DE and needs 2 initial conditions (r(0) and v(0) for instance) to uniquely specify a solution.

FunkHaus said:
But wait a second--why do the positions and velocities of a system yield all the information in classical mechanics? Is Liboff talking about instantaneous positions and velocities, or is he talking about the trajectories (world line functions)

$$\vec x(t)$$ and $$\frac{d \vec x(t)}{dt}$$?

Because, basically, if you have determined the trajectory function

$$\vec x(t)$$ (a map $$\Re^3 \rightarrow \Re$$),

which you can do in mechanics, then you can also find the velocity $$\vec v(t)$$ at all points, the acceleration $$\vec a(t)$$...hell, I could find the 20th derivative of position if I wanted to!
Yes, but this is not practical to make predictions. It is analogue to record all the particle positions in the universe up to its end. At the end, you will know all the functions x = (x(t), t in |R) and all the derivatives of x.

Classical Mechanics proceeds in another way: instead of dealing with the points of the functions (e.g; x(t) and dx/dt(t)) it deals directly with functions (e.g. x and v=dx/dt) and a function of these functions (e.g. the hamiltonian H(q,p)). If the Hamiltonian is sufficiently nice, you have the fundamental result:
Given an initial condition (qo,po) and the Hamiltonian H(q,p), we know the functions q=(q(t),t in |R) and p (p(t), t in |R).

=> we replace the knowledge of the different functions x(t) of a class of particles (having the same Hamiltonian) by one point (qo,po) and one function H(p,q).

Therefore we can say that the position and velocity (at a given instant) defines completely the (the path of a) particle (assuming implicitly an Hamiltonian or if you prefer interactions):

FunkHaus said:
So what's with saying the system is "totally specified" by giving positions and velocities? I would argue that all you need is the position function $$\vec x(t)$$.
By the position and velocity at a given instant (assuming a given Hamiltonian or interactions).

FunkHaus said:
So then that's where my quantum mechanics questions come in. What would the Heisenberg uncertainty principle have to say about, for instance, the uncertainty in position and acceleration ? What about jerk? What about the seventh derivative of position?
In QM, you have the same relations, but with operators: X, V=dX/dt (etc for the other derivatives). However, with operators, things become more complicated: the path concept becomes meaningless. We have no more a classical path if the operators do not commute. The HUP just reflect this property.

Let's take the operators X and P=mV (no em field case). We have [X,P]=/=0 => the two operators have not the same eigenbasis.

=> [operator relation] (V=dX/dt) <=/=> (v=dx/dt) [eigenvalue relation]

where v and x are the eigenvalues of X and V.

=> The classical path concept is no more valid in QM: knowing all the positions at different times do not infer the knowledge of velocity at different times because V=dX/dt=P/m.
=> from an initial position and velocity we are not able to infer the path of a quantum particle.

Seratend.

reilly
Funkhaus -- You said "So what's with saying the system is "totally specified" by giving positions and velocities? I would argue that all you need is the position function."

But you cannot find x(t) without boundary conditions. How long does it take for a ball thrown down, perpendicular to the ground from z=L to z=0, to hit the ground?

Regards,
Reilly Atkinson

Last edited:
Thanks so much for all the replies. Gallileo especially brings up a good point about the second order differential equation of motion. I suppose what I should interpret Liboff to mean is that, if at some instant (ie for some value of the parameter t) you know the position and velocity of a particle, you can determine the entire remaining portion of the trajectory, for all values of t. This is certainly true for any second order diff eq (I'm probably overlooking a mathematical subtlety in claiming such a thing).

But I am still in the dark as to what the quantum mechanical operator for acceleration would be, for instance. I mean I know that
$$\hat x = x$$
and
$$\hat v = -i \hbar \frac{\partial}{\partial x}$$

Seratend you said for instance that
$$V = \frac{dX}{dt}$$
Which I suppose is the same as
$$\hat v = \frac{d\hat x}{dt}$$

But I don't see how
$$V = \frac{dX}{dt}$$
is the same as
$$V = -i \hbar \frac{\partial}{\partial x}$$.
And speaking of which, what would $$\hat a$$ be? Could you enlighten me? Thanks again.

FunkHaus said:
But I am still in the dark as to what the quantum mechanical operator for acceleration would be, for instance. I mean I know that
$$\hat x = x$$
No. You have X|x>=x|x> in the eigenbasis |x>. Do not mix the eigenvalue x with the operator X.
In QM we use functional relations on operators. From these functional relations, we may deduce some relations for a given basis or for a given vector.
If you are a beginner with QM, I advise you to try to formulate first the operator relations before their expression in a peculiar basis: it may avoid a lot of errors (e.g. mixing eigenvalues with operators).

FunkHaus said:
Seratend you said for instance that
$$V = \frac{dX}{dt}$$
Which I suppose is the same as
$$\hat v = \frac{d\hat x}{dt}$$
How can you suppose that (you are mixing operator relations and eigenvalues relations from your above definition)? Try to think a little on what the derivatives of operators could be (how they may be different from the derivatives of simple real valued functions) and their relations with the eignevalues of these operators.
If you develop a little, you will see that we only have this equation (the eigenvalues relation) iff (provided some nice properties on the operators such as "observables"), X=f(S) and V=g(S) where S is another observable. In other words [X,V]=0.

Operators offers a larger set of possibilities than the usual [eigenvalue] relation v(t)=dx(t)/dt.

FunkHaus said:
But I don't see how
$$V = \frac{dX}{dt}$$
is the same as
$$V = -i \hbar \frac{\partial}{\partial x}$$.
And speaking of which, what would $$\hat a$$ be? Could you enlighten me? Thanks again.
I hope you know the heisenberg and schroedinger representation. I have chosen the heisenberg representation of the operators V=dX/dt.
Where A_h= U+ A_s U (A_h operator in the heisenberg representation and A_s in the schroedinger representation, U the unitary time evolution).

I assume you know the time evolution equations for the operators (heisenberg representation): $\frac{d}{dt} A = \frac{i}{\hbar} [H,Q] + \frac{\partial Q}{\partial t}= \frac{i}{\hbar} [H,Q]$ if $\frac{\partial Q}{\partial t}= 0$

Now if you take A=X and H=P^2/2m+V(X), you recover by the definition dX/dt=V, the formula above (V=P/m). In the same way if you introduce dV/dt in the above equation, you get:

mdV/dt=m.Acc=dP/dt=$-\frac{\partial V(X)}{\partial x}$

Seratend.