# Unusual uncertainty relations question

1. Jan 30, 2014

### liometopum

What are the uncertainty relations for the following:

1. position and energy?

2. position and time?

2. Jan 30, 2014

### atyy

The uncertainty relations come about from the commutation relations between canonically conjugate operators such as position and momentum.

The operator corresponding to energy is the Hamiltonian, which is specified in terms of canonically conjugate operators. So you would need to specify the Hamiltonian, derive the commutation relation, then derive an uncertainty principle.

There is no uncertainty relation for position and time, because time is not an observable, and there is no operator corresponding to time.

Once you have the commutation relation, the related uncertainty principle is given by Eq 4.46 of http://www.eng.fsu.edu/~dommelen/quantum/style_a/commute.html#SECTION07353000000000000000 .

Last edited: Jan 30, 2014
3. Jan 30, 2014

### liometopum

Thanks atyy. But a time-energy relation exists, although its "realness" is debated: http://www.math.ucr.edu/home/baez/uncertainty.html

So if a time-energy relation can be stated as h-bar/2 (there are several, at least, independent derivations of it), then I am thinking someone has done this with time and.position.

But what about position and energy? I have values and am looking to check them.

4. Jan 30, 2014

### WannabeNewton

In the Heisenberg picture, the Heisenberg equation of motion for the (time-dependent) position operator $X(t)$ is given by $\frac{dX(t)}{dt} = i[H,X(t)]$; then $V := \frac{dX(t)}{dt}$ is called the velocity operator. The exact form of $V$ will depend on the form of $H$. For example for a free particle we have $[H,X] = -i\frac{P}{m}$ as you would expect.

5. Jan 30, 2014

### atyy

I don't know it off the top of my head. It depends on the system, because the Hamiltonian or energy operator differs from system to system. I indicated how you can derive it in my previous post.

Edit: Check out WannabeNewton's post above.

6. Jan 31, 2014

### kith

The uncertainty principle for two observables A and B is ΔAΔB ≥ |<C>| with C = [A,B]. You cannot expect |<C>| to yield a general value like hbar/2 for arbitrary A and B because it is the expectation value of the operator C and thus depends on the state of the system.

A state-independent value libe hbar/2 can be given only in the case of conjugated variables like position and momentum because there, C is proportional to the identity operator, so its expectation value doesn't depend on the state.

/edit: For uncertainty relations involving time see wikipedia and the corresponding paper by Mandelstam and Tamm.

Last edited: Jan 31, 2014
7. Jan 31, 2014

### liometopum

Let me just share what I calculated, using my own method:

Time-position uncertainty
ΔT×Δx= (Gℏ)/(c⁴) = 8.7114×10⁻⁷⁹ m s

Energy-position uncertainty:
ΔE×Δx=(cℏ)/2= 1.58076×10⁻²⁶ J m

8. Jan 31, 2014

### liometopum

Oh yes, as a method of checking, if we rearrange the two uncertainty equations so that Δx=(cℏ)/(2ΔE) and Δx= (Gℏ)/(c⁴ΔT), set them equal and do the math, we get: ΔE/ΔT=c⁵/2G.
That expression, ΔE/ΔT, is one-half the Planck Power.

9. Jan 31, 2014

### OhYoungLions

What Hamiltonian are you referring to?

10. Feb 1, 2014

### liometopum

I did not use a standard QM approach. I developed my own method.

11. Feb 1, 2014

### ZapperZ

Staff Emeritus
Closed, pending moderation.

Zz.