The Time-Energy Uncertainty Relation

John Baez

March 11, 2000

(Delta Q) (Delta P) >= hbar/2

Now, as you probably know, time is to energy as position is to momentum, so it's natural to hope for a similar uncertainty relation between time and energy. Something like this:

(Delta T) (Delta E) >= hbar/2

There's an energy operator in quantum mechanics, usually called the Hamiltonian and written H. But the problem is, there's no "time operator" in quantum mechanics! This makes people argue a lot about the time-energy uncertainty relation - whether it exists, what it would mean if it did exist, and so on.

A while back on sci.physics.research, Matthew Donald (matthew.donald@phy.cam.ac.uk) wrote something interesting about this subject. I'm editing it a little bit here....

Most treatments of the time-energy uncertainty principle point out that you do have to be careful to consider the meaning of t. t isn't an operator in quantum mechanics.

Uncertainty relations are mathematical theorems as well as physical statements so if we begin with a proof we should end up with an exact definition of what we are trying to understand.

There are probably several forms in which the time-energy uncertainty relation can be proved. Here's one (for the full details, see Messiah ``Quantum Mechanics'' Section VIII.13).

Let H be the (time-independent) Hamiltonian of some non-relativistic system.

Let x be a wavefunction and let A be some other observable.

Write <A> = <x, A x>, sqrt for square root, and define

Delta A = sqrt(<x, (A - <A>)² x>).

Delta A is the standard deviation of the observable A in the state x.

Then, for all real numbers r, <x, (r (A - <A>) + i (H - <H>))(r (A - <A>) - i (H - <H>)) x> is non-negative.

So this quadratic (in r) cannot have two different real roots, and so, (cutting a long but standard story short)

2 (Delta A) (Delta H) >= |< [H,A] >| .

Delta H is the standard deviation of the energy E.

< [H,A] > = <x, [H,A] x>

is i hbar times the time derivative at t = 0 of <x, A x>. (You can see this if you note that the solution to the Schroedinger equation can be written in the form U(t) x = exp(-itH/hbar) x .)

Thus

< [H, A] > = i hbar d <A>/dt

Putting everything together, we have the time-energy uncertainty relation in the form

(Delta A / (|d <A>/dt)|) (Delta H) >= hbar/2.

Here the ``uncertainty'' in time is expressed as the average time taken, starting in state x, for the expectation of some arbitrary operator A to change by its standard deviation.

This is reasonable as a definition for time uncertainty, because it gives the shortest time scale on which we will be able to notice changes by using A in state x.

Suppose you could find an observable T which is canonically conjugate to the Hamiltonian H:

[H,T] = i hbar

Then by one of the formulas you wrote, we'd have

d <T>/dt = 1

so the observable T would function as a "clock" - it would increase at the rate of one second per second. In other words, we could use it as a "time" observable... which is why I called it T.

From your uncertainty relation we then have

(Delta T) (Delta H) >= hbar/2

the famous time-energy uncertainty relation that everyone keeps yearning for!

The problem is, for physically realistic Hamiltonians H one can prove there is no operator T with

[H,T] = i hbar

In other words, there is no time observable!

The reason is this: by the Stone-von Neumann uniqueness theorem, any pair of operators satisfying the canonical commutation relations [H,T] = i hbar can only be a slightly disguised version of the familiar operators p and q. These operators p and q are unbounded below - i.e., their spectra extend all the way down to negative infinity. But a physically realistic Hamiltonian must be bounded below!

(Here I am glossing over some mathematical nuances: if you read the precise statement of the Stone-von Neumann theorem, you'll see how to fill in these details.)

Crudely speaking, this theorem says that it's impossible to construct a clock that works perfectly no matter what its state is. That's not surprising - but it's sort of surprising that you can *prove* it, and it's sort of interesting to see what assumptions you need to prove it.

But what you're saying is: "So what? Let's use any operator A as a clock - we can't make d/dt = 1 in all states, but we can make it close to 1, or even equal to 1, in the state we're interested in! Then we can state the energy-time uncertainty relation even without having a time observable - we just say

(Delta A / (|d<A>/dt)|) (Delta H) >= hbar/2

instead!"

Thanks - you taught me something cool about time, which is one of my favorite subjects, right up there with space.

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