The Time-Energy Uncertainty Relation

John Baez

April 10, 2010

In quantum mechanics we have an uncertainty relation between position and momentum:

(Δq) (Δp) ≥ /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:

(ΔT) (ΔE) ≥ /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 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's Quantum Mechanics Section VIII.13).

Let H be the (time-independent) Hamiltonian of some non-relativistic system. Let ψ be a wavefunction and let A be some other observable. Write

<A> = <ψ, A ψ>

for the expectation value of A in the state ψ, write sqrt for square root, and define

ΔA = sqrt(<ψ, (A - <A>)2 ψ>).

ΔA is the standard deviation of the observable A in the state ψ.

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

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

2 (ΔA) (ΔH) ≥ |<[H,A]>| .

ΔH is the standard deviation of the energy E.

< [H,A] > = <ψ, [H,A] ψ>

is i times the time derivative at t = 0 of <ψ, A ψ>, as you can see if you note that the solution to the Schrödinger equation can be written in the form

U(t) ψ = exp(-itH/) ψ


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

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

(ΔA / (|d <A>/dt)|) (ΔH) ≥ /2.

Here the "uncertainty" in time is expressed as the average time taken, starting in state ψ, 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 ψ.

Hey, that's way cool! For some reason I'd never thought of it that way. But here's something related, which is well-known:

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

[H,T] = i

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

(ΔT) (ΔH) ≥ /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

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 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

(ΔA / (|d<A>/dt)|) (ΔH) ≥ /2


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

Much later, Dmitry A. Arbatsky wrote:

You should mention the paper where the mathematically rigorous formulation of the time-energy uncertainty relation was first given. (It was given there even in nice finite form, not only infinitesimal. In 2005 it was generalized. It turned out that relations for energy and time in Mandelshtam-Tamm formulation, on the one hand, and for coordinate and momentum, on the other hand, are particular consequences of a more general approach.)

With best wishes,
Dmitry A. Arbatsky

Not till we are lost ... do we begin to find ourselves and realize where we are and the infinite extent of our relations. - Thoreau

© 2010 John Baez