## The Time-Energy Uncertainty Relation

#### September 26, 2021

In quantum mechanics we have an uncertainty relation between position and momentum: $$\Delta q \, \Delta p \ge \frac{\hbar}{2}.$$

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: $$\Delta T \, \Delta H \ge \frac{\hbar}{2}.$$

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 — at least, not in the usual formalism. 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 $$\psi$$ be a wavefunction and let $$A$$ be some other observable. Write

$$\langle A \rangle = \langle \psi, A \psi \rangle$$

for the expectation value of $$A$$ in the state $$\psi$$, and define

$$\Delta A = \sqrt{ \langle \psi, (A - \langle A \rangle)^2 \psi \rangle}.$$

$$\Delta A$$ is the standard deviation of the observable $$A$$ in the state $$\psi$$.

Then, for all real numbers $$r$$ $$\langle \psi, \big(r(A - \langle A \rangle) + i(H - \langle H \rangle)\big) \, \big(r(A - \langle A \rangle) - i(H - \langle H \rangle)\big) \psi \rangle \ge 0$$

so this quadratic (in $$r$$) cannot have two different real roots, and so, (cutting a long but standard story short) $$2 (\Delta A)(\Delta H) \ge |\langle [H,A] \rangle|$$ where $$\Delta H$$ is the standard deviation of the energy. On the other hand

$$\langle [H,A] \rangle = \langle \psi , [H,A] \psi \rangle$$

is $$i \hbar$$ times the time derivative at $$t = 0$$ of $$\langle \psi, A \psi \rangle$$, as you can see if you note that the solution to the Schrödinger equation can be written in the form $$U(t) \psi = \exp(-itH/\hbar) \psi.$$ Thus $$\langle [H,A] \rangle = -i \hbar \frac{d \langle A \rangle}{dt}.$$ Putting everything together, we have the time-energy uncertainty relation in the form $$\Delta A \, \Delta H \ge \frac{\hbar}{2} \left| \frac{d \langle A \rangle}{dt} \right|$$ or $$\frac{\Delta A}{|d\langle A \rangle/dt|} \, \Delta H \ge \frac{\hbar}{2}.$$ Here the 'uncertainty' in time is expressed as the average time taken, starting in state $$\psi$$, for the expectation of some arbitrary observable $$A$$ 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 $$\psi$$.

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 $$T$$ canonically conjugate to the Hamiltonian $$H$$: $$[H,T] = i \hbar.$$ Then by an equation above we'd have $$\frac{d\langle T \rangle}{dt} = 1$$

so the observable $$T$$ would function as a perfect '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 time-energy uncertainty relation we'd then have $$\Delta T \, \Delta H \ge \frac{\hbar}{2}$$ 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 well-behaved observable $$T$$ with $$[H,T] = i \hbar.$$ In other words, there is no time observable!

The reason is this: by the Stone–von Neumann theorem, any well-behaved 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! That is, energies can't be be arbitrarily large and negative.

(Here I am glossing over some nuances, which are tucked into the phrase 'well-behaved'. If you read the precise statement of the Stone–von Neumann theorem, you'll see how to fill in these details. You should also note how dropping any of the technical conditions in this theorem lets you come up with interesting counterexamples.)

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. A version of this argument goes back to Pauli's 1933 book Die Allgemeinen Prinzipen der Wellenmechanik

But what Matthew Donald said is: "So what? Let's use any operator $$A$$ as a clock — we can't make $$d\langle A\rangle /dt = 1$$ 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 $$\frac{\Delta A}{|d\langle A \rangle/dt|} \, \Delta H \ge \frac{\hbar}{2}$$ instead!"

So, he 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
• Leonid Mandelstam and Igor Tamm, The uncertainty relation between energy and time in nonrelativistic quantum mechanics, Izv. Akad. Nauk SSSR (ser. Fiz.) 9 (1945) 122-128. (English translation: J. Phys. (USSR) 9 (1945), 249–254.)
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 Mandelstam–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

Finally, I want to say that in some contexts we do have a time operator. Consider for example a particle on the line. Classically, the phase space for this particle is $$\mathbb{R}^2$$, with coordinates $$q,p$$, meaning position and momentum. But the 'extended phase space' for this particle is $$\mathbb{R}^4$$, with coordinates $$q,p,t,E$$. Thus, unlike in the usual formalism where we write the energy $$E$$ as some function of position $$q$$ and momentum $$p$$, here we treat it as an independent coordinate. We also have a coordinate $$t$$ for time.

Now, we can quantize this extended phase space and get a Hilbert space $$L^2(\mathbb{R}^2)$$. A vector in here is a function of position and time, say $$\psi(q,t)$$. On this Hilbert space we have self-adjoint operators $$\hat{q}, \hat{p}, \hat{t}$$ and $$\hat{E}$$. The position and momentum operators are defined in a fairly familiar way: $$(\hat{q} \psi)(q,t) = q \psi(q,t), \qquad (\hat{p} \psi)(q,t) = -i \hbar \frac{\partial}{\partial q} \psi(q,t) .$$ The time and energy operators are defined analogously, but with a slight twist: $$(\hat{t} \psi)(q,t) = t \psi(q,t) , \qquad (\hat{E} \psi)(q,t) = i \hbar \frac{\partial}{\partial t} \psi(q,t) .$$ Each of these pairs of operators obeys a version of the canonical commutation relations: $$[\hat{p},\hat{q}] = -i\hbar , \qquad [\hat{E}, \hat{t}] = i \hbar .$$ This doesn't contradict my earlier claim that in many physically realistic situations it is impossible to define a well-behaved time operator $$T$$ such that $$[H,T] = i\hbar$$. The reason is that $$\hat{E}$$ is unbounded below.

Classically, we can get from the extended phase space to the usual phase space by a process known as symplectic reduction. We impose the constraint $$E = f(q,p)$$ for some function $$f$$, and then mod out by the flow generated by $$E$$. We can do something analogous in the quantum case, at least under good conditions, and get a Hilbert space with a Hamiltonian $$H = f(\hat{q}, \hat{p})$$. But if $$f$$ is bounded below, so will be the operator $$H$$, and thus there can be no time operator $$T$$ on this Hilbert space obeying $$[H,T] = i\hbar$$ and the other conditions of the Stone–von Neumann theorem.