# Understanding E/t uncertainty and $\vec{L}$ uncertainty

1. May 1, 2015

### davidbenari

Warning: I am taking a modern physics course, and haven't taken QM. I know nothing of "commutators, operators, hilbert spaces, etc."

I understand $\Delta E \Delta t >= \hbar /2$ to mean that I can't know the energy of a system and the time at which that energy takes place exactly. These two are fuzzy, so to speak.

However, when deriving the particle in a box wavefunction in 1D or 3D, you get sharp values for the energy of the system. At the same time that wavefunction is clearly a function of time and therefore I can know both $E$ and $t$ with precision. Why isn't this in contradiction with the E/t uncertainty principle?

Also I've seen some proofs that state that $\vec{L}$ can't be known because it would violate the p/x uncertainty principle. But why does this mean that I can't know two components of $\vec{L}$ with precision? I mean, the natural implication of not knowing L with certainty, would be that its 3 components aren't known. I don't see the importance of 2 components here...

Thanks!

2. May 1, 2015

### The_Duck

No, this is not the right statement of the time-energy uncertainty relation.

In the time-energy uncertainty relation, $\Delta E$ is the uncertainty in the energy of a system, and $\Delta t$ is the time it takes the system to change appreciably. These can be defined mathematically.

For example, states of definite energy ($\Delta E = 0$) have "stationary" wave functions where nothing ever changes except for an overall phase ($\Delta t = \infty$).

The canonical example of the time-energy uncertainty relation is the natural width of spectral lines. Since excited atoms return to the ground state by emission of a photon after a certain amount of time $\Delta t$, the energy of the excited state is uncertain by an amount $\Delta E$ which is inversely proportional to its lifetime $\Delta t$. This means that different photons from the same transition will have slightly different energies; that is, spectral lines are not infinitely sharp but have some width. Excited states that decay faster give rise to broader spectral lines.

There is a sort of uncertainty relation between each pair of components of $\vec L$ (we say that the components do not commute), so no two of them can take definite values at the same time (except in a case like $L_x = L_y = L_z = 0$). These uncertainty relations between the components of $\vec L$ can be derived from the uncertainty relations between $x$ and $p_x$, $y$ and $p_y$, and $z$ and $p_z$. You'll see this in a full course on QM.

Last edited: May 1, 2015
3. May 1, 2015

### davidbenari

I think I understood everything except this little bit. What do you mean exactly by this?

Is your interpretation hard to derive from the p/x relation which I know how to derive?

Hmm. How can there be different energies if the same transition occurs? I thought the quantization of energy would imply that same transitions have to have the same energy difference.

4. May 1, 2015

### The_Duck

I mean that since the system isn't changing at all, the "time required for it to change appreciably" is basically infinite.

Well, if you are familiar with the understanding of the p/x uncertainty relation in terms of the Fourier transform, the E/t uncertainty relation is exactly analogous.

Griffiths' QM book derives a rigorous E/t uncertainty relation using operators.

Well, the point is that excited atoms don't have definite energy. If they did, they would stay excited forever because states of definite energy are stationary.

Atoms only have definite discrete energy levels if you neglect their interaction with the electromagnetic field; that is, if you neglect the possibility of emitting photons.

5. May 1, 2015

### davidbenari

Could I relate the last point to a superposition of states? Like: two atoms have same wavefunction and emit a photon. "They underwent the same transition" in the same $\Delta t$ but the E/t relation will give the $\Delta E$ uncertainty.

Is that ok?

I don't get how else you could observe the same transition from two atoms, and observe the same $\Delta t$. Hehe.

Thanks!