A key relation in optical spectroscopy also used to derive natural bandwidth is:

$\Delta E \Delta t=ℏ$

$\Delta E$ is the energy uncertainty of a state, $\Delta t$ should be the uncertainty of the lifetime $τ_b$ of the state. I don't understand why $\Delta t$ can be confounded with the lifetime itself, $τ_b$, and the relation written as

$\Delta E τ_b=ℏ$

  • $\begingroup$ Where does it say $E$ and $t$ are not conjugated? AFAIK you can get the energy-time-uncertainty just by rearranging the position-momentum-uncertainty. $\endgroup$
    – Feodoran
    Commented Feb 19, 2019 at 10:36
  • $\begingroup$ @Alchimista It ($E$ and $t$ not being conjugated) is not part of the question, yes. But since you mentioned it, and I have not heard such statements before, I was hopping you had a reference where you found such statements. $\endgroup$
    – Feodoran
    Commented Feb 19, 2019 at 14:15
  • $\begingroup$ @fedoran here in Chem or Phys SE. I am trying to find it as I was curious already. I feel the discussion boil to be as the situation in Q but I am also going do delete the previous comments as for they can be detrimental and distracting. I could make a question from scratch, if necessary. $\endgroup$
    – Alchimista
    Commented Feb 19, 2019 at 15:14
  • $\begingroup$ Also related the discussion here and link therein physics.stackexchange.com/questions/365873/… $\endgroup$
    – Alchimista
    Commented Feb 19, 2019 at 15:30
  • $\begingroup$ I would understand such a treatment (lifetime = delta t) only assuming the excited state as a very short leaving ones, ie one that would decay "immediately" but it survives just because of uncertainty. $\endgroup$
    – Alchimista
    Commented Feb 19, 2019 at 15:38

3 Answers 3


For practical purposes the relevant aspect of the relation between the uncertainties in time and energy is that it is a Fourier relationship that explains how peak widths in a spectrum (with linewidths in frequency units) relate to the decay rates of time-dependent signals. A decay can be considered the statistical result of many independent events with an average probability of occurring described by a time constant $\tau$. It can further be shown via the Fourier transform that when a signal decays exponentially with time constant $\tau$, this leads to a Lorentzian peak shape with a width at half height proportional to $\tau^{-1}$. It is even simpler to show this in the case of a Gaussian shaped decay with linewidth ~$\sigma_t$. The resulting frequency peak is then Gaussian with linewidth ~$\sigma_t^{-1}$. When you multiply the linewidth parameters of the Gaussian functions you get:


If you use Planck's relation $E=h\nu$ if follows that


Now - as mentioned earlier - you can think of the width of the Gaussian or Lorentzian time-dependent decay as proportional (or described by) a time constant $\tau$ (which is the default description in the case of exponential decays). It follows that

$$\sigma_E~\tau \approx h$$


The time energy uncertainty principle arises from the fact that a function can be described as a sum of terms in a conjugate variable, as is the case of any fourier transform. In this case the variables are frequency and time. The time-energy uncertainty does not seem to have any more fundamental meaning than this, unlike the Heisenberg position-momentum uncertainty principle.

As an excited state decays it is necessary to write the wavefunction in a time dependent form. If a pure state has energy $E$ and wavefuntion $\psi$ we can write the time-dependent (or total) wavefunction as $\displaystyle \varphi=\psi e^{-iE_0/\hbar}$ (where $i=\sqrt{-1}$ and $\hbar =h/2\pi$ ) and is a stationary state because the probability of finding it in that state is $|\varphi|^2=|\psi|^2$ which is a constant.

However, if the state decays, because of interactions between electrons/nuclei within a molecule or with nearby molecules (collisions for example), and if the lifetime is $\tau$ then the probability of being in the excited state at time $t$ is $\displaystyle |\varphi|^2=|\psi|^2 e^{-t/\tau}$ which means that

$$ \varphi(t)=\psi e^{-iE_0t/\hbar-t/2\tau}$$

expressing this as a product $\displaystyle \varphi=\psi e^{-iE_0t/\hbar}e^{-t/2\tau}$ and using the mathematical relationship $e^{ix}=\cos(ix)+i\sin(ix)$ shows that the real part of $\varphi$ oscillates due to the cosine term and decays away due to the exponential term in $\tau$. i.e a damped oscillation.

The calculation is easier by making some substitutions, for example the frequency is $\omega_0=E_0/\hbar$ and decay rate constant $\gamma=1/\tau$ then $\displaystyle \varphi=\psi \cos(\omega_0 t)e^{-\gamma/2}$.

As the excited state is decaying the frequency of the emitted radiation is not monochromatic, which it would be if the state did not decay, but has a frequency distribution that is related to the exponential decay by a fourier transform. Put another way, the exponential decay can be constructed out of a set of waves each with a different frequency and amplitude, just as any normal function can be constructed as a fourier series. This means that we can find a function $A$ such that

$$\varphi(t) =\int A(\omega)e^{i\omega t}d\omega$$


$$A(\omega) \approx \frac{1}{i(\omega-\omega_o)+\gamma/2}+\frac{1}{i(\omega+\omega_o)+\gamma/2}$$

The second fraction approaches zero as $\omega \to \omega_0$, i.e close to the transition frequency (or transition energy) and so is insignificant compared to the first term where the frequency difference tends to zero. The intensity of the transition is $A(\omega)^*A(\omega)$ where * indicates the complex conjugate giving,

$$I(\omega)= \frac{\gamma/2}{(\omega-\omega_0)^2+(\gamma/2)^2}$$

where the intensity is normalised to unity over all frequencies. The full width at half maximum of the transition is

$$ \Delta\omega =\gamma , \text{ or} \qquad \Delta E\tau =\hbar $$

which is the 'time-energy' uncertainty after substituting back for the energy.


I've finally found the answer in Griffiths' Quantum Mechanics Ch.3; The $\Delta t$ in the energy-time uncertainty principle is not the standard deviation of a collection of time measurements (like $\Delta E$) but it is the time the system takes to change substantially;

As a mesure of how fast the system is changing, let us compure the time derivative of the expectation value of an Oobservable $Q(x,p,t)$:

$\frac{d}{dt} \langle Q \rangle = \frac{d}{dt} \langle \Psi | Q \Psi \rangle = \langle \frac{\partial \Psi}{\partial t} | Q \Psi \rangle +\langle \Psi| \frac{\partial Q}{\partial t} \Psi \rangle + \langle \Psi | Q \frac{\partial \Psi}{\partial t} \rangle$

Now, the Schroedinger equation says

$i\hbar\frac{\partial\Psi}{\partial t} = H \Psi$


$\frac{d}{dt} \langle Q \rangle = -\frac{1}{i\hbar} \langle H \Psi | Q \Psi \rangle +\frac{1}{i\hbar} \langle \Psi | H Q \Psi \rangle + \langle \frac{\partial Q}{\partial t}\rangle$

Given the fact that H is hermitian $\langle H\Psi | Q \Psi \rangle = \langle \Psi |HQ\Psi \rangle$, hence

$\frac{d}{dt} \langle Q \rangle = \frac{i}{\hbar} \langle [\hat A,\hat B] \rangle + \langle \frac{\partial Q}{\partial t}\rangle $

If you consider the operator $\hat{Q}$ time independent (Operators that depend explicitly by $t$ are quite rare), and the general form of the Uncertainty principle:

$\sigma_A^2 \sigma_B^2 ≥ (\frac{1}{2i} \langle [\hat A,\hat B] \rangle)^2$

If A=H and B=Q you have eventually:

$\sigma_H^2 \sigma_Q^2 ≥ (\frac{1}{2i} \langle [\hat H,\hat Q] \rangle)^2 = \left(\frac{1}{2i} \frac{\hbar}{i} \frac{d\langle Q \rangle}{dt}\right)^2 =\left(\frac{\hbar}{2}\right)^2 \left(\frac{d \langle Q \rangle}{dt}\right)^2$

and more simply

$\sigma_H^2 \sigma_Q^2 ≥ \frac{\hbar}{2}\left|\frac{d \langle Q \rangle}{dt}\right| $

By defining $\Delta E \equiv \sigma_H$ and

$\Delta t \equiv \frac{\sigma_Q}{d \langle Q \rangle/dt}$


$\Delta E \Delta t ≥ \frac{\hbar}{2}$

The important point to understand (which confused me) is that $\Delta t$ represents the amount of time it takes the expectation value of Q to change by one standard deviation (thus to $\tau_b$) and not the standard deviation of a collection of time measurements

  • $\begingroup$ Regarding your last sentence, this is like distinguishing between homogenenous versus inhomogenenous sources of line broadening. In fact, for a collection of time measurements both will occur, but the linewidth will reflect the lifetime you derive with the Heisenberg relation only in the limit of no inhomogeneous broadening mechanisms. $\endgroup$
    – Buck Thorn
    Commented Mar 2, 2019 at 10:55

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