# Heisenberg uncertainty principle and lifetime of the excited state

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

$$\Delta E τ_b=ℏ$$

• 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. – Feodoran Feb 19 at 10:36
• @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. – Feodoran Feb 19 at 14:15
• @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. – Alchimista Feb 19 at 15:14
• Also related the discussion here and link therein physics.stackexchange.com/questions/365873/… – Alchimista Feb 19 at 15:30
• 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. – Alchimista Feb 19 at 15:38

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:

$$\sigma_f\sigma_t=1$$

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

$$\sigma_E\sigma_t=h$$

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

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

where

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

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

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

so

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

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

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

then

$$\Delta E \Delta t ≥ \frac{\hbar}{2}$$

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

• 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. – Buck Thorn Mar 2 at 10:55