In all of my posts on the Uncertainty Principle, I left a few points open or rather vague, and that was usually because I didn’t have a clear understanding of them. As I’ve read some more in the meanwhile, I think I sort of ‘get’ these points somewhat better now. Let me share them with you in this and my next posts. This post will focus on the Uncertainty Principle for time and energy.
Indeed, most (if not all) experiments illustrating the Uncertainty Principle (such as the double-slit experiment with electrons for example) focus on the position (x) and momentum (p) variables: Δx·Δp = h. But there is also a similar relationship between time and energy:
ΔE·Δt = h
These pairs of variables (position and momentum, and energy and time) are so-called conjugate variables. I think I said enough about the Δx·Δp = h equation, but what about the ΔE·Δt = h equation? Indeed, we can sort of imagine what ΔE stands for, but what about Δt? It must also be some uncertainty: about time obviously–but what time are we talking about?
I found one particularly appealing explanation in a small booklet that I bought–long time ago– in Berlin: the dtv-Atlas zur Atomphysik. First, note that the uncertainty about the position (Δx) of our ‘wavicle’ (let’s say an electron) is to be related to the length of the (complex-valued) wave-train that represents the ‘particle’ (or ‘wavicle’ if you prefer that term) in space (and in time). In turn, the length of that wave-train is determined by the spread in the frequencies of the component waves that make up that wave-train, as illustrated below. [However, note that the illustration assumes the amplitudes are real-valued only, so there’s no imaginary part. I’ll come back to this point in my next post.]
Now, we can use the de Broglie relation (λ = h/p) to relate the uncertainty about the position to the spread in the wavelengths (and, hence, the frequencies) of the component waves:
p = h/λ and, hence, Δp = Δ(h/λ) = hΔ(1/λ)
In case you wonder why I can simply take h out of the brackets, i.e. why I can write Δ(h/λ) = hΔ(1/λ), just remember that the delta symbol here (Δ) refers to a measure like the standard deviation of a variable, so Δx represents σx. Now, one can prove the following:
- The standard deviation of some constant function is 0: Δ(k) = 0
- The standard deviation is invariant under changes of location: Δ(x + k) = Δx
- The standard deviation scales with the scale of the variable: Δ(kx) = |k |Δ(x)
It’s obviously the last rule that we’re using here.
Now, Δx equals h/Δp according to the Uncertainty Principle—if we take it as an equality, rather than as an inequality, that is. Therefore, Δx must equal:
Δx = h/Δp = h/[Δ(h/λ)] =h/[hΔ(1/λ)] = 1/Δ(1/λ)
That’s obvious, but so what? We cannot write Δx = Δλ, because there’s no rule that says that Δ(1/λ) = 1/Δλ and, therefore, h/Δp ≠ Δλ. Indeed, suppose we define Δλ as an interval or a length defined by the difference between its upper bound and its lower bound. Then we can write Δλ as Δλ = λ2 – λ1 and, hence, we can then write Δp as Δp = Δ(h/λ) = h/λ1 – h/λ2 = h(1/λ1 – 1/λ2) = h[λ2 – λ1]/λ1λ2. Now, that’s obviously something very different than h/Δλ = h/(λ2 – λ1). So we should surely not write that Δp = h/Δλ. Never ever. Having said that, the Δx = 1/Δ(1/λ) = λ1λ2/(λ2 – λ1) relationship that emerges here is quite interesting. I encourage you to explore it yourself, as I need to move on here.
So… We’re kinda stuck. What to do? How do we get that energy-time relationship? The de Broglie relation tells us that E = hν, so we can write that ΔE = Δ(hν) = hΔν. But we need to get ΔE = Δ(hν) = hΔν = h/Δt. How do we get Δν = 1/Δt, which is – obviously – the relationship that we need to get ΔE = h/Δt?
To get the answer to that question, we need to ask ourselves another one: what’s Δt here? What are we talking about?
The answer is remarkably mundane: Δt is the measurement time. What measurement time? Relax. You’ll understand in a moment. Let’s go through it.
We know there’s a universal relationship between the propagation speed of a wave (which I’ll denote by c for the time being, but don’t confuse this variable with the speed of light: it can be any speed) and the wavelength and frequency. More specifically, c = λν, and hence, 1/λ = ν/c. So we can now write Δ(1/λ) as Δ(ν/c) = Δ(ν)/c. We also know that the frequency of the wave is the reciprocal of the so-called period of the wave, i.e. the time that’s needed to go through one oscillation: τ = 1/ν and, hence, ν = 1/τ. Hence, we can write Δ(ν) = Δ(1/τ).
OK. That’s stating the obvious. So what? Where do we go from here?
First, note that, for a wavetrain, there’s no precise frequency or period, nor is there any precise number of oscillations. That’s the essence of the Uncertainty Principle in its most ubiquitous form (Δx = h/Δp). But so we can try to measure. Now, to measure something, we need some time. More in particular, to measure the frequency of a wave, we’ll need to look at that wave and register (i.e. measure) at least a few oscillations, as shown below.
I took the image from the above-mentioned German booklet and, hence, the illustration incorporates some German. However, that should not deter you from following the remarkably simple argument, which is the following:
- The error in our measurement of the frequency (i.e. the Meβfehler, denoted by Δν) is related to the measurement time (i.e. the Meβzeit, denoted by Δt in the diagram above). Indeed, if τ represents the actual period of the oscillation – which is the reciprocal of the frequency: τ = 1/ν) (both τ and ν are obviously unknown to us: otherwise we wouldn’t be trying to measure the frequency), then we can write Δt as some multiple of τ. More specifically, in the example above we assume that Δt ≈ 4τ = 4/ν. [Note that we use an almost equal to sign (≈) rather than an equality sign (=) because we don’t know τ (or ν). That’s the whole point about it, indeed.]
- During that time, we measure four oscillations in our example and, hence, we are tempted to write that ν = 4/Δt. However, because of the measurement error, we should interpret the value for our measurement not as 4 exactly but as 4 plus or minus one: 4 ± 1. Indeed, it’s like measuring the length of something: if our yardstick has millimeter marks, then we’ll measure someone’s length as some number plus or minus 1 mm. Here we are counting the number of oscillations. Hence, the result of our measurement should be written as ν ± Δν = (4 ± 1)/Δt = 4/Δt ± 1/Δt. If you have trouble following the argument, just put in some numbers in order to gain a better understanding. For example, imagine an oscillation of 100 Hz (i.e. 100 oscillations per second), and a measurement time of four hundredths of a second (i.e. Δt = 4×10–2 s). Suppose, then, we do indeed measure 4 ± 1 oscillations during that time. Then the frequency of this wave must be equal to ν ± Δν = (4 ± 1)/Δt = 4/(4×10–2 s) ± 1/(4×10–2 s) = 100 ± 25 Hz. In other words, we here accept that we have a measurement error of Δν/ν = 25/100 = 25%. That’s a relatively large error because the measurement time was relatively short, [Note that ‘relatively short’ means ‘short as compared to the actual period of the oscillation’. Indeed, 4×10–2 s is obviously not short in any absolute sense: in fact, it is like an eternity when we’re talking light waves, which have frequencies measured in terahertz.]
- The example makes it clear that Δν, i.e. the error in our measurement of the frequency, is related to the measurement time as follows: Δν = 1/Δt. Hence, if we double the measurement time, we halve the error in the measurement of the frequency. The relationship is quite straightforward indeed: let’s take the example of that 100 Hz wave once again and assume that our measurement time Δt is equal to Δt = 10τ = 10×10–2 s = 10–1 s. In that case, we get Δν = 1/10–1 s = 10 Hz. Hence, the measurement error is now Δν/ν = 10/100 = 10%.
- How long should the measurement time be in order to get a 1% error only? Let’s write the error as a percentage first: Δν/ν = x % = x/100. But Δν = 1/Δt. Hence, we have Δν/ν = (1/Δt)/ν = 1/(Δt·ν) = x/100 or Δt = 100/(x·ν). So, for x = 1 (i.e. an error of 1%), we get Δt = 100/(1·100) = 1 second; for x = 5 (i.e. an error of 5%), we get Δt = 100/(5·100) = 0.2 seconds. Finally, for x = 25 (i.e. an error of 25%), we get Δt = 100/(25·100) = 0.04 seconds, or 4×10–2 s, which is what this example started out with.
You’ll say: so what? We’re still nowhere… Well… No. We’ve got a formula with the frequency variable here, so we can now derive the Uncertainty Principle for time and energy from the other de Broglie relation (E = hν), which relates the energy of a ‘wavicle’ to the de Broglie frequency. Hence, the uncertainty about the energy about the energy must be related to the measurement time as follows:
E = hν ⇒ ΔE = Δ(hν) = hΔν = h(1/Δt) = h/Δt ⇔ ΔE·Δt = h
So, what this expression of the Uncertainty Principle says is the following: if we increase the measurement time, we’ll reduce the uncertainty in our knowledge of the energy of our ‘wavicle’. Conversely, if we only have a very short measurement time, we’ll not be able to say much about its energy.
A final note needs to be made on the value of h: it’s very tiny. Indeed, a value of (about) 6.6×10−34 J·s or, using the smaller eV unit for energy, some 4.1×10−15 eV·s is unimaginably small, especially because we need to take into account that the energy concept as used in the de Broglie equation includes the rest mass of a particle. Now, anything that has any rest mass has enormous energy according to Einstein’s mass-energy equivalence relationship: E = mc2. Let’s consider, for example, a hydrogen atom. Its atomic mass can be expressed in eV/c2, using the same E = mc2 but written as m = E/c2, although you will usually find it expressed in so-called unified atomic mass units (u). The mass of our hydrogen atom is approximately 1 u ≈ 931.5×106 eV/c2. That means its energy is about 931.5×106 eV. In plain language, that’s 931.5 million eV. Hence, if we’d be happy with an uncertainty of plus or minus one million eV, then it’s obvious that even very small values for Δt (i.e. very short measurements) will give us what we want. However, it is likely that we’ll want to reduce the measurement error to much less than plus or minus one million eV, so that means that our measurement time Δt will have to go up. Having said that, the point is still quite clear: we don’t need much time to measure the mass (or the energy) of this hydrogen atom very accurately.
The corollary of this is that the de Broglie frequency f = E/h of such particle is very high. To be precise, the frequency will be in the order of (931.5×106 eV)/(4.1×10−15 eV·s) = 0.2×1024 Hz. In practice, this means that the wavelength is so tiny that there’s no detector which will actually measure the ‘oscillation’: any physical detector will straddle most – in fact, I should say: all – of the wiggles of the probability curve. All these facts basically state the same: a hydrogen atom occupies a very precisely determined position in time and space. Hence, we will see it as a ‘hard’ particle, not as a ‘wavicle’.
That’s why the interference experiment mentions electrons, rather than hydrogen atoms or other ‘big stuff’, even if I should immediately add that interference patterns have been observed using much larger particles as well. However, I wrote about that before, so I won’t repeat myself here. The point was to make that energy-time relationship somewhat more explicit, and I hope I’ve been successful at that at least. You can play with some more numbers yourself now. 🙂
Post scriptum: The Breit-Wigner distribution
The Uncertainty Principle applied to time and energy has an interesting application: it’s used to assign a lifetime to very short-lived particles. In essence, the ‘spread’ around their mean energy (ΔE) is used to calculate their lifetime through the ΔEΔt = ħ/2 equation. I won’t say much about this, because Georgia University’s Hyperphysics website gives an excellent quick explanation of this, and so I just copied that below.