# The uncertainty principle and the measurement postulate

1. Apr 30, 2013

### la6ki

Hi everybody.

I have a quick question regarding the relationship between the uncertainty principle and the measurement postulate. According to the former, the higher our certainty is about the position of a particle, the lower our certainty is regarding its momentum, and vice versa. This means that our uncertainty about the momentum would be maximum if we knew the position of the particle exactly. Now, according to the measurement postulate, if we made a measurement and found the particle at a particular position, the wave function will collapse into a delta function, which means that the probability distribution for the position of the particle is now just that delta function. Furthermore, we know that if we were to measure its position again a very short time after the first measurement, we are bound to find the particle at the same position (the wave function remains collapsed for some small amount of time).

But this is where I'm confused. If we know where the particle is now and that we are going to find it at the exact same position after some short Δt, doesn't this mean that the particle just hasn't moved at all and therefore its momentum must be zero? Clearly this can't be true, since that would mean we know both the position and the momentum exactly.

I would really appreciate clarifying comments!

2. Apr 30, 2013

### jk22

if the particle is measured after a time delta t again it can be everywhere since the wavefunction is at that time a gaussian. (if we use schroedinger equ.)
The problem could arise when we take the limit delta t towards 0 but then it means that both measurement happened at the same time hence it is only one measurement.

3. Apr 30, 2013

### Fredrik

Staff Emeritus
In reality, no position measurement has infinite accuracy. So we will just be making approximate position measurements, and every time we do, it leaves the particle not in a delta function state, but in a state represented by a wavefunction that's peaked around the position indicated by the detector, and close to zero elsewhere. If the measurement is very accurate, the peak is very sharp, and the Fourier transform of the wavefunction is spread out over a large region.

An approximate momentum measurement is a series of approximate position measurements, that are inaccurate enough to not spread out the wave function's Fourier transform too much.

4. Apr 30, 2013

### jk22

If you use delta you can look under propagator in wikipedia.

5. Apr 30, 2013

### Jano L.

Hi la6ki,

your question is a good one. The paradox is due to [literal application of the] projection postulate.

The first problem is that there is no valid wave function that would correspond to one definite position. "Delta function", which you suggest, is associated with only one point, which is good, but is not really a function. It can be understood as distribution, but then it does not fit into the set of $\psi$ functions which have the Born interpretation, namely that the square of their modulus gives the probability density. This is because the square of $\delta$ is not a valid mathematical object.

Similarly, the function $e^{i\mathbf p\cdot\mathbf r/\hbar}$ is not normalizable wave function.

The proper conclusion from this is that a knowledge that particle has definite position or definite momentum cannot be transcribed into the description based on wave functions. Admissible wave functions have to be somewhere in between, i.e. they have to be not localized too much, but also not spread too much.

So let's assume we work with such "peak" functions instead.

If somehow (not trivial!) we perform what you describe at time $t_1$, then after the measurement we obtain the position $x_1$, and we assume that after the measurement the wave function is a peak centered at $x_1$.

In the interval $t_1..t_2$, it spreads out.

At time $t_2$, we do the same thing and obtain $x_2$.

We can calculate probable momentum as

$$p_x = m \frac{x_2 - x_1}{\Delta t}$$

and assign it to the particle and the time interval $(t_1, t_2)$. If we assume the particle moved uniformly, we can ascribe position and this momentum to it for any time in the interval $(t_1,t_2)$, although this is rather speculative as we do not know whether the particle moved from $x_1$ to $x_2$ uniformly (consider what happens during the Brownian motion). Heisenberg was aware of this and he did not deny it in his book Principles of Quantum Theory.

What Heisenberg tried to say is that when the position of the particle is measured at time $t_1$, the value $p$ known from before becomes unreliable, because the measurement of $x$ involves an interaction of the detector with the electron, which changes electron's momentum $p_x$ in an uncontrollable way. He showed that this happens in such a way that the product of statistical spread in $x_1$ and statistical spread in the conjugated momentum $p_x$ just after the measurement at time $t_1$ (calculated for a set of many measurements) is a quantity close to Planck constant.

So there is no direct contradiction with the Heisenberg principle, because our average $p_x$ for time $t_1..t_2$ does not correspond to $p_x(just~after~ t_1)$ which Heisenberg had in mind. In order to break HP, we would have to find somehow the momentum just after the measurement of position at $t_1$, but our measurement finds only average momentum during time $t_1..t_2$. We may find the momentum after $t_2$ by calculation, but this is only a posteriori and hypothetical.

(Soon after Heisenberg's lectures, Einstein, Tolman and Podolsky in their paper

"Knowledge of Past and Future in Quantum Mechanics", Physical Review, 1931

came to conclusion that retrodiction of past motion has to have similar uncertainty, so the assumption of uniform motion is dubious.)

It remains to ask: is there a contradiction with the mathematical uncertainty relation?

The answer is again no. In UR, the probable momentum and its uncertainty is not calculated from positions, but directly from the wave function, at one time.

In all instants of time, if we assume we have some admissible wave function, the product of uncertainties is calculated to be greater than $\hbar / 2$.

6. May 2, 2013

### la6ki

Thanks for the clarifying comments Fredrik & Jano!

Yeah, I think it makes sense now, the idea that what people mean when they talk about uncertainty in the momentum is at the exact moment at which the position is measured, and not after a particular time interval after the measurement.

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