# Differences in measurement between quantum physics and classical physics?

I know that there are differences in measurement due to quantisation, probability, and collapse in quantum physics, but I am having trouble explaining how these ideas relate to measurement in an understandable manner.

In what ways does measurement in quantum mechanics differ from measurement in classical mechanics?

• Commented Aug 21, 2015 at 16:12

In quantum phyiscs we have a state.

In classical physics we have a state.

The former is function from configuration space at each time and the later is a point in configuration space for each time.

What we call a measurement in classical physics reveals as much information as we want about the state without changing the state.

What we call measurement in quantum physics changes the state (to orthogonally project it onto an eigenspace of an operator). So in general it changes the state (unless it was already an eigenvector of that operator).

In classical physics we can control how precisely we learn the results of the measurement.

In classical physics the state of the measurement device can give information about the state of the subject as precisely as we want.

In quantum physics the state of the measurement device coevolves with the state of the subject in a such a way as give the different projections onto the different eigenspaces in a way that is not solely determined by the subject's state but only the frequencies of getting the projections is determined solely by the subject's state. Frequencies of results if you do the measurement again and again on different but identically prepared subject states. The relative frequencies are the relative size of the squares of the orthogonal projections.

All the essential elements are already mentioned in this answer and the linked post in comments. So I'll try to give you more intuition with examples.

In principle all measurements performed on a system, will disturb it in some way, be it a classical or quantum system. In the former case, the disturbance for the most part is negligible. Telescope observations are based on catching photons from the Sun (or other light sources) that have bounced off a comet or planet. By bouncing it is clearly implied that the photons exchanged momentum with the object (comet, planet, ...) and thus disturbed it (goes without saying that the object would be equally disturbed regardless of whether the photons went unnoticed by our telescope or not). But without the photons and the involved exchange, we wouldn't be able to measure the position of the object. This was an example where you can entirely ignore the disturbance that is caused, so if you will, a classical measurement.

Another nice classical example would be measuring the voltage across a resistor in a circuit. One way or another you will have to set up another circuit that bypasses the resistor, e.g. by measuring the current through a galvanometer connected to the resistor. In any way your measurement is inevitably disturbed as your reading of the voltage is reduced due to the secondary circuit.

So it is clear that even in classical systems a measurement involves some form of interaction with the measured object. So what's different for quantum systems? Short answer would be that in quantum system, the caused disturbance is almost never negligible anymore. For example instead of our earlier macroscopic setup of comet-telescope, consider now a tiny particle and a microscope. The precision with which we can define the position of the particle is limited by the resolution of our microscope which is $\alpha \lambda$ where $\alpha$ is some numerical factor (that depends e.g. on the diameter of the objective) and $\lambda$ the wavelength of the light used. So in order to improve our precision, $\lambda$ has to be made as small as possible, but this means ever higher energetic photons! The resulting scattering between the energetic photons and the particle will involve huge jolts to the system, entirely changing its momentum state.

But more generally, the act of measuring the system will jog its original state onto another one. A bit more formally, if our system is originally in some coherent superposition of its eigenstates, the measurement will result in the system falling onto one of its eigenstates, with some probability. So if you consider the state of your system being $|\phi\rangle$ before the measurement, and that of the measurement device being $|M\rangle,$ the current uncoupled state of our global system will look like:

$$|\psi_{input} \rangle = |\phi\rangle |M\rangle = (a|\psi_1\rangle + b |\psi_2\rangle)|M\rangle$$ Notice that the input state can be factored as a product of its constituents. Now the output state will be an entanglement between the device's state and our system, in the form (with $\theta$ a non-zero real number):

$$|\psi_{output}\rangle = a |e^{i\theta} M \rangle |\psi_1\rangle + b |e^{-i\theta} M \rangle |\psi_2\rangle$$ Comparing the input and output states, you will hopefully be able to convince yourself of the consequence of the type of interaction involved between device-system during a measurement.

On a last note, a main fundamental difference between measurements in quantum systems (or more correctly a difference in measurability of system properties) and classical ones is the fact that in the former there are conjugate pairs of observables such as position and momentum that are not measurable simultaneously, not even in principle, whereas classically the state of the system corresponds to a point in its phase space at any instant in time $t,$ which by definition is a tuple of position and momentum $(\vec{r}(t),\vec{p}(t)).$ If you are interested, the book by J.J. Binney, The physics of Quantum Mechanics is filled with intuitive explanations and examples.