Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I'm studying Quantum Mechanics in my spare time from a general point of view (no technical details) so some fundamental question came into my mind:

How is it possible to detect a single photon without making any change to it?

How does observation is related to the total energy of a system and does the act of performing measurement alter the internal energy of the observed system as a general rule?

share|improve this question
add comment

2 Answers 2

up vote 4 down vote accepted

How is it possible to detect a single photon without making any change to it?

In general, if you have detected a photon in your experimental apparatus, you have changed it drastically. It may have disappeared completely, as in this bubble chamber picture:

enter image description here enter image description here

The colored diagram shows the photon in the picture that has materialized into an electron positron pair. How to we know there was a photon( the yellow line) when it does not leave a trace in the chamber? Because we know that it started from an electron interacting with the liquid in the chamber: the curly bit is the electron that has given most of its energy to the photon.

In general in particle physics if you detect a particle, you have disturbed it, except if you detect it as a missing energy and momentum and can thus identify its mass.

How does observation is related to the total energy of a system and does the act of performing measurement alter the internal energy of the observed system as a general rule?

It depends on what you count as total energy. Energy is conserved, it just changes phases. Subsystems give up energy or absorb the energy of the photon, as in the picture: the pair production took energy, that the photon gave up. The electron interacting with matter lost energy giving it to the photon.

And what about measurement in macroscopic scales; Can a measurement to a close system be without any consequences?

In a closed system whether classical or quantum mechanical, energy is conserved. Different equations for the conservation are necessary for the two regimes.

share|improve this answer
    
Thanks for your complete answer. So can we, conclude about measurement (observation) even in microscopic or macroscopic world, that it is an active (not passive) action? In a close system, i.e. a closed bottle of warm water, we cannot get information about system's inside without changing it's total energy, can we? –  Xaqron Apr 7 '13 at 5:06
    
in your example, the energy of the bottle continually falls as it radiates according (more or less) to the black body radiation.From that radiation we can get the temperature of the bottle fitting to the black body curve. It is not an isolated system. If you think of a Thermos bottle en.wikipedia.org/wiki/Dewar_flask ( a better approximation to an isolated system) yes, you will not know what is happening inside unless you interact with it ( a thermometer will bring in some energy and take out some energy). –  anna v Apr 7 '13 at 5:26
    
Yes I meant a closed one like your example. Can we conclude that getting information from a closed system always is at cost of decrease in it's internal energy? It seems to me, the act of observation is amplification of some part of energy of a system and if the system is closed (I mean something else, but thermodynamic concept of a close-system is OK to me as an analogy), then we notify that change because boundaries of system are well defined. Is that true for a closed system in that context? –  Xaqron Apr 7 '13 at 6:03
    
Well, it is true that one cannot get information from a closed system without disturbing it, and since information carries energy, some energy does come out, but it could be that the instrumentation for the measurement introduces more energy in the system than it takes out. Classically there will be a change in entropy but it will depend on the measurement setup what exactly happens. –  anna v Apr 7 '13 at 6:15
    
Thanks for your kindness and patience. brainyquote.com/quotes/quotes/a/alberteins383803.html –  Xaqron Apr 7 '13 at 6:38
add comment

The act of measurement causes a quantum system to collapse into an eigenstate of the operator associated with the measurement. So unless the photon wave function is in an eigenstate of the detection operator, it will be changed.

Unless a system is in an eigenstate of the hamiltonian it will not have a definite energy. If one measures the energy of the system, the wave function will collapse into a definite energy state.

share|improve this answer
    
Thanks but I need do research to understand your answer. And what about measurement in macroscopic scales; Can a measurement to a close system be without any consequences? –  Xaqron Apr 7 '13 at 3:04
    
I am not sure what you mean by measurement on macroscopic scales. The usual picture is that you have a quantum system and you use a macroscopic measuring device. I am also not sure what you mean by a closed system. I suspect you are thinking of more refined treatments where you treat both the measuring device and quantum system using quantum mechanics. One needs to start looking at the theory of quantum decoherence in that case which is starting to go quite a bit beyond the simple (but often adequate) treatment I was referring to in my answer. –  physicsphile Apr 8 '13 at 4:25
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.