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The Wikipedia says on the page for the uncertainty principle:

Mathematically, the uncertainty relation between position and momentum arises because the expressions of the wave function in the two corresponding bases are Fourier transforms of one another (i.e., position and momentum are conjugate variables).

Does that mean that position and momentum are just 2 different measurements of the same wave function? I.e., it is the same thing that is being measured, just in two different ways? Meaning, they are not really two different things, but two different views on the same thing?

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2 Answers 2

Can I provide a pedestrian answer? When you measure the position of a quantum, you project or force it to commit to a unique position, and from Fourier Analysis, this commitment requires all possible momenta. Think of focusing a quantum wave to single location (ala Dirac Delta), this would require a wave generator to combine all momenta (and therefore no unique momentum). On the other hand, a measured momenta, would hold for the wave throughout all space, and make its position totally arbitrary. Also, when you do measure a quantum system, you do change it as it changes your measurement apparatus, unless you just performed the very same measurement a moment before. A quantum measurement is generally not objective, since your apparatus and the quantum wave are both involved. Measuring is an active process in Quantum Mechanics, changing your lab (its measuring dials) and the quantum wave.

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Any measurement in physics is in general described by a probability distribution of different outcomes. This distribution depends both on the state of the system being measured and on the measurement apparatus, which are two different things. In quantum mechanics states are described by vectors in Hilbert space $\left|\psi\right>$ (wavefunctions may be seen as their coordinates in some basis), and measurements by Hermitian operators $\hat{A}$ acting on this space (this is the simpliest case, actually the formalizm is a bit more complicated). Probability distribution of measurement outcomes is given by eigenvalues of these operators, and average values of measured quantities by $\hat{\left<A\right>}=\left<\psi\right|\hat{A}\left|\psi\right>$.

Position and momentum measurements are described by two different operators $\hat{x}$ and $\hat{p}$, such that $\hat{x}\hat{p}-\hat{p}\hat{x}=i\hbar$. Their noncommutativity leads to Heisenberg uncertainty relations for variances of corresponding measurements, as described in wikipedia. So the answer is no, they are different things, measured with different apparatus, but if their are done on a system in a given state, their variances turn out to be related.

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Don't you need to use these operators on the wave function to obtain a given measurement? I.e. you would be observing the same wave function, from the point of view of two different operators, i.e. the wave function itself contains all the information about the system. Then you look at one or another aspect of this system through the operators? –  Frank Jul 20 '12 at 16:00
    
Yes, the idea is right. You have different ways to look at the system, and outcomes of your measurements depend both on the measurement you are doing and on the state of the system you are measuring. I was just confused with position and momentum being "the same thing that is being measured, just in two different ways", that is not correct, they are different phisical quantities. Of course nothing prevents you from measuring them for the same system, and the outcomes of these measurements will obey uncertainty relations. –  straups Jul 20 '12 at 17:53
    
Well, the image I have is that there is 1 "vector" and that you project it either on one base, or on another base, to get p or x. Is that correct? If so, I'm surprised that just changing the basis in which you project actually leads to two physical quantities of a different nature. For example, by "changing basis" again, can you get p from x or vice-versa? (I have a very simplistic image in mind of a 2D vector being projected on a basis, then on another basis rotated 45 degrees). –  Frank Jul 20 '12 at 17:58
    
That is exactly what you want - knowing the state you can infer the outcomes of any measurement. Your vector may be expressed in any basis. Squares of its coordinates in the eigenbasis of some observable, say position or momentum, are the probabilities to find the system in that state. And finding the system in an eigenstate means obtaining a definite value of corresponding observable. Quantum mechanics is in some sense all about changing basis, evolution of the system is described by a unitary evolution operator, which is nothing but a change of basis parametrized by time. –  straups Jul 20 '12 at 18:34
    
I see. But once I have projected the vector onto a basis, say the position basis, if I know how the momentum basis relates to the position basis, I can go straight from the position to the momentum, just by changing basis? I.e., I take the projections of my naive 2D vector and re-project them directly onto the second basis, that is, the data of the position and the basis change formulas are all I need to get the momentum? –  Frank Jul 20 '12 at 18:46

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