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As the title says, does a particle lose its location wavefunction if its location is measured exactly (I know this would be impossible in reality)? Also, in reality, if one measures a particle, does the wavefunction of a particle become something different from original afterwards? |
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The wavefunctions form a vector space (the Hilbert space). In fact, from the Hilbert space, we keep only the wavefunctions with norm equal to $1$ - the directions. Please note that the directions are in the Hilbert space, which is a space of functions (the wavefunctions), so in general it has nothing to do with the space directions. The state of the wavefunction changes after the measurement. In terms of the Hilbert space, each type of measurement has associated a basis in the Hilbert space (or at least a decomposition of the Hilbert space in orthogonal subspaces). This decomposition of the Hilbert space depends on the property we measure. After the measurement is performed, the particle ends up in one of these directions in which is decomposed the Hilbert space. In other words, the final state is the projection of the initial state on one of these directions. So, the original wavefunction is changed. If one measures the position, the wavefunction gets concentrated in a small region. This corresponds, in the Hilbert space, to a projection. The initial wavefunction is radically changed. Added I'll give you a fictional example. Think instead of the Hilbert space, at the space of vectors in plane. We take the vectors of unit length. Let's also consider a fixed orthonormal basis. Let's invent a property called "A". A vector has the property A if it is vertical or horizontal with respect to that basis. That is, one of its coordinates vanishes. Hence, the other one is $\pm 1$. Let's consider another property, $B$. A vector has the property B if it is "diagonal", that is, its coordinates are $\pm\frac{1}{\sqrt 2}$. If the original state has the property A, and we want to measure the property B, we project it on one of the diagonal directions. You can't tell what the original vector was, just from its projections. Returning to a space of functions over the physical space $\mathbb R^3$, these functions have a scalar product, defined by using integrals. With this space we can project them. What could be a basis in a space of functions? Think at a power series expansion - in this case, the basis is given by power functions $x^n$. Or a Fourier series expansion - the basis would be made of functions of the form $\sin(nx)$. Another basis can be of Dirac functions, which are zero everywhere except at a point a, where they are infinite in such a way that the integral is equal to 1. Each other function can be expanded in one of these bases. A projection on one of these bases will give a function which belongs to the basis. Similar things happen in the Hilbert space used in quantum mechanics. |
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The particle doesn't "lose" its location wavefunction, rather its wavefunction changes to one which is sharply peaked around the location which results from the measurement. In the Copenhagen interpretation, the wavefunction changes smoothly with time until a measurement, such as this one, is performed, after which it abruptly changes to a new one. The wavefunction represents the particle state, and in the language of states, the state evolves unitarily until a measurement is performed, at which time it abruptly changes to an eigenstate of the operator representing the quantity being measured. Note that there are different pictures to this Copenhagen one, in particular ones involving decoherence, which do a more thorough job of describing what happens to the wavefunctions of the measuring apparatus, measured system and environment. These show how the appearance of a sudden change in the particle wavefunction can arise, whilst in reality, everything continues to evolve unitarily. |
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