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What does state mean in quantum mechanics? 
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#1
Jan2912, 04:07 AM

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I'm currently self learning quantum mechanics from Griffith's book. I'm in the first chapter and he's using phrases like "a particle in state Ψ", but he doesn't explain what it means by "state".
Can someone please enlighten me? 


#2
Jan2912, 04:30 AM

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#3
Jan3012, 03:09 AM

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PF Gold
P: 2,912

In ordinary probability theory a state is simply a list of non negative numbers that adds up to one and gives the probability of getting its corresponding outcome. It's similar in QM  the outcomes are the complex vector elements of the QM state space relevant to your problem and a state allows you to calculate the probability of those outcomes. Specifically it is a positive definite operator P of trace 1 and Tr(P u><u) gives the probability of getting the vector u> (u> normalized).
I share your frustration  the above probably uses notation Griffith has not introduced  at least from what I seem to recall when reading it ages ago myself. Its really 'bad' when they don't explain a concept and its true explanation requires concepts beyond what they have imparted. I got frustrated with that sort of stuff all the time. Thanks Bill 


#4
Jan3012, 03:53 AM

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PF Gold
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What does state mean in quantum mechanics?
The state is what you need to know to be able to assign probabilities to all possible* results of all measurements. In particular, if you know the state and what measuring device is being used, you will be able to assign probabilities to the possible results of the measurement you're doing right now.
If you know the preparation procedure that the particle has been subjected to, you have enough information, but that doesn't mean that preparation=state, because it's possible that many different preparation procedures are associated with the same probability assignments. So a state should be thought of as an equivalence class of preparation procedures. There are many ways to represent a state mathematically. In a book like Griffiths (which I have only read a very small part of), the only mathematical representation of a state is a wavefunction. In classical mechanics, a state is usually represented by a pair (x,v) or (x,p). x is position, v is velocity and p is momentum. The reason for that is that the theory is built up around a differential equation of the form x''(t)=f(x'(t),x(t),t), which has a unique solution x for each initial condition. An initial condition is a pair of equalities that specify the values of the two functions at a single point in their domain: x(t_{0})=x_{0}, p(t_{0})=p_{0}. If you know the solution x, you can compute v and p. So if you know the value of (x(t),v(t)) or (x(t),p(t)) at one time, you can determine it at all times. Note that this is consistent with the more general definition of "state" that I opened with, because knowing the state of a classical system allows you to assign probabilities to each possible* result of each measurement. We don't usually talk about classical mechanics in this way, because the probabilities are always 0 or 1 when we have the maximum amount of obtainable information about the system. (The maximum amount of information that would be obtainable in principle in a universe that's exactly as described by this classical theory). By the way, if we have all the relevant obtainable information, the state is said to be pure. If we can only assign a probability to each preparation procedure that may have been used, the state is said to be mixed. Wavefunctions in QM represent pure states. Mixed states are represented by density operators. *) By "possible results of measurements", I mean those numbers that the measuring device is capable of telling us is the result. I don't mean that every possible results can actually happen in a realistic situation. 


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