I've done an intro course on QM and I'm now hoping to understand exactly how to use probability theory rigorously in solving problems. My question is: How do I do the same thing, or the closest thing possible, in quantum mechanics?

To make my question more concrete, it's best to use an example - when solving problems with classical probability theory, I find you can set up problems using the following pattern

  • Random experiment: Toss two coins
  • Example of an outcome: $10 = (Heads, Tails)$
  • Sample space: $S = {11,10,01,00}$, $|S| = 4$
  • Examples of events: 2 Heads $= 2H = \{11\}$, $|2H| = 1$, $1H = \{10,01\}$, $|1H| = 2$, $0H = \{00\}$, $|0H| = 1$
  • Random variable: "Number of heads in $\omega$, $X(10)=1$

Translating this to quantum mechanics, my best attempt is:

  • Random experiment: Modelled by operators, Energy $\mathcal{H}$, Spin $S_x$,... Position or momoentum,
  • Example of an outcome: The result of applying an operator to a state vector, expressible as a linear combination of basis vectors.
  • Sample space: Set of all normalized l.c.'s of basis state vectors
  • Examples of events: Defining $\mathcal{H}|\phi_1>=\frac{\sqrt{2}}{3}|\phi_1>$, $\mathcal{H}|\phi_2>=\frac{\sqrt{3}}{3}|\phi_2>$ and $\mathcal{H}|\phi_3>=\frac{2}{3}|\phi_3>$ implies that $|\phi> = \frac{\sqrt{2}}{3}|\phi_1> + \frac{\sqrt{3}}{3}|\phi_2> + \frac{2}{3}|\phi_3>$ is a valid normalized state.
  • Random variable: a function assigning an outcome to it's eigenvalue, i.e. dual vector functionals, so $<\phi_1|$ is such that $<\phi_1|\mathcal{H}|\phi_1>= <\phi_1| \frac{\sqrt{2}}{3}|\phi_1>= \frac{\sqrt{2}}{3}<\phi_1|\phi_1> = \frac{\sqrt{2}}{3}$

It seems like all the work of quantum mechanics goes into constructing the domain on which the operator in your random experiment is operating on, so to me it is a complete mystery as to what the sample space actually is, thus the potentially a more concrete question: How do I fit the construction of quantum mechanical sample spaces containing state vectors into the framework of classical probability theory as closely as possible?

To clarify: any time I'm solving a problem like the harmonic oscillator or a new problem I'd like to be able to think about the mathematical principles behind what I'm doing, before solving the Schrodinger equation or anything like that. In classical probability you're just using set theory and you know what everything is going to look like the second you've figured out what your random experiment is, what is the analogue in QM?

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    $\begingroup$ You can use the density matrix formalism. In the case of a classical probabilistic problem, the density matrix has only diagonal terms. $\endgroup$ – Trimok Jun 30 '14 at 8:20
  • $\begingroup$ My experience with density matrices is only really in statistical mechanics, but I don't mean classical statistical mechanical probability theory of multiple particles, I'm referring to classical mathematical (i.e. Kolmogorov) probability theory of single particles (or coins, die, etc... not position-velocity phase space states on energy shells etc...). Can you phase classical math prob theory in terms of density matrices? Can you translate my coin problem into that language in a way that allows me to do it for all problems, including continuous distributions? $\endgroup$ – bobby Jun 30 '14 at 8:28
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    $\begingroup$ Your head/tail classical problem could be described by the diagonal density matrix $\rho = Diag (\frac{1}{4}, \frac{1}{4},\frac{1}{4},\frac{1}{4})$. More generally, for a classical probabilistic problem, every probability $p_i$ is translated into a diagonal element $\rho_{ii}$ of a diagonal density matrix. You will use diagonal classical operators, for instance, diagonal position operator $X(t)$, with the correspondance $(x_i(t)\to X_{ii}(t))$, the mean value would be $\langle x(t)\rangle = Tr(\rho X(t)) = \sum \rho_{ii} X_{ii}(t) = \sum p_i x_i(t)$. $\endgroup$ – Trimok Jun 30 '14 at 8:42
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    $\begingroup$ ...For quantum problems, there are non-zero non-diagonal terms in the density matrix and (hermitian) operators like the position operator , but one always use the same formalism :$\langle x(t)\rangle = Tr(\rho X(t))$ $\endgroup$ – Trimok Jun 30 '14 at 8:43
  • $\begingroup$ Okay cool, thanks, so "diagonal matrices correspond to classical probability distributions on the set of basis vectors" books.google.ie/… and this books.google.ie/… nicely shows how to get from one to the other! Regarding the sample space, this is my attempt, any problems comments I'd love to hear (for example, examples of events and what they really mean): $\endgroup$ – bobby Jun 30 '14 at 9:34

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