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 world:ch01 [2020/07/27 14:47]talipovm created world:ch01 [2020/07/28 08:14] (current)talipovm 2020/07/28 08:14 talipovm 2020/07/27 14:47 talipovm created 2020/07/28 08:14 talipovm 2020/07/27 14:47 talipovm created Line 3: Line 3: ===== Binary systems ===== ===== Binary systems ===== - Consider a collection of non-interacting particles, where each particle can be in one of two states ​quantum states. As an example, we will deal with //N// elementary magnets that could have their spins oriented up or down (another example: flipping ​coins). The magnetic moment of an individual magnet is //+m// if the magnet is pointing up and //-m// if it is pointing down. The total number of configurations is $2^N$ while the total number of possible momenta of the entire system is $N+1$ (that is, $-Nm, -(N-2)m, ..., +Nm$). + Consider a collection of non-interacting particles, where each particle can be found in one of two quantum states. As an example, we will deal with //N// elementary magnets that could have their spins oriented up or down (another example: flipping ​a coin). The magnetic moment of an individual magnet is //+m// if the magnet is pointing up and //-m// if it is pointing down. The total number of configurations is $2^N$ while the total number of possible momenta of the entire system is $N+1$ (that is, $-Nm, -(N-2)m, ..., +Nm$). + + The spin excess of the system is equal to: + + $N_{up} - N_{down} = 2s$, + + where $N_{up}$ and $N_{down}$ are the number of magnets with spin up and down, respectively. Spin excess is defined as $2s$ instead of $s$ for convenience. It is easy to show that: + + $N_{up} + N_{down} = N$ + + $N_{up} =N/2 + s$ + + $N_{down} =N/2 - s$ + + The number of possible states with a specific spin excess can be evaluated using a simple formula from combinatorics:​ + + $g(N,s) = \frac {N!} {N_{up}! N_{down}!}.$ + + The function //g// is called a multiplicity function. The multiplicity function forms a Gaussian distribution for //s//: + + $g(N,s) \approx g(N,0) exp(-2s^2/​N)$ + + $g(N,0) = \frac {N!} { (N/2)! (N/2)! } \approx \sqrt \frac 2 {\pi N} 2^N$ + + ​ + We will use the Stirling approximation:​ + + $ln (N!) \approx 1/2 ln (2\pi) + (N+1/2) ln N - N$ + + ​ -