meaning of of the mean square deviation in physics

Discussion in 'Physics' started by Omega, May 10, 2009.

  1. Omega

    Thread Starter Member

    Feb 23, 2009
    15
    0
    Hi every one

    when I calculated (using the density matrix) the average value for the third pauli matrix (σ_{z}) for an spin in a magnetic field parralel to the z_direction at low temperatures (T \rightarrow 0) I found it equalls one and the man square deviation equalls zero. I thought it is a resonable results since when the temerature is low and then the corresponding thermal energy is low enough compared to the magnetic energy the z_direction is the only preferred direction for the spin so the average value of the third pauli matrix equalls one and the mean square deviation which is a measure of the mean square error equalls zero am I right till now? and when I did the same calculation when T \rightarrow ∞ I found <σ_{z}>=0 (seems resonable also) but I found the man square deviation equalls one! so I am confused any help?
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    Is there some reason why you selected an unusually tiny font? How about reposting in 10 point or so?
     
  3. Omega

    Thread Starter Member

    Feb 23, 2009
    15
    0
    Hi every one

    when I calculated (using the density matrix) the average value for the third pauli matrix (σ_{z}) for an spin in a magnetic field parralel to the z_direction at low temperatures (T \rightarrow 0) I found it equalls one and the man square deviation equalls zero. I thought it is a resonable results since when the temerature is low and then the corresponding thermal energy is low enough compared to the magnetic energy the z_direction is the only preferred direction for the spin so the average value of the third pauli matrix equalls one and the mean square deviation which is a measure of the mean square error equalls zero am I right till now? and when I did the same calculation when T \rightarrow ∞ I found <σ_{z}>=0 (seems resonable also) but I found the man square deviation equalls one! so I am confused any help?
    I hope the font is better now
     
  4. Papabravo

    Expert

    Feb 24, 2006
    10,144
    1,791
    A variance that approaches zero is another way of saying that the mean of a sample or a population is nearly constant. A variance that approaches large numbers is another way of saying the mean of a sample or a population is indeterminate or not constant.


    Intuitively the low thermal energy is consistent with the view that we have a situation in which the quantity of interest is very nearly constant. As a practical matter I did not think it was possible to achieve absolute zero; it can only be approached asymptotically.
     
  5. Omega

    Thread Starter Member

    Feb 23, 2009
    15
    0
    so we can say the high value of the variance at high temperature case means that the situation is completely random and the mean of the z-component of the spin can not be measured accurately.

    about the zero temperature I guess we do not mean here actual zero but very low thermal energy just enouph to make the z_direction is the preferred one for the most of the time.
     
  6. Papabravo

    Expert

    Feb 24, 2006
    10,144
    1,791
    I would condition that by saying that I would expect a mean value close to zero with a small variance for the x-direction and the y-direction, to support the conclusion that the z-direction was preferred. Is the z-direction constrained by external forces in this problem or is there some other way of knowing which direction it is in an absolute sense?
     
  7. Omega

    Thread Starter Member

    Feb 23, 2009
    15
    0
    actualy I calculated the average value for the pauli matrices in x- an y-direction (at the both limits of temperature 0 and ∞) and they are all equall zero which is ok and seems compatible with the situation.

    The z-direction is the preferred one because we have an external applied magnetic field in the z-direction and actually if there is no external applied field in specified direction the situation will be completely random just like the case when there is an applied field but the thermal agitation overcomes it.
     
  8. Papabravo

    Expert

    Feb 24, 2006
    10,144
    1,791
    I think you have the situation well in hand.
     
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