Understanding Perturbation Theory and Symmetry in Quantum Mechanics

[eigenpost53]

could someone explain the following sentence to me?



"First-order perturbation correction is often
precluded because of a symmetry principle operating
for the state under consideration."

"In second-order, on the other hand, the perturbation
has to connect the given state with some other state
and back, and therefore, its contribution is no longer
inhibited by symmetry."


I can comprehend the mathematical skills of the perturbation theory.
However, depending on the circumstance I do not understand the physical
basis how first-order perturbation can become zero due to symmetry.




2.
peopel say the cause of degeneracy is related with symmetry. then how is
the symmetry related with degeneracy ?

Since I am a Korean, please explain your answer in qualitative simple phrases.

 

[heardie]

I'm not too sure with 1, but the first order perbutation correction is given by the mean of the perubtation, so if the state is symmetrical about 0, the mean (and therefore the permutation) will be zero.

 

[turin]

1. Why is the standard deviation based on the average of the squared difference rather than the average of the difference itself? Why is RMS value used to describe a sine wave rather than just the straightforward average displacement? How can we always get away with using the HO potential to approximate a system undergoing small oscillations?

2. As an example, the free particle sees the world identically in all directions. Regardless of the direction it travels, the magnitude of its momentum will determine its energy. There is an infinitude of different directions that a free particle can travel with the same kinetic energy because the direction is symetric. Perhaps it is simpler to consider the 1-D case. A particle can travel to the left with momentum = +p or to the right with momentum = -p, but the energy is energy = +p² in either direction. It is symmetric about p = 0 and therefore the energy state is degenerate (it could be +p or -p).

 

[selfAdjoint]

To answer your first question, suppose you have a sample of three values, and say they happen to be 1, 2, and 3. Then the mean, or average is (1+2+3)/3 = 6/3 = 2.
Now you want to calculate the standard deviation. Let's do it your way first. Take the difference of each datum minus the average, as you suggested. We have 1-2 = -1, 2-2 = 0, and 3 -2 = +1. Following your suggestion we add these up, and find that they add up to zero. In fact no matter what your data numbers are, they will always add up to zero, just from the way the average is defined, there is always just as much difference less than average as there is greater than average.

You might think to fix this problem by using the absolute value of the difference. But the absolute value is an awkward function to work with, since it isn't differentiable at zero. So statisticians went to the other simple way to make all the addends positive, they squared them. And then to get back to the same units they had to take the squaare root of the average squared deviation. There's your RMS (root mean square, for the lurkers - it just tells you what to do: take the ROOT of the MEAN of the SQUARE deviations).

So the standard deviation of our example goes like this:

Square the deviations, they are -1, 0, and +1, and the squares are +1. 0. and +1.

Take the average. (1 + 0 + 1)/3 = 2/3

Take the square root. \sqrt{2/3} = .816

This is the standard deviation. (I have ignored the shift for small samples)