My study of Quantum Mechanics have brought me to perturbation theory. I'm here talking about the non-degenerate type.(adsbygoogle = window.adsbygoogle || []).push({});

My questions relate to the math behind it, and the power series expansion that we do.

[itex]H = H^0 + \lambda H'[/itex] (Eq. 1)

Question 1:

So in equation 1 I think I understand that, in order for us to write this expression, we assume that the hamiltonian would to a "good" approximation be able to be written as a known hamiltonian,plusa small perturbation.

Or is it in fact, from a mathematical perspective, such that it is actually possible topreciselywrite any hamiltonian in the form of equation 1. Meaning that it is not an approximation, but it is atrueequal sign?

Question 2:

Moving on from the previous, when they in the books (Griffiths in my case), write the following power series in lambda:

[itex] E_n = E_n^{(0)} + \lambda E_n^{(1)} + \lambda^2 E_n^{(2)} + \cdots[/itex] (Eq. 2)

[itex] \psi_n = \psi_n^{(0)} + \lambda \psi_n^{(1)} + \lambda^2 \psi_n^{(2)} + \cdots[/itex] (Eq. 3)

This question is actually quite the same as the previous. Is this anexactrepresentation of the energy E_n and wavefunction Psi_n? Or can we only approximate these in this sense? Mathematically speaking, as I see it, it should be perfectly possible to precisely write ANY energy (which is just a number) in the form of one known number, plus a second number?

The same goes with the functions psi. Is it a precise representation or an approximation?

Note:I see the relationship with the taylor expansion, and from wikipedia I think I saw that perhaps the validity of equation 2 and 3 comes from an argument from Taylor expansion?

Question 3:

Well, this question is kinda the all over question I guess. When you use the equations in the SchrÃ¶dinger, we get the equation

[itex]H^0\psi_n^{(0)} + \lambda(H^0 \psi_n^{(1)} + H' \psi_n^{(0)}) + \lambda^2(H^0 \psi_n^{(2)} + H'\psi_n^{(1)}) + \cdots \\ = E_n^{(0)}\psi_n^{(0)} + \lambda(E_n^{(0)}\psi_n^{(1)} + E_n^{(1)}\psi_n^{(0)}) + \lambda^2(E_n^{(0)}\psi_n^{(2)} + E_n^{(1)}\psi_n^{(1)} + E_n^{(2)}\psi_n^{(0)}) + \cdots [/itex] (Eq. 4)

Nowifwhat I talked about with regards to preciseness is true, then I understand that this is a perfectly valid form of the schrÃ¶dinger equation for our hamiltonian, actually precisely describing the solution. And it is from this equation that we choose in which "order" we want to perturbate the system.

Now comes the funny part. If we choose to perturbate the system to first order, I can see how we easily get to the equation

[itex] H^0 \psi_n^{(1)} + H' \psi_n^{(0)} = E_n^{(0)}\psi_n^{(1)} + E_n^{(1)}\psi_n^{(0)} [/itex] (Eq. 5)

And from this we can calculate stuff. However, in the textbook (Griffiths) and other places Im sure, it is then stated that if we do second-order perturbation, we also get the equation

[itex] H^0 \psi_n^{(2)} + H'\psi_n^{(1)} = E_n^{(0)}\psi_n^{(2)} + E_n^{(1)}\psi_n^{(1)} + E_n^{(2)}\psi_n^{(0)} [/itex] (Eq. 6)

This makes sense in a way,except that it doesn't to me... Because, if we have equation 4 and then say"lets discard all orders greater than 2", then we're left with the equation

[itex] \lambda(H^0 \psi_n^{(1)} + H' \psi_n^{(0)}) + \lambda^2(H^0 \psi_n^{(2)} + H'\psi_n^{(1)}) = \lambda(E_n^{(0)}\psi_n^{(1)} + E_n^{(1)}\psi_n^{(0)}) + \lambda^2(E_n^{(0)}\psi_n^{(2)} + E_n^{(1)}\psi_n^{(1)} + E_n^{(2)}\psi_n^{(0)}) [/itex] (Eq. 7)

In my curious yet unknowing mind, I cannot see how you get from equation 7 to equation 6. In my mind, you cant just say that because of the orders of lambda, the things inside the brackets must be equal. To me its a bit like the following:

[itex] x(a+b) + x^2(c+d) = x(e+f) + x^2(g+h+i)[/itex]

In the above example, is it generally true that because of their orders (a+b) must equal (e+f) and (c+d) must equal (g+h+i) ??

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# Perturbation theory (the math)

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