Why is the Hamiltonian written as H|x> = E|x> instead of H|x> = |x>E?

  • Thread starter Thread starter quietrain
  • Start date Start date
  • Tags Tags
    Matrix
quietrain
Messages
648
Reaction score
2

Homework Statement



i have matrix A which is diagonalisable

by doing an example on wiki under the section " how do diagonalise a matrix"
http://en.wikipedia.org/wiki/Diagonalizable_matrix

i realize that

A x is not equal to \lambda x , where x are eigen vectors of A , \lambda is eigen values

instead

x A = \lambda x

and i think

A x = x \lambda

QUESTION 1)
if this is so, why do they always write the hamiltonian as H |x> = E |x> ? shouldn't it be H |x> = |x> E ?

if i remembered correctly, for matrix multiplication AB =/= BA right?

but i read wiki and it says something like (i can't remember the exact phrasing)

"it is equal if both A and B are diagonalisable matrix , and are both n by n matrix. "


QUESTION 2)
also, for PT A P = \lambda

if i want to "bring over" the P, is it like this

A = P \lambda PT

but why is it like this?



thanks!
 
Physics news on Phys.org
You are right that matrices don't commute, however the multiplication of a matrix with a scalar (real or complex number) always commutes. So λx = xλ. What you write above is also not correct. Ax is not at all the same as xA, the wiki page is correct.

Question 2):

this works for so called orthogonal matrices where P^T * P = P*P^T = 1 (the identity matrix). It is not true for general matrices.
 
You do not have to choose orthonormal vectors for eigenvalues but you can. If you do then the matrix P is "orthogonal"- in particular P^T= P^{-1}. Without P being orthogonal, that is not true but it is still true that P^{-1}AP= D so, multiplying on the left by P and on the right by P^{-1},
P(P^{-1}AP)P^{-1}= PDP^{-1}
(PP^{-1})A(PP^{-1})= IAI= A= PDP^{-1}

In the case that P is "orthogonal", You can replace P^{-1} with P^T.
 
oh shucks ... for question 1 i realize i multiplied the P ,matrix of eigen vectors, with the diagonal matrix eigen values

but in fact i think they are talking about the individual eigen value and eigen functions , not the combined P.

so yup i get it thanks everyolne!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top