zedya
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In the coordinate representation of a quantum mechanical system, is it always true that the Hamiltonian of the system is diagonal? If so, can someone explain to me why this is true?
zedya said:In the coordinate representation of a quantum mechanical system, is it always true that the Hamiltonian of the system is diagonal? If so, can someone explain to me why this is true?
zedya said:In the coordinate representation of a quantum mechanical system, is it always true that the Hamiltonian of the system is diagonal? If so, can someone explain to me why this is true?
samalkhaiat said:So, when the potential is local, the Hamiltonian is diagonal in the sense that it can be written as
\langle x| \hat{H}| y \rangle = H(x,-i\partial_{x}) \delta(x-y)
where
H(x,-i\partial_{x}) = -\frac{\partial^{2}}{\partial x^{2}} + V(x)
zedya said:So, one more question (thanks a lot for this reply by the way, beautiful explanation)... how would you prove the above? I always get confused with notation stuff... how would you prove that
\langle x| \hat{H}| y \rangle = H(x,-i\partial_{x}) \delta(x-y)
is true mathematically?
Thanks again.
In the X-representation, and for an arbitrary ket |\Psi\rangle, we have;
\langle x|\hat{X}|\Psi \rangle = x \langle x| \Psi \rangle \equiv x\Psi(x)
\langle x|\hat{P}|\Psi\rangle = -i \partial_{x}\langle x|\Psi\rangle \equiv\ –i \partial_{x}\Psi(x)
Now letting |\Psi\rangle be the position eigenket |y\rangle, we get
\langle x |\hat{X}| y \rangle =x \langle x |y \rangle = x \delta (x-y)
\langle x |\hat{P}| y \rangle = -i \partial_{x}\delta(x-y)
Next, write
<br /> \langle x |\hat{P}^{2} | \Psi \rangle = \int \ dy \ \langle x |\hat{P} | y \rangle \langle y | \hat{P} | \Psi \rangle<br />
and use the above relations to obtain
<br /> \langle x |\hat{P}^{2} | \Psi \rangle = (-i)^{2} \partial_{x} \int \ dy \delta(x-y) \partial_{y} \langle y | \psi \rangle = - \frac{\partial^{2}}{\partial x^{2}} \langle x | \Psi \rangle<br />
Again, let |\Psi \rangle = |y \rangle to get
\langle x | \hat{P}^{2} | y \rangle = -i \frac{\partial^{2}}{\partial x^{2}} \delta (x-y)
Now if you put the above in the Hamiltonian matrix ( 2m = 1),
\langle x| \hat{H} |y \rangle = \langle x | \hat{P}^{2}| y\rangle + \langle x | \hat{V}| y \rangle
you will get what you wanted.
Regards
sam