- #1
zedya
- 4
- 0
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
[tex]\langle x| \hat{H}| y \rangle = H(x,-i\partial_{x}) \delta(x-y)[/tex]
where
[tex]H(x,-i\partial_{x}) = -\frac{\partial^{2}}{\partial x^{2}} + V(x)[/tex]
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
[tex]\langle x| \hat{H}| y \rangle = H(x,-i\partial_{x}) \delta(x-y)[/tex]
is true mathematically?
Thanks again.
In the X-representation, and for an arbitrary ket [itex]|\Psi\rangle[/itex], we have;
[tex]\langle x|\hat{X}|\Psi \rangle = x \langle x| \Psi \rangle \equiv x\Psi(x)[/tex]
[tex]\langle x|\hat{P}|\Psi\rangle = -i \partial_{x}\langle x|\Psi\rangle \equiv\ –i \partial_{x}\Psi(x)[/tex]
Now letting [itex]|\Psi\rangle[/itex] be the position eigenket [itex]|y\rangle[/itex], we get
[tex]\langle x |\hat{X}| y \rangle =x \langle x |y \rangle = x \delta (x-y)[/tex]
[tex]\langle x |\hat{P}| y \rangle = -i \partial_{x}\delta(x-y)[/tex]
Next, write
[tex]
\langle x |\hat{P}^{2} | \Psi \rangle = \int \ dy \ \langle x |\hat{P} | y \rangle \langle y | \hat{P} | \Psi \rangle
[/tex]
and use the above relations to obtain
[tex]
\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
[/tex]
Again, let [itex] |\Psi \rangle = |y \rangle [/itex] to get
[tex]\langle x | \hat{P}^{2} | y \rangle = -i \frac{\partial^{2}}{\partial x^{2}} \delta (x-y)[/tex]
Now if you put the above in the Hamiltonian matrix ( 2m = 1),
[tex] \langle x| \hat{H} |y \rangle = \langle x | \hat{P}^{2}| y\rangle + \langle x | \hat{V}| y \rangle [/tex]
you will get what you wanted.
Regards
sam
The Hamiltonian being diagonal in the coordinate representation is important because it simplifies the equations of motion and makes it easier to solve for the energy levels and wavefunctions of a system. It also allows for a clearer understanding of the physical properties of the system.
No, the Hamiltonian is not always diagonal in the coordinate representation. It depends on the specific system and the choice of coordinates. In some cases, it may be more convenient to use a different representation, such as the momentum representation, where the Hamiltonian may be diagonal.
The Hamiltonian being diagonal in the coordinate representation can be determined by looking at the potential energy function of the system. If the potential energy is independent of position, then the Hamiltonian will be diagonal. However, if the potential energy depends on position, then the Hamiltonian will not be diagonal.
Using the coordinate representation for the Hamiltonian allows for a more intuitive understanding of the system and its physical properties. It also simplifies the equations of motion and makes it easier to solve for the energy levels and wavefunctions of the system.
Yes, it is possible to transform the Hamiltonian to a diagonal form in the coordinate representation through a change of basis. This can be done by using a linear combination of the original basis states that diagonalize the Hamiltonian, known as eigenstates.