- #1
soviet1100
- 50
- 16
Hello!
On p.145 of Shankar's Principles of Quantum Mechanics, the author derives the general propagator for the Schrodinger equation in the following manner.
Shankar's working
Expanding the state vector in the energy basis,
[itex] |\psi(t)\rangle = \sum_{E} |E\rangle \langle E| \psi(t) \rangle \equiv \sum_{E} a_{E}(t)|E\rangle [/itex]
then acting on both sides with [itex]\,(\imath\hbar\frac{\partial}{\partial t} - H) [/itex]
he gets [itex]\,\,\,0 = \sum(\imath\hbar\dot{a_{E}}-Ea_{E})|E\rangle [/itex]
and so [itex]\,\,\imath\hbar\dot{a_{E}} = Ea_{E} [/itex]
from this [itex]\,\, a_{E}(t) = a_{E}(0)\cdot e^{\frac{-\imath Et}{\hbar}} [/itex]
My working
[itex] |\psi(t)\rangle = \sum_{E} |E\rangle \langle E| \psi(t) \rangle \equiv \sum_{E} a_{E}(t)|E\rangle [/itex]
then acting on both sides with [itex]\,(\imath\hbar\frac{\partial}{\partial t} - H) [/itex]
I get [itex]\,\,\,0 = \sum(\imath\hbar\dot{a_{E}}|E\rangle + \imath\hbar a_{E}|\dot{E}\rangle - a_{E}E|E\rangle). [/itex] Note the extra (middle) term that I get here; this is missing in Shankar's corresponding step.
But [itex] \,\imath\hbar a_{E}|\dot{E}\rangle - a_{E}E|E\rangle \,\,[/itex] is the schrodinger equation with an eigenket (of H, of course) substituted in and multiplied throughout by [itex] \dot{a_E}[/itex]. So this term equals zero.
and so [itex]\,\,\,\dot{a_{E}} = 0 [/itex]
Clearly, this is at odds with Shankar's result, whose first time derivative is not zero. Is there a mistake anyone can spot in either of the workings above?
Thanks,
soviet1100
On p.145 of Shankar's Principles of Quantum Mechanics, the author derives the general propagator for the Schrodinger equation in the following manner.
Shankar's working
Expanding the state vector in the energy basis,
[itex] |\psi(t)\rangle = \sum_{E} |E\rangle \langle E| \psi(t) \rangle \equiv \sum_{E} a_{E}(t)|E\rangle [/itex]
then acting on both sides with [itex]\,(\imath\hbar\frac{\partial}{\partial t} - H) [/itex]
he gets [itex]\,\,\,0 = \sum(\imath\hbar\dot{a_{E}}-Ea_{E})|E\rangle [/itex]
and so [itex]\,\,\imath\hbar\dot{a_{E}} = Ea_{E} [/itex]
from this [itex]\,\, a_{E}(t) = a_{E}(0)\cdot e^{\frac{-\imath Et}{\hbar}} [/itex]
My working
[itex] |\psi(t)\rangle = \sum_{E} |E\rangle \langle E| \psi(t) \rangle \equiv \sum_{E} a_{E}(t)|E\rangle [/itex]
then acting on both sides with [itex]\,(\imath\hbar\frac{\partial}{\partial t} - H) [/itex]
I get [itex]\,\,\,0 = \sum(\imath\hbar\dot{a_{E}}|E\rangle + \imath\hbar a_{E}|\dot{E}\rangle - a_{E}E|E\rangle). [/itex] Note the extra (middle) term that I get here; this is missing in Shankar's corresponding step.
But [itex] \,\imath\hbar a_{E}|\dot{E}\rangle - a_{E}E|E\rangle \,\,[/itex] is the schrodinger equation with an eigenket (of H, of course) substituted in and multiplied throughout by [itex] \dot{a_E}[/itex]. So this term equals zero.
and so [itex]\,\,\,\dot{a_{E}} = 0 [/itex]
Clearly, this is at odds with Shankar's result, whose first time derivative is not zero. Is there a mistake anyone can spot in either of the workings above?
Thanks,
soviet1100