- #1
buffordboy23
- 548
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I am having trouble with this derivation. This is where I am at:
[tex] \frac{d \left< p \right>}{dt} = \int \left[ \left( -\frac{\hbar^{2}}{2m}\frac{\partial^{2} \Psi^{*}}{\partial x^{2}} + V\Psi^{*} \right) \frac{\partial \Psi}{\partial x} + \frac{\partial \Psi^{*}}{\partial x}\left( \frac{\hbar^{2}}{2m}\frac{\partial^{2} \Psi}{\partial x^{2}} - V\Psi \right) \right] dx + \int \Psi^{*}\left(-\frac{\partial V}{\partial x} \right) \Psi dx[/tex]
[tex] \frac{d \left< p \right>}{dt} = \int \left( H\Psi^{*}\frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^{*}}{\partial x}H\Psi \right) dx + \int \Psi^{*}\left(-\frac{\partial V}{\partial x} \right) \Psi dx[/tex]
I am guessing that this is correct so far; it looks like the terms under the first integral will cancel out nicely somehow and the second integral has the term of interest. If this is incorrect, then I will post my previous steps of the derivation.
Is the Hamiltonian commutative? At this point I am confused and do not know how to proceed.
EDIT: I just thought of something. Can I say that since [tex] H\Psi = E\Psi [/tex], where [tex]E[/tex] represent the energy and is a constant. Therefore, I can pull this constant outside the integral and apply an integration by parts to get
[tex] \frac{d \left< p \right>}{dt} = E \int \left(\Psi^{*}\frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^{*}}{\partial x}\Psi \right) dx + \left<-\frac{\partial V}{\partial x} \right> = \left E\Psi^{*}\Psi \right|^{+ \infty}_{- \infty}+ \left<-\frac{\partial V}{\partial x} \right>= \left<-\frac{\partial V}{\partial x} \right>[/tex]
since [tex]\Psi[/tex] goes to zero at infinity.
EDIT2: This shouldn't work either because of the minus sign between the terms in the first integral.
[tex] \frac{d \left< p \right>}{dt} = \int \left[ \left( -\frac{\hbar^{2}}{2m}\frac{\partial^{2} \Psi^{*}}{\partial x^{2}} + V\Psi^{*} \right) \frac{\partial \Psi}{\partial x} + \frac{\partial \Psi^{*}}{\partial x}\left( \frac{\hbar^{2}}{2m}\frac{\partial^{2} \Psi}{\partial x^{2}} - V\Psi \right) \right] dx + \int \Psi^{*}\left(-\frac{\partial V}{\partial x} \right) \Psi dx[/tex]
[tex] \frac{d \left< p \right>}{dt} = \int \left( H\Psi^{*}\frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^{*}}{\partial x}H\Psi \right) dx + \int \Psi^{*}\left(-\frac{\partial V}{\partial x} \right) \Psi dx[/tex]
I am guessing that this is correct so far; it looks like the terms under the first integral will cancel out nicely somehow and the second integral has the term of interest. If this is incorrect, then I will post my previous steps of the derivation.
Is the Hamiltonian commutative? At this point I am confused and do not know how to proceed.
EDIT: I just thought of something. Can I say that since [tex] H\Psi = E\Psi [/tex], where [tex]E[/tex] represent the energy and is a constant. Therefore, I can pull this constant outside the integral and apply an integration by parts to get
[tex] \frac{d \left< p \right>}{dt} = E \int \left(\Psi^{*}\frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^{*}}{\partial x}\Psi \right) dx + \left<-\frac{\partial V}{\partial x} \right> = \left E\Psi^{*}\Psi \right|^{+ \infty}_{- \infty}+ \left<-\frac{\partial V}{\partial x} \right>= \left<-\frac{\partial V}{\partial x} \right>[/tex]
since [tex]\Psi[/tex] goes to zero at infinity.
EDIT2: This shouldn't work either because of the minus sign between the terms in the first integral.
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