# Thread: Time Independant Schrodinger Equation: Hamiltonian Query

1. ## Time Independant Schrodinger Equation: Hamiltonian Query

Dear Folks,

I am in the process of determing the expectation value of H the total kinetic energy...I am stuck on the following following lines....see attached. I understand the rest of the derivation except these 2 lines

THanksAttachment 19495

2. The Hamiltonian, considered as an operator, has two parts: one with a second-order spatial derivative, and one with simple, regular multiplication by $\displaystyle V(x).$ So, the second-to-last line you could think of as this:

$\displaystyle \displaystyle\hat{H}\varphi(t)=\left(-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}+V(x)\right)\varphi(t)$ (substitution)

$\displaystyle \displaystyle=-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}\,\varphi(t)+V(x)\varphi(t)$ (distribute)

$\displaystyle \displaystyle=V(x)\varphi(t)\qquad$ ($\displaystyle \varphi(t)$ does not depend on $\displaystyle x$, hence its spatial derivative is zero.)

So there's the second-to-last line.

For the last line, just add another function in there:

$\displaystyle \displaystyle\hat{H}[\psi(x)\varphi(t)]=\left(-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}+V(x)\right)[\psi(x)\varphi(t)]$ (substitution)

$\displaystyle \displaystyle=-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}\,[\psi(x)\varphi(t)]+V(x)[\psi(x)\varphi(t)]$ (distribute)

$\displaystyle \displaystyle=-\frac{\hbar^{2}}{2m}\varphi(t)\frac{\partial^{2}}{ \partial x^{2}}\,\psi(x)+\varphi(t)V(x)\psi(x)$ (again, $\displaystyle \varphi(t)$ doesn't depend on $\displaystyle x$)

$\displaystyle \displaystyle=\varphi(t)\left(-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}+V(x)\right)\psi(x)$ (left- and right- factor: operators can be tricky!)

$\displaystyle \displaystyle=\varphi(t)\hat{H}\psi(x).$ (substitute).

Make sense?

3. Originally Posted by Ackbeet

For the last line, just add another function in there:

$\displaystyle \displaystyle\hat{H}[\psi(x)\varphi(t)]=\left(-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}+V(x)\right)[\psi(x)\varphi(t)]$ (substitution)

$\displaystyle \displaystyle=-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}\,[\psi(x)\varphi(t)]+V(x)[\psi(x)\varphi(t)]$ (distribute)

$\displaystyle \displaystyle=-\frac{\hbar^{2}}{2m}\varphi(t)\frac{\partial^{2}}{ \partial x^{2}}\,\psi(x)+\varphi(t)V(x)\psi(x)$ (again, $\displaystyle \varphi(t)$ doesn't depend on $\displaystyle x$)

$\displaystyle \displaystyle=\varphi(t)\left(-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}+V(x)\right)\psi(x)$ (left- and right- factor: operators can be tricky!)

$\displaystyle \displaystyle=\varphi(t)\hat{H}\psi(x).$ (substitute).

Make sense?
Hi,

Yes that makes perfect sense thanks. If i was to tackle the following line I would probably have made the mistake of swapping the phi and psi therfore making the first term go to zero. Can it be explained why this would be wrong? IT sounds like a stupid question i know but I cant pinpoint why.

$\displaystyle \displaystyle=-\frac{\hbar^{2}}{2m}\psi(x)\frac{\partial^{2}}{\pa rtial x^{2}}\,\varphi(t)+\varphi(t)V(x)\psi(x)$ (again, $\displaystyle \varphi(t)$ doesn't depend on $\displaystyle x$)

4. Yes, it's wrong because of the way that second derivative operator works. Cardinal rule of operators: they ALWAYS look to the right, and only ever to the right. They also have "tunnel vision" (pun intended): they only see as far as the next addition or subtraction symbol. Thus, in the expression

$\displaystyle \dfrac{d}{dx}f(x)+g(x),$

the derivative operator sees and operates on $\displaystyle f(x),$ but not $\displaystyle g(x).$

Operators do not, in general, commute. In particular, the multiplication operator and the derivative operator don't commute. Proof:

$\displaystyle f(x)\dfrac{d}{dx}[y(x)]=f(x)y'(x).$

$\displaystyle \dfrac{d}{dx}f(x)[y(x)]=\dfrac{d}{dx}[f(x)y(x)]=f'(x)y(x)+f(x)y'(x)\not=f(x)y'(x).$

So, when you have a spatial derivative operator like in your case, it must always act on anything that depends on that spatial variable and is within its "tunnel vision".

In reality, the result of an operator acting on a function is another function. If the operator doesn't act on a function, then your expression is still an operator. Your expression

$\displaystyle \displaystyle-\frac{\hbar^{2}}{2m}\psi(x)\frac{\partial^{2}}{\pa rtial x^{2}}\,\varphi(t)+\varphi(t)V(x)\psi(x)$ is equal to

$\displaystyle \displaystyle-\frac{\hbar^{2}}{2m}\psi(x)\varphi(t)\frac{\partia l^{2}}{\partial x^{2}}+\varphi(t)V(x)\psi(x),$

because the spatial derivative operator doesn't act on $\displaystyle \varphi(t).$ Hence, the resulting expression is an operator, which doesn't match up with the function on the RHS.

Make sense?

5. Originally Posted by Ackbeet

.......$\displaystyle \displaystyle-\frac{\hbar^{2}}{2m}\psi(x)\frac{\partial^{2}}{\pa rtial x^{2}}\,\varphi(t)+\varphi(t)V(x)\psi(x)$ is equal to

$\displaystyle \displaystyle-\frac{\hbar^{2}}{2m}\psi(x)\varphi(t)\frac{\partia l^{2}}{\partial x^{2}}+\varphi(t)V(x)\psi(x),$

because the spatial derivative operator doesn't act on $\displaystyle \varphi(t).$ Hence, the resulting expression is an operator, which doesn't match up with the function on the RHS.

Make sense?
So the second line is just a reorder of the first but both are wrong and should be as per your third last line in post #2... ie

$\displaystyle \displaystyle=-\frac{\hbar^{2}}{2m}\varphi(t)\frac{\partial^{2}}{ \partial x^{2}}\,\psi(x)+\varphi(t)V(x)\psi(x)$

I think that makes perfect sense :-) Thanks alot Ackbeet I appreciate it!

6. Yes. All correct. You're very welcome!

Thanks for posting such a fun problem. We don't see all that much quantum mechanics on the MHF, and it's my absolute favorite area of physics.

7. Originally Posted by Ackbeet
.... We don't see all that much quantum mechanics on the MHF, and it's my absolute favorite area of physics.
I am new to QM and I find it very interesting and mind boggling to say the least. Its amazing how Schrodingers equation can represent any physical system from atoms to macroscopic. :-)

Thanks!!!

8. You're welcome. Have a good one!