Does the Equation of Motion Satisfy the Commutator Relation?

[jeebs]

I have this density operator \rho(t) = \sum_a |\psi_a(t)\rangle P_a \langle\psi_a(t)| and I am supposed to be showing that "the equation of motion satisfies i\hbar\frac{\partial\rho(t)}{\partial t} = [H,\rho(t)].
I'm not making much progress though, this is all the info I'm given.

I'm thinking I have to use the product rule here, ie. \frac{\partial\rho(t)}{\partial t} = \sum_a (\frac{\partial}{\partial t}|\psi_a(t)\rangle) P_a \langle\psi_a(t)| + \sum_a |\psi_a(t)\rangle P_a \frac{\partial}{\partial t}\langle\psi_a(t)|

also if H = i\hbar\frac{\partial}{\partial t} then H\rho = i\hbar(\sum_a (\frac{\partial}{\partial t}|\psi_a(t)\rangle) P_a \langle\psi_a(t)| + \sum_a |\psi_a(t)\rangle P_a \frac{\partial}{\partial t}\langle\psi_a(t)|

and also I know the commutator is just [H,\rho] = H\rho - \rho H
so that gives me [H,\rho] = i\hbar(\sum_a (\frac{\partial}{\partial t}|\psi_a(t)\rangle) P_a \langle\psi_a(t)| + \sum_a |\psi_a(t)\rangle P_a \frac{\partial}{\partial t}\langle\psi_a(t)| - \sum_a |\psi_a(t)\rangle P_a \langle\psi_a(t)|i\hbar\frac{\partial}{\partial t}

but I can't see how I'm supposed to get any further. I mean, I don't see what's wrong with saying i\hbar\frac{\partial\rho}{\partial t} = H\rho. I don't see where the commutator comes from at all, unless for some reason we can say that \rho H = 0

 

[fzero]

If

<br /> H |\psi \rangle = i\hbar\frac{\partial}{\partial t} |\psi \rangle ,<br />

can you relate <br /> \frac{\partial}{\partial t} \langle \psi |

to an expression involving H?

Also, your expression for H\rho is incorrect. While \hat{H} is related to the time derivative by Schrödinger's equation, it doesn't itself act like a derivative.

\hat{H} (| \psi \rangle \langle \psi | ) = (\hat{H} | \psi \rangle ) \langle \psi |.

 

[jeebs]

If

<br /> H |\psi \rangle = i\hbar\frac{\partial}{\partial t} |\psi \rangle ,<br />

can you relate <br /> \frac{\partial}{\partial t} \langle \psi |

to an expression involving H?

well if H |\psi \rangle = i\hbar\frac{\partial}{\partial t} |\psi \rangle

then \langle \psi |H^* = -i\hbar\frac{\partial}{\partial t} \langle \psi | = \langle \psi |H
since H is Hermitian? so
\frac{i}{\hbar}\langle \psi |H = \frac{\partial}{\partial t} \langle \psi |

i'll try and see where this gets me...

Also, your expression for H\rho is incorrect. While \hat{H} is related to the time derivative by Schrödinger's equation, it doesn't itself act like a derivative.

\hat{H} (| \psi \rangle \langle \psi | ) = (\hat{H} | \psi \rangle ) \langle \psi |.

so you mean doing the product rule is incorrect?
if that is the case, then how do I get to use the \frac{\partial}{\partial t} \langle \psi | you suggested? how else would I get H acting on a bra?

 

[fzero]

so you mean doing the product rule is incorrect?
if that is the case, then how do I get to use the \frac{\partial}{\partial t} \langle \psi | you suggested?

Yes, if you have \hat{H} acting on a general object, the product rule is incorrect. The way to think about it is the following. A system is described by its Hamiltonian, which can be written as the operator \hat{H}. The particular kinetic term and potential tells us how the Hamiltonian acts on states and operators.

Now the quantum states of the system |\psi\rangle are solutions to the Schrödinger equation

\hat{H} |\psi\rangle = i \hbar \frac{\partial}{\partial t} |\psi\rangle . (*)

It's only on these states that the Hamiltonian has a representation as \hat{H} |\psi\rangle = i \hbar \partial/\partial t; in general it doesn't. The point of your problem is to obtain a relationship between the action of the Hamiltonian and the time derivative for the density matrix and it will be slightly more complicated than (*).

As for the conjugate state \langle \psi |, you obtain the conjugate version of (*) because |\psi\rangle is a solution to the SE.

 

[jeebs]

actually I think I've got it but it did involve use of the product rule...

H\rho = \sum_a P_a H|\psi_a\rangle\langle \psi_a|

\rho H = \sum_a P_a |\psi_a\rangle\langle \psi_a|H

H\rho - \rho H = \sum_aP_a (H|\psi\rangle\langle \psi_a| - |\psi\rangle\langle \psi_a|H) = [H,\rho]

\frac{\partial}{\partial t}|\psi\rangle = \frac{-i}{\hbar}H|\psi\rangle

\frac{\partial}{\partial t}\langle \psi| = \frac{i}{\hbar}\langle\psi|H
i\hbar\frac{\partial \rho}{\partial t} = i\hbar\sum_aP_a(\frac{\partial}{\partial t}|\psi\rangle\langle \psi_a| + |\psi\rangle\frac{\partial}{\partial t}\langle \psi_a|) = i\hbar\sum_aP_a(\frac{-i}{\hbar}H|\psi\rangle\langle \psi_a| + |\psi\rangle\frac{i}{\hbar}\langle \psi_a|H) = \sum_aP_a (H|\psi\rangle\langle \psi_a| - |\psi\rangle\langle \psi_a|H) = [H,\rho]

So what's the lesson here? when I've got say, \frac{\partial}{\partial t}|\psi\rangle\langle\psi| I product rule it like (\frac{\partial}{\partial t}|\psi\rangle)\langle\psi| + |\psi\rangle\frac{\partial}{\partial t} \langle\psi| but when I've got an operator with a derivative in it, I only apply it to the thing directly to the right of it?

 

[fzero]

So what's the lesson here? when I've got say, \frac{\partial}{\partial t}|\psi\rangle\langle\psi| I product rule it like (\frac{\partial}{\partial t}|\psi\rangle)\langle\psi| + |\psi\rangle\frac{\partial}{\partial t} \langle\psi| but when I've got an operator with a derivative in it, I only apply it to the thing directly to the right of it?

It's valid to use the product rule when the operator is a pure derivative. Hamiltonians generally have a potential term, for which the product rule is not valid. You can check this by representing

\hat{H} = - \hbar^2 \frac{\partial^2}{\partial x^2} + V

and acting on the product fg of some arbitrary functions f and g.

 

[jeebs]

ahh, right, got it. cheers.