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

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

can you relate $$\frac{\partial}{\partial t} \langle \psi |$$

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

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

can you relate $$\frac{\partial}{\partial t} \langle \psi |$$

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.