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Proving an equation of motion?

  • Thread starter jeebs
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  • #1
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I have this density operator [tex] \rho(t) = \sum_a |\psi_a(t)\rangle P_a \langle\psi_a(t)|[/tex] and I am supposed to be showing that "the equation of motion satisfies [tex]i\hbar\frac{\partial\rho(t)}{\partial t} = [H,\rho(t)][/tex].
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. [tex]\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)|[/tex]

also if [tex]H = i\hbar\frac{\partial}{\partial t}[/tex] then [tex] 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)| [/tex]

and also I know the commutator is just [tex] [H,\rho] = H\rho - \rho H [/tex]
so that gives me [tex] [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} [/tex]

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

  • #2
fzero
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If

[tex]
H |\psi \rangle = i\hbar\frac{\partial}{\partial t} |\psi \rangle ,
[/tex]

can you relate


[tex]
\frac{\partial}{\partial t} \langle \psi | [/tex]

to an expression involving [tex]H[/tex]?

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

[tex] \hat{H} (| \psi \rangle \langle \psi | ) = (\hat{H} | \psi \rangle ) \langle \psi |. [/tex]
 
  • #3
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If

[tex]
H |\psi \rangle = i\hbar\frac{\partial}{\partial t} |\psi \rangle ,
[/tex]

can you relate


[tex]
\frac{\partial}{\partial t} \langle \psi | [/tex]

to an expression involving [tex]H[/tex]?
well if [tex] H |\psi \rangle = i\hbar\frac{\partial}{\partial t} |\psi \rangle [/tex]

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

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

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

[tex] \hat{H} (| \psi \rangle \langle \psi | ) = (\hat{H} | \psi \rangle ) \langle \psi |. [/tex]
so you mean doing the product rule is incorrect?
if that is the case, then how do I get to use the [tex] \frac{\partial}{\partial t} \langle \psi | [/tex] you suggested? how else would I get H acting on a bra?
 
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  • #4
fzero
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so you mean doing the product rule is incorrect?
if that is the case, then how do I get to use the [tex] \frac{\partial}{\partial t} \langle \psi | [/tex] you suggested?
Yes, if you have [tex]\hat{H}[/tex] 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 [tex]\hat{H}[/tex]. The particular kinetic term and potential tells us how the Hamiltonian acts on states and operators.

Now the quantum states of the system [tex]|\psi\rangle [/tex] are solutions to the Schrodinger equation

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

It's only on these states that the Hamiltonian has a representation as [tex] \hat{H} |\psi\rangle = i \hbar \partial/\partial t[/tex]; 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 [tex]\langle \psi |[/tex], you obtain the conjugate version of (*) because [tex] |\psi\rangle[/tex] is a solution to the SE.
 
  • #5
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actually I think I've got it but it did involve use of the product rule...

[tex] H\rho = \sum_a P_a H|\psi_a\rangle\langle \psi_a| [/tex]

[tex] \rho H = \sum_a P_a |\psi_a\rangle\langle \psi_a|H [/tex]

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

[tex]\frac{\partial}{\partial t}|\psi\rangle = \frac{-i}{\hbar}H|\psi\rangle [/tex]

[tex]\frac{\partial}{\partial t}\langle \psi| = \frac{i}{\hbar}\langle\psi|H [/tex]



[tex] 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] [/tex]

So what's the lesson here? when I've got say, [tex] \frac{\partial}{\partial t}|\psi\rangle\langle\psi| [/tex] I product rule it like [tex] (\frac{\partial}{\partial t}|\psi\rangle)\langle\psi| + |\psi\rangle\frac{\partial}{\partial t} \langle\psi|[/tex] 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?
 
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  • #6
fzero
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So what's the lesson here? when I've got say, [tex] \frac{\partial}{\partial t}|\psi\rangle\langle\psi| [/tex] I product rule it like [tex] (\frac{\partial}{\partial t}|\psi\rangle)\langle\psi| + |\psi\rangle\frac{\partial}{\partial t} \langle\psi|[/tex] 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

[tex]\hat{H} = - \hbar^2 \frac{\partial^2}{\partial x^2} + V[/tex]

and acting on the product [tex] fg[/tex] of some arbitrary functions [tex] f[/tex] and [tex]g[/tex].
 
  • #7
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ahh, right, got it. cheers.
 
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