QFT, more a QM Question, Hamiltonian relation time evolution

[binbagsss]

Homework Statement

Question attached here:

I am just stuck on the first bit. I have done the second bit and that is fine. This is a quantum field theory course question but from what I can see this is a question solely based on QM knowledge, which I've probably forgot some of.

Homework Equations

I believe the only relevant equation to use is:

$\hat{H}\Lambda = i\frac{h}{2\pi}\frac{\partial H}{\partial t}$

The Attempt at a Solution

First of all I thought I should log the expression to get, obviously I need to be careful to keep order:

$\hat{H}\hat{a(k)} = \hat{a(k)} (\hat{H}-\omega_k)$

I'm a bit thrown of by the fact that the Schrödinger equation acts on a state, the ladder operator itself isn't a state until acting upon a vacuum,, so perhaps if I included this it would make the algebra a bit easier as a 'test state' sort of thing with the operators, but I have omitted it for now).

I then conclude for the left hand side from the Schrödinger equation I get:

$i\frac{h}{2\pi} \frac{\partial(\hat{H}\hat{a(k)}}{\partial t} = \hat{a(k)} (\hat{H}-\omega_k)$

$\hat{a(k)}$ does not depend on time, $w_k$ does, and so I have a first order linear differential equation, and could identify an integrating factor, however I don't think this approach is valid since my integrating involves the exponential of $\hat{a(k)}$ and order matters, also when working backward with the product rule of differentiation there is no order on $\frac{d}{dt}(uv)=vdu/dt+udv/dt$ and so if i turn u or v to operators, i don't think this integrating factor method will work?

I'm unsure what to do next/ how to approach if I am totally off. Other ideas I had where to
1) drop the 'test state' in the Schrödinger equation and raise the operators to the power of n, but this seems unnecessary when you can just log
2) take the conjugate of the Schrödinger equation, still treating $\hat{a(k)}$ as the 'test state' as said before, in order that I can plug something into the right hand side, once expanded, which ofc has opposite order of the operators to the lhs : $\hat{a(k)}\hat{H}$ instead.

Many thanks

 

[Orodruin]

Just write down the series expansion of the exponent and apply the commutator relation given.

 

[binbagsss]

Just write down the series expansion of the exponent and apply the commutator relation given.

Just write down the series expansion of the exponent and apply the commutator relation given.

I haven't been given an explicit expression for the Hamiltonian in terms of the ladder operators in order to apply the commutator relations between the ladder operators a(k) ?. Of which exponent lhs or rhs or both?

thanks

 

[Orodruin]

I haven't been given an explicit expression for the Hamiltonian in terms of the ladder operators in order to apply the commutator relations between the ladder operators a(k) ?

You have been given an explicit commutation relation. $H^n a_k = a_k (H - \omega_k)^n$

Of which exponent lhs or rhs or both?

Start with one of them and work your way to the other.

 

[binbagsss]

You have been given an explicit commutation relation. $H^n a_k = a_k (H - \omega_k)^n$Start with one of them and work your way to the other.

how is this going to introduce $t$? its not in the commutator relation

 

[binbagsss]

how is this going to introduce $t$? its not in the commutator relation

we have not been asked to derive this result btw

 

[Orodruin]

how is this going to introduce $t$? its not in the commutator relation

So what?

 

[binbagsss]

Start with one of them and work your way to the other.

are your responses to show the relation(that we have not been asked to derive), rather than the second stated equality?

 

[Orodruin]

are your responses to show the relation(that we have not been asked to derive), rather than the second stated equality?

No.

 

[binbagsss]

No.

No.

so why would you randomly iintroduce an $i$ and a $t$. the exponential relation?

 

[Orodruin]

so why would you randomly iintroduce an $i$ and a $t$. the exponential relation?

What do you mean randomly? It is a part of the relation you want to show. The exponential of $iHt$ is the time evolution operator.

 

[binbagsss]

What do you mean randomly? It is a part of the relation you want to show. The exponential of $iHt$ is the time evolution operator.

omg got it at last, thank you