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Help with second quantization practice problem

  • Thread starter CMJ96
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Moved from a technical forum, so homework template missing
Hi, to help further my understanding of the second quantization for one of my modules I would like to show that the following expressions
$$ \hat{H} = \Sigma_{ij} \langle i| \hat{T} | j \rangle \hat{a_i }^{\dagger} \hat{a_j} $$
$$\hat{\psi}(r,t)= \Sigma_k \psi_k(r) \hat{a}_k(t)$$
Obey the Heisenberg equation
$$i \hbar \frac{d\hat{a}_k}{dt}=\Sigma_j \langle k | \hat{T} |j \rangle \hat{a}_j$$
I started by subbing ##\hat{\psi}## and ##\hat{H}## into the Heisenberg equation of motion for a generic operator
$$ i \hbar \frac{\partial \hat{O}_H}{\partial t}= \left[\hat{O}_H,\hat{H} \right] $$
This gave me
$$\Sigma_k \psi_k(r) \hat{a}_k(t) \Sigma_{ij} \langle i | \hat{H} | j \rangle \hat{a}_i ^{\dagger} \hat{a}_j - \Sigma_{ij} \langle i | \hat{H} | j \rangle \hat{a}_i ^{\dagger} \hat{a}_j \Sigma_k \psi_k(r) \hat{a}_k(t)=\Sigma_k \psi_k (r) \frac{\partial \hat{a}_k (t)}{\partial t}$$
I cancelled the ##\Sigma_k \psi_k(r) ## to give me the following but I'm not sure how to proceed after that
$$\ \hat{a}_k(t) \Sigma_{ij} \langle i | \hat{H} | j \rangle \hat{a}_i ^{\dagger} \hat{a}_j - \Sigma_{ij} \langle i | \hat{H} | j \rangle \hat{a}_i ^{\dagger} \hat{a}_j \hat{a}_k(t)= \frac{\partial \hat{a}_k (t)}{\partial t}$$
 
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TSny

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This gave me
$$\Sigma_k \psi_k(r) \hat{a}_k(t) \Sigma_{ij} \langle i | \hat{H} | j \rangle \hat{a}_i ^{\dagger} \hat{a}_j - \Sigma_{ij} \langle i | \hat{H} | j \rangle \hat{a}_i ^{\dagger} \hat{a}_j \Sigma_k \psi_k(r) \hat{a}_k(t)=\Sigma_k \psi_k (r) \frac{\partial \hat{a}_k (t)}{\partial t}$$
For reference, label this equation as (eq. 1).
(1) Should ##\hat{H}## be replaced by ##\hat{T}## in the two terms on the left of (eq. 1)?
(2) I believe a factor of ##i \hbar## is missing on the right side.
(3) You can replace the partial derivative with respect to ##t## on the right side by an ordinary derivative with respect to ##t## since ##\hat{a}_k(t)## is a function of a single variable, ##t##.

I cancelled the ##\Sigma_k \psi_k(r) ## to give me the following but I'm not sure how to proceed after that
$$\ \hat{a}_k(t) \Sigma_{ij} \langle i | \hat{H} | j \rangle \hat{a}_i ^{\dagger} \hat{a}_j - \Sigma_{ij} \langle i | \hat{H} | j \rangle \hat{a}_i ^{\dagger} \hat{a}_j \hat{a}_k(t)= \frac{\partial \hat{a}_k (t)}{\partial t}$$
Call this (eq. 2).

To get to (eq. 2), you can't actually cancel ##\Sigma_k \psi_k(r) ## on both sides of (eq. 1). But you can get (eq. 2) by multiplying both sides of (eq. 1) by an arbitrarily chosen ##\psi_n^*(r)## and integrating over all ##r##. Using the orthonormality of the ##\psi_k(r)##'s, you will end up with (eq. 2) except your index ##k## in (eq. 2) would be ##n##. However, at that point, you could rename ##n## as ##k## and therefore have (eq. 2) as you wrote it. Maybe all of this is what you actually meant when you said that you "canceled the ##\Sigma_k \psi_k(r) ## ".

So, your (eq. 2) actually looks OK to me except for fixing up points (1), (2), and (3) that I listed in regard to (eq. 1).

Since the various ##\langle i | \hat{T} | j \rangle## 's are just numbers, you can move ##\hat{a}_k(t)## in the first term of (eq. 2) inside the summation to obtain
$$\ \Sigma_{ij} \langle i | \hat{T} | j \rangle \hat{a}_k \hat {a}_i ^{\dagger} \hat{a}_j - \Sigma_{ij} \langle i | \hat{T} | j \rangle \hat{a}_i ^{\dagger} \hat{a}_j \hat{a}_k= i \hbar \frac{d\hat{a}_k }{dt}$$
Now try to rearrange the order of the ##\hat a## operators in the first term so that they match the order of the operators in the second term. That is, try to move the operator ##\hat{a}_k## past the operators ##{a}_i ^{\dagger} \hat{a}_j## in the first term. You will need to invoke the commutation relations obeyed by these operators. See what you end up with.
 
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Thank you for the help and I apologise for taking so long to reply
Your help was very useful, I used the commutation relation
$$\left[a_k,a_i ^{\dagger} \right] = a_k a_i ^{\dagger} - a_i ^{\dagger} a_k = \delta_{ki} $$
Rearranging this and replacing ##a_k a_i ^{\dagger} ## (and using the identity ##\langle i | j \rangle =\delta_{ij}## ) I ended up with the following
$$\sum_{ij} \langle i | T | j \rangle \langle k | i \rangle a_j + \sum_{ij} \langle i | T | j \rangle a_i ^{\dagger} a_k a_j - \sum_{ij} \langle i | T | j \rangle a_i ^{\dagger} a_j a_k = i \hbar \frac{d a_k}{dt}$$
The last two terms on the left hand side cancel off as ##a_j## and ##a_k## commute and since ##\langle i | i \rangle =1## the first term becomes what I was looking for
$$\sum_{j} \langle k | \hat{T} | j \rangle \hat{a_j} = i \hbar \frac{d\hat{a_k}}{dt}$$
Thank you!
 

TSny

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I used the commutation relation
$$\left[a_k,a_i ^{\dagger} \right] = a_k a_i ^{\dagger} - a_i ^{\dagger} a_k = \delta_{ki} $$
Rearranging this and replacing ##a_k a_i ^{\dagger} ## (and using the identity ##\langle i | j \rangle =\delta_{ij}## ) I ended up with the following
$$\sum_{ij} \langle i | T | j \rangle \langle k | i \rangle a_j + \sum_{ij} \langle i | T | j \rangle a_i ^{\dagger} a_k a_j - \sum_{ij} \langle i | T | j \rangle a_i ^{\dagger} a_j a_k = i \hbar \frac{d a_k}{dt}$$
The last two terms on the left hand side cancel off as ##a_j## and ##a_k## commute and since ##\langle i | i \rangle =1## the first term becomes what I was looking for
$$\sum_{j} \langle k | \hat{T} | j \rangle \hat{a_j} = i \hbar \frac{d\hat{a_k}}{dt}$$
OK. Personally, I wouldn't bother to bring in the identity ##\langle i | j \rangle =\delta_{ij}##. From the commutation relation, we have ##\ a_k a_i ^{\dagger} = a_i ^{\dagger} a_k + \delta_{ki} ##. Using this yields $$\sum_{ij} \langle i | T | j \rangle \delta_{ki}a_j+ \sum_{ij} \langle i | T | j \rangle a_i ^{\dagger} a_k a_j - \sum_{ij} \langle i | T | j \rangle a_i ^{\dagger} a_j a_k = i \hbar \frac{d a_k}{dt}$$The first term then reduces immediately to $$\sum_j \langle k| T | j \rangle a_j$$
 
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OK. Personally, I wouldn't bother to bring in the identity ##\langle i | j \rangle =\delta_{ij}##. From the commutation relation, we have ##\ a_k a_i ^{\dagger} = a_i ^{\dagger} a_k + \delta_{ki} ##. Using this yields $$\sum_{ij} \langle i | T | j \rangle \delta_{ki}a_j+ \sum_{ij} \langle i | T | j \rangle a_i ^{\dagger} a_k a_j - \sum_{ij} \langle i | T | j \rangle a_i ^{\dagger} a_j a_k = i \hbar \frac{d a_k}{dt}$$The first term then reduces immediately to $$\sum_j \langle k| T | j \rangle a_j$$
This is a more concise way of doing it, how exactly does it immediately reduce?
 

TSny

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This is a more concise way of doing it, how exactly does it immediately reduce?
When summing over ##i## in ##\sum_{ij} \langle i | T | j \rangle \delta_{ki}a_j##, the only value of ##i## that will contribute is when ##i = k##, due to the Kronecker delta ## \delta_{ki}##.
 
50
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Ahhh yes, I understand now, thank you very much, you have been very helpful
 

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