Help with second quantization practice problem

[CMJ96]

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

 

[TSny]

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 canceled 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.

 

[CMJ96]

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]

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 $$\sum_j \langle k| T | j \rangle a_j$$

 

[CMJ96]

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]

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 $\delta_{ki}$.

 

[CMJ96]

Ahhh yes, I understand now, thank you very much, you have been very helpful