- 50

- 0

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

$$\ \hat{a}_k(t) \Sigma_{ij} \langle i | \hat{H} | j \rangle \hat{a}_i ^{\dagger} \hat{a}_j - \Sigma_{ij} \langle i | \hat{H} | j \rangle \hat{a}_i ^{\dagger} \hat{a}_j \hat{a}_k(t)= \frac{\partial \hat{a}_k (t)}{\partial t}$$

$$ \hat{H} = \Sigma_{ij} \langle i| \hat{T} | j \rangle \hat{a_i }^{\dagger} \hat{a_j} $$

$$\hat{\psi}(r,t)= \Sigma_k \psi_k(r) \hat{a}_k(t)$$

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

$$\ \hat{a}_k(t) \Sigma_{ij} \langle i | \hat{H} | j \rangle \hat{a}_i ^{\dagger} \hat{a}_j - \Sigma_{ij} \langle i | \hat{H} | j \rangle \hat{a}_i ^{\dagger} \hat{a}_j \hat{a}_k(t)= \frac{\partial \hat{a}_k (t)}{\partial t}$$

Last edited: