# How is the Bloch equation derived?

I am reading a textbook on many-electron theory, which gives a Bloch equation in its generalized form:

$$\tag{1} [\Omega,H_0]P=Q(V\Omega-\Omega V_{eff})P,$$

where $$P$$ denotes the projection from exact state to a model state:

$$\tag{2} P\mid \psi^{(\alpha)}\rangle=\mid \psi_0^{(\alpha)}\rangle,$$

and $$\Omega$$ denotes a mapping from the model state to the exact state:

$$\tag{3} \Omega\mid \psi_0^{(\alpha)}\rangle=\mid \psi^{(\alpha)}\rangle,$$

and $$Q=1-P$$.

The textbook I use has an ambiguous derivation of the Bloch equation with many steps skipped. I would appreciation if someone could explain to me how it is derived.

It's eq. 55 on pg 21 of the textbook.

• It would be helpful for others if you cite the textbook from which you got the equation. – Rashid Rafeek Oct 26 '20 at 12:58
• +1. But I agree with Rashid. This is not the notation in which Bloch equations are typically written. I personally wouldn't even call this the Bloch equation, unless there's something I'm missing, which is probably the case, since without the textbook I can't see the context. – Nike Dattani Oct 26 '20 at 16:17
• Unfortunately I can see everything up to pg 20, and everything from pg 22 onwards, but pg 21 says "image not available". – Nike Dattani Oct 26 '20 at 22:07
• Really? That is weird. I can see up to page 30?@NikeDattani – Paulie Bao Oct 26 '20 at 22:55
• @PaulieBao unfortunately, google book previews, like Google searches, tend to be customized to the particular user. – Tyberius Oct 27 '20 at 1:32

## 1 Answer

They start with the time-independent Schrödinger equation:

$$\tag{1} H|\psi^\alpha\rangle = E^\alpha |\psi^\alpha\rangle.$$

Then they define $$|\psi_0^\alpha\rangle$$ to be what they call a "model state", which is an approximation to the true state $$|\psi^\alpha\rangle$$ but more easily accessible, and is related to the true state by a Møller operator $$\Omega$$, and to the true energy $$E^\alpha$$ by $$H_{\textrm{eff}}$$:

$$\tag{2} |\psi^\alpha\rangle \equiv \Omega |\psi_0^\alpha\rangle ~~~, ~~~ H_{\textrm{eff}}|\psi_0^\alpha\rangle\equiv E^\alpha |\psi_0^\alpha\rangle.$$

We can multiply the second part of Eq 2 by $$\Omega$$ on both sides of the equation, then re-write it using the first part of Eq 2:

\begin{alignat}{2} \tag{3} \Omega H_{\textrm{eff}}|\psi_0^\alpha\rangle &= E^\alpha |\psi^\alpha\rangle. \end{alignat}

Likewise we can simply substitute the first part of Eq 2 into Eq 1 and get:

\begin{alignat}{3} \tag{4} H \Omega |\psi_0^\alpha\rangle &= E^\alpha |\psi^\alpha\rangle. \end{alignat}

Since the right-side of Eqs 3 and 4 are the same, we can equate the left sides too:

\begin{alignat}{3} \tag{5} \Omega H_{\textrm{eff}}|\psi_0^\alpha\rangle &= H \Omega |\psi_0^\alpha\rangle. \end{alignat}

Then they say that since Eq 5 applies for all model states (i.e. all values of $$\alpha$$) we can replace $$|\psi_0^\alpha\rangle$$ by $$P$$:

\begin{alignat}{3} \tag{6} \Omega H_{\textrm{eff}}P &= H \Omega P. \end{alignat}

By partitioning $$H=H_0 + V$$ we get:

\begin{alignat}{3} \tag{6} \Omega H_{\textrm{eff}}P &= H_0 \Omega P + V \Omega P. \end{alignat}

What happens next is not explained in the text, but they get:

\begin{alignat}{3} \tag{7} H_{\textrm{eff}} &= PH_0 P + PV \Omega P. \end{alignat}

To get the last term, I would left-multiply all terms by $$P$$, but something else must have happened for the two other terms to be the way they are. Anyway, they make the second term in Eq. 7 to be defined as $$W$$ and remove the $$H_{\textrm{eff}}$$ to get the equation at the beginning of your question. They cite [113,117,124] but with your Google Books link, I can't see pages 61 to 405 and it seems the bibliography is at the end (in those missing pages). Do you happen to know what [113,117,124] are?

Anyway, they finally say that in the case where all model states have the same energy $$E_0$$, we get the original Bloch equation that they say was derived in the late 1950s by the French nuclear physicist Claude Bloch, (not the much more famous Bloch equation first presented in 1948 by the Swiss-American physicist Felix Bloch and his PhD student Roald Wangsness):

$$\tag{8} \left(E_0 - H_0 \right)\Omega P = V \Omega P - \Omega W.$$

Edit: After a lot more searching I have found that the author of that book explained this same derivation in a lot more detail in this 1974 paper. Pay attention to what he says about $$P$$ and $$\Omega$$ not being "true" inverses of each other, and what happens when you combine them. It is also quite amazing that the notation is almost the same in the original 1958 paper of Claude Bloch. The paper is in French, but it's not too hard to follow the equations (he shows quite a lot of steps!).

• Thanks for your explanation. Actually I follow all the step of derivation up to eq(6) (second eq(6)) but get stuck from equation 6 to equation 7. I will read in details on the reference . – Paulie Bao Oct 27 '20 at 7:53
• Actually, I manage to prove that if the model space is complete degenerate with a single energy $E_0$, my equation(1) will be reduce to the original form of Bloch equation (your equation 8) – Paulie Bao Oct 27 '20 at 9:42