Hello! I have some question about CPS collapse in the Minimally entangled typical thermal state algorithms.

Assume we have know the projection operator Pi(m), which project the ith site state to |m>. When we collapse from the first site of a METTS, the probability is p1(m) = (psi|P1(m)|psi).

after a collapse, the wave function should be updated as follow:(NJP, 12 (2010) 055026, Annals of Physics 326(2011) 96-192 sec. 8.3)

```
A_{1N}^{s_1} = <s_1|m>
A_{2N}^{s_2}= <m|s1> A_{1O}^{s1} A_{2O}^{s2}/sqrt(p1(m))
Latex:
A_{1N}^{s_{1}}=\left\langle s_{1}|m\right\rangle
A_{2N}^{s_{2}}=p_{1m}^{-\frac{1}{2}}\left\langle m|s_{1}\right\rangle A_{1O}^{s_{1}}A_{2O}^{s_{1}}
```

lower index N and O represent new and old respectively, and summation is assumed.

In the finite-T tutorial of ITensor 3.1.3 , the related code is:

```
ITensor jstate = (st==1) ? upState : downState;
if(j < N)
{
auto newA = psi(j+1)*(dag(jstate)*psi(j));
newA /= norm(newA);
psi.set(j+1,newA);
}
//Set site j tensor
psi.set(j,jstate);
```

here newA is not divided by the square root of the probability, newA /= norm(newA) normalize the projected state. But I noticed that the square root of the probability occurs here. For newA = psi(j+1)*(dag(jstate)*psi(j)), should it be like this newA = psi(j+1)*(dag(jstate))*psi(j) ?

For the absence of the square root of the probability here, can I understand like this:

when we divide the newA by the square root of the probability, we just multiply the wave function by a real number. It will not affect the collapse probability for the next step(if we do a normalization). And for the measurements, because of the normalization there is also no affect?