The h Bar

Silly Goose

02:29

I am confused about how Sakurai introduces time-independent nondegenerate perturbation theory

sakurai's approach is to invert the operator $O$ on the LHS. okay. as a linear transformation, $O$ is singular on $\lvert n_0 \rangle$. okay. let's restrict the domain and codomain of $O$ to be $\mathcal{H} \backslash \{\lvert n_0 \rangle \}$. this is what I would think to do

instead, sakurai defines a complement projector to allow $O$ to be defined over the whole $\mathcal{H}$ while also not having to ever act on $\lvert n_0 \rangle$.

I don't understand why sakurai's approach (which later on allows you to simplify the first order correction of perturbed eigenstate) should be inequivalent to my approach

especially when $\lvert n \rangle$ is proved to not have a component in $\lvert n_0 \rangle$. hence, we should be able to simply restrict Hilbert space for the purpose of this calculation instead of inserting projections

the above is what I wish to do

oh whoops ignore the "sakurai's approach" in my first sentence above

this is the particular step in which sakurai's method succeeds and my approach fails

because we get an expression in which $O^{-1}$ is acting on a vector that is not contained in its domain or codomain

perhaps an argument can be made that because $\lvert n \rangle$ actually has no component in $\lvert n_0 \rangle$, we can safely assume that any term which contains $\lvert n_0 \rangle$ is cancelled at some point in the expansion

wait disregard everything i said lol

naturallyInconsistent

04:06

@SillyGoose This proof is actually rather remarkable. Most treatments would have failed to point this out.

Also, you have multiple computation errors

$\left|n_1\right>=O^{-1}(H_1-\Delta_1)\left|n_0\right>$

Silly Goose

oh yes is forgot to drop the $\epsilon$ on the RHS and include a $\Delta_n$ in the term that gets destroyed for one reasno or anothehr

naturallyInconsistent

$\forall k\in\mathbb Z^+\ :\qquad \left<n_0|n_k\right>=0$

Relativisticcucumber

HONK

naturallyInconsistent

H O N K

Silly Goose

@naturallyInconsistent okay so this is what justifies ignoring the second $\lvert n_0 \rangle$ term then?

Relativisticcucumber

04:14

Silly Goose

Because I would like to restrict the (co)domain of $O$ instead of using a complement projector like Sakurai

Relativisticcucumber

what is the difference between the operators in heisenberg vs interaction picture?

naturallyInconsistent

@SillyGoose He proved it. There is no ignoring? It is rigorously arrived at, no ignoring anything at all.

@SillyGoose I think these are the same thing in different statements, but I am worried of your tiny mistakes

Silly Goose

but the math does not work out that way unless you do what sakurai does (or other)

Relativisticcucumber

@SillyGoose keep calm and HONK

Silly Goose

04:17

or unless you enforce it as a condition, which is what I mean by just throwing it away

naturallyInconsistent

In Heisenberg picture, the wavefunction is purely independent of time. In the interaction picture, the wavefunction changes only via the interaction.

In the Heisenberg picture, the operators change in time due to the full Hamiltonian. In the Interaction picture, the operators change in time due only to the free Hamiltonian.

@SillyGoose I am quite sure that the maths would have worked out if you do it correctly.

Relativisticcucumber

@naturallyInconsistent if this is in the context of time dep potentials, is the free ham the time ind one and the interaction is the time dep potential?

naturallyInconsistent

@Relativisticcucumber we will almost always pick it to be so, but there is no reason why we cannot have a time dependent free Hamiltonian. Whatever is exactly solvable and close enough to the problem we want solved, we can use it as the free Hamiltonian

Relativisticcucumber

@naturallyInconsistent okay and i feel i am missing the point of the interaction picture. do you have insights into why this approach/split is useful?

Silly Goose

hm well i fixed the two errors i saw but are you saying i made other mistakes in this

oh the $\Delta_n$ should be a $\Delta_1$ in the last line

Relativisticcucumber

04:25

also, wait, in the above picture, the time evolution operator only has $H_0$ in it, so how can the entities (states and operators) be impacted by the interaction if it doesn't appear?

naturallyInconsistent

Well, I am actually working on Sakurai in precisely that chapter, so should we start over and get this sorted out?

Silly Goose

sounds good

naturallyInconsistent

@Relativisticcucumber Under what context are you looking at this? We often only use these things in QFT, so are you learning QFT?

Relativisticcucumber

in sak ch 5 as well XD

i am learning qft but rn im doing time dep perturbation theory/interaction picture intro in quantum (sak)

so in the context of qm

naturallyInconsistent

In the context of QM, you should not need to leave Schrödinger picture. You gain not much from even interaction picture.

It is in QFT that you gain a lot.

Silly Goose

04:28

this is the conclusion that i am at atm

naturallyInconsistent

@SillyGoose There are mistakes in this, which I have to comb through. Please wait a little.

@Relativisticcucumber In QFT, we want to be able to say that, for free particles that are travelling inwards or outwards of the interaction zone for infinitely far away and infinite time to do so, those states are approximately time-independent. That is where the separation of the Hamiltonian into the free part and the interacting part is useful, because we can just focus upon the interesting interacting part, ignoring the infinite phase factors from the free part just free-time-evolution

Relativisticcucumber

@naturallyInconsistent oh i see

naturallyInconsistent

@SillyGoose, let us start with Sakurai Equation (5.19) $$\tag{5.19}(H_0+\lambda H_1)\left|n\right>{}_{exact}=E_n{}^{exact}\left|n\right>{}_{exact}$$ which I have chosen to match your notation

Silly Goose

sounds good

naturallyInconsistent

To be particularly clear, we should insert a few steps to clarify what Sakurai did to get Equation (5.21). Namely, he did $$(\lambda H_1-E_n^{\text{exact}}+E_n)\left|n\right>_{\text{exact}}=(E_n-H_0)\left|n\right>_{\text{exact}}$$

Silly Goose

04:39

right

naturallyInconsistent

He had annoyingly defined $$\tag{5.20}\Delta_n\equiv E_n^{\text{exact}}-E_n$$ which hides the fact that this is really proportional to $\lambda$. So I am going to define $$\Delta_n=\lambda D_n$$ for convenience, so that my Equation (5.21) is going to be $$\tag{5.21 modified}(E_n-H_0)\left|n\right>_{\text{exact}}=\lambda(H_1-D_n)\left|n\right>_{\text{exact}}$$

The $\left|n_0\right>$ argument is this: apply $\left<n_0\right|$ on the left on Equation (5.21)$$\begin{align}\left<n_0\right|E_n-H_0\left|n\right>_{\text{exact}}&=\lambda\left<n_0\right|H_1-D_n\left|n\right>_{\text{exact}}\\
0&=\lambda\left<n_0\right|H_1-D_n\left|n\right>_{\text{exact}}\end {align}$$

This is our version of Equation (5.22). The important thing to note here is that this is an identity that holds for all $\lambda$ within some unspecified region of convergence (that we hope isn't zero, and might well fail in this hope).

Now, $\left<n_0\right|$ and $H_1$ are independent of $\lambda$, but $D_n$ and $\left|n\right>_{\text{exact}}$ are secretly functions of $\lambda$. This means that there is no way for higher orders of $\lambda$ in their expansions to cancel any dependence. i.e. the only way for Equation (5.22) to hold, is that all corrections to the wavefunctions to be orthogonal to $|n_0\rangle$

Silly Goose

okay that i believe now

hm well then im not sure where my calculation goes wrong

naturallyInconsistent

04:57

Our version of Equation (5.34) reads as $$\tag{5.34 modified}\left|n\right>_{\text{exact}}=\left|n_0\right>+\lambda O^{-1}(H_1-D_n)\left|n\right>_{\text{exact}}$$ and $$\tag{5.35 modified}D_n=\left<n_0\right|H_1\left|n\right>_{\text{exact}}$$

Now we do the wretched computation Equation (5.38) to obtain Equation (5.39)$$\begin{align}\left|n_0\right>+\lambda\left|n_1\right>+\cdots&=\left|n_0\right>+\lambda O^{-1}(H_1-D_n^0-\cdots)(\left|n_0\right>+\cdots)\\\left|n_1\right>&=O^{-1}(H_1-D_n^0)\left|n_0\right>\\&=O^{-1}H_1\left|n_0\right>\end {align}$$

And I am sure it is this last step that you had not internalised. Sakurai was being annoying for omitting such a crucial clarifying statement here.

@SillyGoose That is what you wrote as "justify vanishing" earlier

Silly Goose

is $D_n^0 = 0$, i..e the 0th order change in energy?

naturallyInconsistent

@SillyGoose ooops, maybe I should have called it $D_n^1$ instead.

Silly Goose

oh i see okay

yes this last step i am unsure of

naturallyInconsistent

$$\left|n\right>_{\text{exact}}=\sum_{k=0}^\infty\lambda^k\left|n_k\right>$$ $$D_n=\sum_{r=1}^\infty\lambda^{r-1}D_n^r$$

@SillyGoose It is trivial. As you yourself wrote, $D_n^1$ is a pure number, so it commutes with $O^{-1}$ as you noted, and then $O^{-1}$ acts on $|n_0\rangle$, which it is explicitly orthogonal to.

Silly Goose

05:13

but the argument i want to make is to start with $O: \mathcal{H} \rightarrow \mathcal{H}$, restrict its domain and codomain so that $O: \mathcal{H} \backslash \{\lvert n_0 \rangle \} \rightarrow \mathcal{H} \backslash \{\lvert n_0 \rangle \}$, which removes the singularity and allows us to actually define an inverse of $O$. But if I make this restriction, then it does not make sense to write $O^{-1}\lvert n_0 \rangle$. because $\lvert n_0 \rangle$ is not in the domain of $O^{-1}$

however, it is doubly confusing because we can prove that $\lvert n_1 \rangle$ has no components in $\lvert n_0 \rangle$ with a proof that is totally independent of whatever we want to do with $O$

naturallyInconsistent

@SillyGoose So, maybe you can just abandon your restricted domain and codomain thingy and just accept that Sakurai is correct?

It is, after all, the simplest way to proceed.

Silly Goose

hm but it seems like an uncomfortably arbitrary choice

to choose one of 1) restrict domain and codomain, 2) redefine the operator by assigning (essentially a constant value) to the singularity

Relativisticcucumber

@SillyGoose u hate arb choices

Silly Goose

and the second option (which sakurai does) seems really arbitrary

@Relativisticcucumber they nasty

naturallyInconsistent

You ought to note that in Equations (5.41), (5.44) and (5.48), Sakurai explicitly shows that there are more and more terms appearing from the $D_n$ term. It is only vanishing in the first step. I kinda think this is coming from the normalisation choice, but without explicit computation I cannot be sure.

Silly Goose

05:19

H O N K

naturallyInconsistent

@SillyGoose No, it is not at all arbitrary. Sakurai showed above that the exact answer would never have any contributions in the $|n_0\rangle$ part, so it is perfectly legitimate to handle the singularity by defining the projection to discard them as zero.

@SillyGoose H O N K

Silly Goose

but if we proved that there shouldn't be any $\lvert n_0 \rangle$ anywhere, then there shouldn't be. but we just showed that there is and we have to ourselves actually redefine an operator to agree with this fact.

naturallyInconsistent

@SillyGoose no, we showed that there is not; whatever terms that appear to seem like it will produce a $|n_0\rangle$ term, has to be fake. They must be discarded.

Note that in Equation (5.24) it is explicitly defined that this reciprocal operator has no $|n_0\rangle$ term.

It is literally well-defined.

Silly Goose

I agree that $\phi_nO^{-1}$ is well defined

naturallyInconsistent

We did not "redefine" any operator. It is well-defined from the beginning, because Sakurai was careful enough to analyse that particular singularity from the beginning.

Silly Goose

05:27

but the operator is $O^{-1}$ not $\phi_n O^{-1}$ unless I am misunderstanding

naturallyInconsistent

Well, Sakurai only ever had $\frac{\phi_n}{E_n-H_0}$ operator, so, if your desired alternative route is actually going into a dead end, there is no good reason to be avoiding Sakurai's well-defined route.

Silly Goose

well maybe my experience is biased

if I have an equation $x = x^2$, then I can divide both sides by $x$ given that I restrict the values that $x$ can take on--nothing more nothing less--so $1 = x$ for $x \neq 0$ is valid.

naturallyInconsistent

@SillyGoose It is actually more supposed to be $x-x^2=0=x(1-x)$ so that either $x=0$ or $x=1$ (and in this case there is no way to have both).

Silly Goose

well for this particular case yes there is a nice way to solve it avoiding all problems and introductions of additional structure, but i was trying to do the dividing way so as to evoke a situation closer to what seems problematic to me and how i would handle such a problematic situation in a less complicated setting

naturallyInconsistent

I think what Sakurai is doing here is pretty standard. You will see it again in degenerate perturbation theory. You should have seen it before in classical mechanics when defining normal modes. Quite a lot of singularity handling is handled by defining things so that the numerator goes to zero faster than the denominator.

Things like Cramer's rule for normal modes, critically depends upon such shenanigans.

Silly Goose

05:37

cripes to the shenanigans

naturallyInconsistent

i.e. even if you are purely in maths land, you might well have to face these shenanigans too.

I am actually very happy with how Sakurai even bothered to treat the singularity here. Most textbooks would have just glossed over such headaches.

hmm, now that I am looking at this again, I am annoyed that we are defining terms this way. We are almost always interested in finding the energy eigenfunctions from ground state upwards, so we should try to arrange things so that as many things are positive as possible, with finite numbers of negatives. We should have $H_0 - E_n$ rather than $E_n-H_0$, so $D_n-H_1$

Slereah