Advanced Dirac Notation Question

[Jeremiah Givens]

Hello everyone,

I have been working through some research papers on a topic that really interests me, but I believe I am misunderstanding a few things about Dirac Notation. I have expressions that read:
\begin{align*}
&< \psi_n \mid g(H - E_{n+1}) \mid \psi_n> \text{,} \\
&< \psi_n \mid (H - E_{n})g(H - E_{n+1}) \mid \psi_n> \text{, and} \\
&< \psi_n \mid (H - E_{n})g(H - E_{n+1})g(H - E_{n}) \mid \psi_n>\text{,}
\end{align*}
where $H$ is the hamiltonian of the system, $E_{i}$ are just scalars, and $g$ is a function of position.

Now I thought that these expressions would be expanded out for calculations as follows:
\begin{align*}
< \psi_n \mid g(H - E_{n+1}) \mid \psi_n> &= \int \psi_n^* g(H - E_{n+1}) \psi_n d \tau \\
&= \int \psi_n^* g(H\psi_n - E_{n+1}\psi_n) d \tau \text{, and} \\
< \psi_n \mid (H - E_{n})g(H - E_{n+1}) \mid \psi_n> &= \int \psi_n^* (H - E_{n})g(H - E_{n+1}) \psi_n d \tau \\
&= \int \psi_n^* (H - E_{n})g(H\psi_n - E_{n+1}\psi_n) d \tau \\
&= \int \psi_n^* (H(g(H\psi_n - E_{n+1}\psi_n)) - E_{n}(g(H\psi_n - E_{n+1}\psi_n))) d \tau \text{, and finally}\\
< \psi_n \mid (H - E_{n})g(H - E_{n+1})g(H - E_{n}) \mid \psi_n> &= \int \psi_n^* (H - E_{n})g(H - E_{n+1})g(H - E_{n}) \psi_n d \tau \\
&= \int \psi_n^* (H - E_{n})g(H - E_{n+1})g(H\psi_n - E_{n}\psi_n) d \tau \\
&= \int \psi_n^* (H - E_{n})g(H(g(H\psi_n - E_{n}\psi_n)) - E_{n+1}(g(H\psi_n - E_{n}\psi_n))) d \tau \\
&= \int \psi_n^* (H(g(H(g(H\psi_n - E_{n}\psi_n)) - E_{n+1}(g(H\psi_n - E_{n}\psi_n)))) - E_{n}(g(H(g(H\psi_n - E_{n}\psi_n)) - E_{n+1}(g(H\psi_n - E_{n}\psi_n)))))d \tau \\
\end{align*}

So, essentially, my question is whether this is the proper way to expand these out? I have been trying to do calculations with these expressions, and I cannot get them to work, and I believe this is where my mistake is.

One of my professors said that he believes you're supposed to take the operator on the left and operate on the wave function on the left, and take the one one the right and operate on the wave function on the right, which seams a simpler, but I have been unable to find a text explaining why it is done this way. This also raises another question: what about when there are three operators? Do I act on the right with the right most operator, left with the left most, and then have a choice for the middle? Does the hamiltonian ever operate on the $g$ function?

If anybody could help me clear this up, it would be greatly appreciated!

Thanks,
Jere

 

[PeterDonis]

@Jere, please note, the delimiter for inline LaTeX is two pound signs, not a dollar sign. I have used magic moderator powers to edit your post accordingly.

 

[Jeremiah Givens]

@Jere, please note, the delimiter for inline LaTeX is two pound signs, not a dollar sign. I have used magic moderator powers to edit your post accordingly.

Thanks for letting me know!

 

[DrClaude]

Operating on the left or on the right are the same. It is often a question of which is more convenient for a particular case (especially if the operator is hermitian)
$$
\begin{align*}
\langle \phi | \hat{A} | \psi \rangle &= \langle \phi | \left(\hat{A} | \psi \rangle \right) \\
&= \left( \langle \phi | \hat{A} \right) \psi \rangle
\end{align*}
$$
with
$$
\langle \phi | \hat{A} = \left( \hat{A}^\dagger | \phi \rangle \right)^\dagger
$$
When there are three operators, you can split them how you want, acting on the left or on the right.In your particular case, I don't understand what the Dirac bracket means. You say that $g$ is a function of position, but then use it as a function of an operator.

 

[A. Neumaier]

The integral signs and the $d\tau$ are spurious and don't belong there. Also, in general, $g$ and $H$ do not commute, so you cannot move one past the other. Finally, what is the point of expanding these expressions? They don't simplify, and you don't do anything else with them.

 

[Jeremiah Givens]

What do you mean spurious? Also, could you be more precise on what you mean by $g$ and $H$ cannot move past one another? You are correct in that they certainly do not commute in the cases I am dealing with. The point of expanding these expressions is to calculate the scalar that is obtained when the integrals are computed. For my purposes, I know what $\psi$, $g$, and $H$ are, and the integrals should converge. I will reply to Dr. Claude with some more exact questions and a description of where and how these arise, and perhaps that will help you make my questions more clear.

 

[Jeremiah Givens]

Operating on the left or on the right are the same. It is often a question of which is more convenient for a particular case (especially if the operator is hermitian)
$$
\begin{align*}
\langle \phi | \hat{A} | \psi \rangle &= \langle \phi | \left(\hat{A} | \psi \rangle \right) \\
&= \left( \langle \phi | \hat{A} \right) \psi \rangle
\end{align*}
$$
with
$$
\langle \phi | \hat{A} = \left( \hat{A}^\dagger | \phi \rangle \right)^\dagger
$$
When there are three operators, you can split them how you want, acting on the left or on the right.In your particular case, I don't understand what the Dirac bracket means. You say that $g$ is a function of position, but then use it as a function of an operator.

To be honest, I'm having a hard time understanding what the Dirac bracket means as well. I will explain where these expressions are coming from, and perhaps that will give you more insight as to what everything is supposed to mean.

I'm working through a series of papers on a new numerical technique for solving the Schrodinger Equation, and I am trying to recreate the calculations found in the papers (and apply it to new systems once I've got the grasp of it). So here is the basic idea:

We have the scaled Schrodinger Equation:
\begin{align*}
g(H - E)\psi = 0 \text{,}
\end{align*}
where the scaling function $g$ is a function of electron coordinates. It is a multiplicative operator and does not commute with the Hamiltonian. It is always positive or always negative except at the singularity points(like when $r=0$ in the coulombic potential terms that blow up when $r = 0$). Even at the singularity point $r_0$, the scaling function satisfies
\begin{align*}
\lim_{r \to r_0} gV \not= 0 \text{,}
\end{align*}
where $V$ is the potential operation in the Hamiltonian. The Scaled Schrodinger Equation (SSE) is equivalent to the original Schrodinger Equation, as the author proves.

Now the whole numerical technique relies upon the variational formula, which for the SSE is written as
\begin{align*}
\langle \delta \psi \mid g(H-E) \mid \psi \rangle = 0 \text{,}
\end{align*}
for some arbitrary $\delta \psi$.

Now I've been told there is a very similar variational formula used for finding the best possible wave function within a given functional form that is associated with the original Schrodinger Equation, but this is not how variational theory was presented to me in my classes, nor can I find a text that describes how to use it. If anyone knows of a text that covers the variational method using this notation, please let me know.

The energy (known as the scaled energy, which is different from the Ritz energy), is defined by
\begin{align*}
E = \dfrac{\langle \psi \mid gH \mid \psi \rangle}{\langle \psi \mid \psi \rangle} \text{.}
\end{align*}
Now we have two H-square theorems for the SSE that are valid only for wave functions that satisfy the SSE or SE, and they are written as
\begin{align*}
&\langle \psi \mid (H - E)g \cdot g(H - E) \mid \psi \rangle = 0 \text{, and} \\
&\langle \psi \mid (H - E)g(H - E) \mid \psi \rangle = 0 \text{.}
\end{align*}

Now when a $\psi$ including only one variable $C$ satisfies
\begin{align*}
\frac{d \psi}{d C} = g(H - E) \psi \text{,}
\end{align*}
it has the structure of the exact wavefunction (which is the really cool part of this technique!). Now when we have $\psi$ with the exact structure of the wavefunction, we can put it into the ordinary variational formula or the SSE variational principle to solve for the unknown variable $C$.

Now the main method I'm trying to use now is known as the simplest ICI theory(SICI), and is based on the SSE. The SICI based on the SSE is defined by the recursion formula
\begin{align*}
\psi_{n+1} = [1 + C_n g(H - E_n)]\psi_n \text{.}
\end{align*}
The unknown variable $C_n$ is determined by using either the ordinary variational principle (given by $\langle \delta \psi \mid H-E \mid \psi \rangle = 0$), or by using the SSE variational principle. When we use the SSE variational principle, we obtain the two-dimensional secular equation (if anyone can actually show the math to obtain these, I would greatly appreciate it, as it is not clear to me)
\begin{align*}
&\langle \psi_n \mid g(H - E_{n+1}) \mid \psi_n \rangle C_{0,n} + \langle \psi_n \mid (H - E_{n+1})g \cdot g(H - E_n) \mid \psi_n \rangle C_{1,n} = 0 \text{, and} \\
&\langle \psi_n \mid (H - E_{n})g \cdot g(H - E_{n+1}) \mid \psi_n \rangle C_{0,n} + \langle \psi_n \mid (H - E_{n})g \cdot g(H - E_{n+1})g(H-E_n) \mid \psi_n \rangle C_{1,n} = 0 \text{.}
\end{align*}

When we use the ordinary variational principle, we obtain (again, if anyone can show me how to get this, that would be great)

\begin{align*}
&\langle \psi_n \mid (H - E_{n+1}) \mid \psi_n \rangle C_{0,n} + \langle \psi_n \mid (H - E_{n+1})g(H - E_n) \mid \psi_n \rangle C_{1,n} = 0 \text{, and} \\
&\langle \psi_n \mid (H - E_{n})g(H - E_{n+1}) \mid \psi_n \rangle C_{0,n} + \langle \psi_n \mid (H - E_{n})g(H - E_{n+1})g(H-E_n) \mid \psi_n \rangle C_{1,n} = 0 \text{,}
\end{align*}
where the coefficient $C_n$ is given by $C_n = C_{1,n}/C_{0,n}$.

I believe, that once these equations are obtained, you can throw these two sets of equations into a matrix, set the determinant equal to zero, and solve for the value(s) of $E_{n+1}$. Then the value of $E_{n+1}$ can be plugged back into one of the equations to obtain the ratio $C_{1,n}/C_{0,n}$.

However, I need to know how to properly expand those equations out into integral form, in order to perform any of the calculations. My main question is, how exactly do those equations expand out into integral form? Does $g$ ever get operated on, and if so, where? Thank you very much for taking the time to help me with this.

 

[DrClaude]

Could you please cite the papers you are looking at?

 

[Jeremiah Givens]

General method of solving the Schrödinger equation of atoms and molecules by Hiroshi Nakatsuji. Attached is a pdf of this paper. If you would like anymore of his papers on the subjects, I have pdf's of almost all of them.

<Moderator's note: attachment deleted>

 

[DrClaude]

Unfortunately, I had to delete the attachment, as it can't be posted here for copyright reasons.

The reference is H. Nakatsuji, Phys. Rev. A 72, 062110 (2005).

I'll have a look and get back to you.

 

[Jeremiah Givens]

Thank you very much for taking the time to help me.