How does $ (\hat H_1 + \hat H_2)(\Psi_1 + \Psi_2) =\hat H_1 \Psi_1 + \hat H_2 \Psi_1 + \hat H_1 \Psi_2 + \hat H_2 \Psi_2 $?

Question

How does

$$ (\hat H_1 + \hat H_2)(\Psi_1 + \Psi_2) =\hat H_1 \Psi_1 + \hat H_2 \Psi_1 + \hat H_1 \Psi_2 + \hat H_2 \Psi_2 $$

and what are its implications?

Answer 1

$$\Psi = k\sin (na\pi x)$$

is an exact solution to the constraint equation (for any integer n)

$$-\frac{d^2}{dx^2}\Psi + v_0\Psi = E\Psi$$

since

$$-\frac{d^2}{dx^2}\Psi + V_0\Psi $$ $$= -\frac{d^2}{dx^2}k\sin (na\pi x) + V_0\Psi$$ $$= (na\pi)^2k\sin(na\pi x) + V_0\Psi$$ $$=[(na\pi)^2 + V_0]\Psi$$ $$=E\Psi$$

---

Consider the slightly modified constraint equation $$\left[\color{red}{-\frac{d^2}{dx^2} + v_0} + \color{orange}{\lambda \delta \left(x - \frac{a}{2} \right)} \right]\color{blue}{\Psi}= \color{purple}{E}\color{blue}{\Psi}$$

Assuming there exist $\Psi, E$ of the form,

$$\color{blue}{\Psi} = \Psi_0 + \Psi_1$$ $$\color{purple}{E} = E_0 + E_1$$

then

$$\left[\color{red}{-\frac{d^2}{dx^2} + v_0} + \color{orange}{\lambda \delta \left(x - \frac{a}{2} \right)} \right]\color{blue}{(\Psi_0 + \Psi_1)}= \color{purple}{(E_0 + E_1)}\color{blue}{(\Psi_0 + \Psi_1)}$$

$$ \left[-\frac{d^2}{dx^2} + v_0 + \lambda \delta \left(x - \frac{a}{2} \right) \right]\Psi_0 + \left[-\frac{d^2}{dx^2} + v_0 + \lambda \delta \left(x - \frac{a}{2} \right) \right]\Psi_1 = (E_0 + E_1)\Psi_0 + (E_0 + E_1)\Psi_1$$

$$ \color{red}{\left[-\frac{d^2}{dx^2} + v_0\right]}\Psi_0 + \color{orange}{\left[ \lambda \delta \left(x - \frac{a}{2} \right) \right]}\Psi_0 + \color{red}{\left[-\frac{d^2}{dx^2} + v_0\right]}\Psi_1 + \color{orange}{\left[ \lambda \delta \left(x - \frac{a}{2} \right) \right]}\Psi_1 $$ $$= \color{purple}{E_0}\Psi_0 + \color{purple}{E_1}\Psi_0 + \color{purple}{E_0}\Psi_1 + \color{purple}{E_1}\Psi_1$$

Hence

$$\boxed{\color{red}{H_0}\Psi_0 + \color{orange}{H_1}\Psi_0 + \color{red}H_0\Psi_1 + \color{orange}{H_1}\Psi_1 = \color{purple}{E_0}\Psi_0 + \color{purple}{E_1}\Psi_0 + \color{purple}{E_0}\Psi_1 + \color{purple}{E_1}\Psi_1}$$

is the natural expansion of

$$\boxed{(H_0 + H_1)(\Psi_0 + \Psi_1) = (E_0 + E_1)(\Psi_0 + \Psi_1)}$$

by normal distributive rule.

---

Assuming identities

$$\boxed{H_0\Psi_0 = E_0\Psi_0\\ H_0\Psi_1 = E_0\Psi_1}$$

$$\boxed{\int_{-\infty}^\infty \Psi_0 \Psi_0 = 1\\ \int_{-\infty}^\infty \Psi_0 \Psi_1 = 0}$$

we obtain

$$\color{pink}{E_0\Psi_0} + H_1\Psi_0 + \color{pink}{E_0\Psi_1} + H_1\Psi_1 = \color{pink}{E_0\Psi_0} + E_1\Psi_0 + \color{pink}{E_0\Psi_1} + E_1\Psi_1$$

$$H_1\Psi_0 = E_1\Psi_0 + (E_1-H_1)\Psi_1$$

Assuming $(E_1-H_1)$ should be a small quantity,

$$H_1\Psi_0 \approx E_1\Psi_0$$

Multiplying by $\Psi_0$ and integrating $\int_{-\infty}^\infty$,

$$\int_{-\infty}^\infty \Psi_0 H_1\Psi_0 \approx \int_{-\infty}^\infty \Psi_0 E_1\Psi_0$$

$$E_1 \approx \int_{-\infty}^\infty \Psi_0 H_1\Psi_0 $$

since $\int_{-\infty}^\infty \Psi_0 \Psi_0 = 1$.

---

Assuming that $\Psi_0$ is an linear combination of 3 functions:

$$\Psi_0 = \alpha \Psi_a + \beta \Psi_b + \gamma \Psi_c$$

Then

$$H_1\Psi_0 \approx E_1\Psi_0$$ $$H_1(\alpha \Psi_a + \beta \Psi_b + \gamma \Psi_c) \approx E_1(\alpha \Psi_a + \beta \Psi_b + \gamma \Psi_c)$$

1. Multiplying by $\Psi_a$ and integrating $\int_{-\infty}^\infty$,

$$\int_{-\infty}^\infty \Psi_a H_1(\alpha \Psi_a + \beta \Psi_b + \gamma \Psi_c) \approx \int_{-\infty}^\infty \Psi_aE_1(\alpha \Psi_a + \beta \Psi_b + \gamma \Psi_c)$$

$$\boxed{\alpha \int_{-\infty}^\infty \Psi_a H_1\Psi_a + \beta \int_{-\infty}^\infty \Psi_a H_1\Psi_b + \gamma \int_{-\infty}^\infty \Psi_a H_1\Psi_c = \alpha E_1}$$

since $$\int_{-\infty}^\infty \Psi_a \Psi_a = 1,\int_{-\infty}^\infty \Psi_a \Psi_b = 0, \int_{-\infty}^\infty \Psi_a \Psi_c = 0$$

2. Analogously multiplying by $\Psi_b$ and integrating $\int_{-\infty}^\infty$,

$$\boxed{\alpha \int_{-\infty}^\infty \Psi_b H_1\Psi_a + \beta \int_{-\infty}^\infty \Psi_b H_1\Psi_b + \gamma \int_{-\infty}^\infty \Psi_b H_1\Psi_c = \beta E_1}$$

3. Analogously multiplying by $\Psi_c$ and integrating $\int_{-\infty}^\infty$,

$$\boxed{\alpha \int_{-\infty}^\infty \Psi_c H_1\Psi_a + \beta \int_{-\infty}^\infty \Psi_c H_1\Psi_b + \gamma \int_{-\infty}^\infty \Psi_c H_1\Psi_c = \gamma E_1}$$

This results in a linear simultaneous equations in variables $\alpha, \beta, \gamma$

$$ \begin{bmatrix} \int_{-\infty}^\infty \Psi_a H_1\Psi_a & \int_{-\infty}^\infty \Psi_a H_1\Psi_b & \int_{-\infty}^\infty \Psi_a H_1\Psi_c \\ \int_{-\infty}^\infty \Psi_a H_1\Psi_a & \int_{-\infty}^\infty \Psi_b H_1\Psi_b & \int_{-\infty}^\infty \Psi_b H_1\Psi_c \\ \int_{-\infty}^\infty \Psi_c H_1\Psi_a & \int_{-\infty}^\infty \Psi_c H_1\Psi_b & \int_{-\infty}^\infty \Psi_c H_1\Psi_c \end{bmatrix} \begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix} \approx E_1 \begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix}$$

which may be solved for $E_1$ using standard matrix techniques.

---

Eg.

$$ \Psi = k\sin(a\pi n_1 x) \sin(a\pi n_2 y) \sin(a\pi n_3 z)$$

is an exact solution to the constraint equation (for any integer n)

$$ -\left(\frac{d^2}{dx^2} - \frac{d^2}{dy^2} - \frac{d^2}{dz^2}\right)\Psi = E\Psi$$

since

$$ \left(-\frac{d^2}{dx^2} - \frac{d^2}{dy^2} - \frac{d^2}{dz^2}\right)\Psi$$

$$=-\frac{d^2}{dx^2}k\sin(a\pi n_1 x) \sin(a\pi n_2 y) \sin(a\pi n_3 z) \\- \frac{d^2}{dy^2}k\sin(a\pi n_1 x) \sin(a\pi n_2 y) \sin(a\pi n_3 z) \\- \frac{d^2}{dz^2}k\sin(a\pi n_1 x) \sin(a\pi n_2 y) \sin(a\pi n_3 z)$$ $$=a^2\pi^2(n_1^2 + n_2^2 + n_3^2)\Psi$$ $$=E\Psi$$

Due of three-way additive partitioning of $E$, 3 distinct combinations

$$H_0\Psi_{112} = E_0 \Psi_{112}$$ $$H_0\Psi_{121} = E_0 \Psi_{121}$$ $$H_0\Psi_{211} = E_0 \Psi_{211}$$

will have identical $E_0 = 6a^2\pi^2$ and cannot be told apart.

Suppose now that a small $H_1$ is introduced to the constraint, then an ensemble of $$\Psi = \alpha \Psi_{112} + \beta \Psi_{121} + \gamma \Psi_{211}$$ will behave according to

$$H_1(\alpha \Psi_{112} + \beta \Psi_{121} + \gamma \Psi_{211}) \approx E_1(\alpha \Psi_{112} + \beta \Psi_{121} + \gamma \Psi_{211})$$

$$ \begin{bmatrix} \int_{-\infty}^\infty \Psi_{112} H_1\Psi_{112} & \int_{-\infty}^\infty \Psi_{112} H_1\Psi_{121} & \int_{-\infty}^\infty \Psi_{112} H_1\Psi_{211} \\ \int_{-\infty}^\infty \Psi_{112} H_1\Psi_{112} & \int_{-\infty}^\infty \Psi_{121} H_1\Psi_{121} & \int_{-\infty}^\infty \Psi_{121} H_1\Psi_{211} \\ \int_{-\infty}^\infty \Psi_{211} H_1\Psi_{112} & \int_{-\infty}^\infty \Psi_{211} H_1\Psi_{121} & \int_{-\infty}^\infty \Psi_{211} H_1\Psi_{211} \end{bmatrix} \begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix} \approx E_1 \begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix}$$

There are at most 3 distinct values of $E_1$ that satisfy this system of simultaneous equations for all $\alpha, \beta, \gamma$. Consistency over all ensembles of different $\alpha, \beta, \gamma$ numbers requires that at most 3 specific values of $E_1$ are observed.

The introduction of $H_1$ constraint has split the $E = E_0+E_1$ values between the distinct $\Psi_{112}, \Psi_{121}, \Psi_{211}$ partition compositions.

---

Eg.

$$\Psi_{111} = ke^{ar}$$

is the lowest $E$ hydrogen atom solution satisfying constraint

$$-k_0\nabla^2 \Psi - \frac{Zq_e}{r} \Psi = E \Psi$$

Suppose an improved (special relativistic) expansion of $\nabla^2$ is introduced as follows:

$$ \boxed{\nabla^* = - k_0\nabla^2 + k_1(\nabla^2)^2 - k_2 (\nabla^2)^3 + ...}$$

Then

$$\left( -k_0\nabla^2 + k_1(\nabla^2)^2 - k_2 (\nabla^2)^3 + ... \right)\Psi - \frac{Zq_e}{r} \Psi = E \Psi$$

$$\color{orange}{-k_0\nabla^2\Psi + ... - \frac{Zq_e}{r} \Psi} + \color{red}{k_1(\nabla^2)^2\Psi} = E \Psi$$

Therefore

$$H_1 = k_1(\nabla^2)^2$$

and evaluating

$$ E_1 \approx \int \int \int \Psi_{111} H_1 \Psi_{111} d\tau$$ $$ =k_1 \int (\Psi_{111} \nabla^2) (\nabla^2 \Psi_{111}) (r^2 dr)$$

$$ =\int_0^\infty\left(a^2 e^{ar} + \frac{2a}{r}e^{ar} \right)\left(a^2 e^{ar} + \frac{2a}{r}e^{ar} \right) (r^2 dr)$$

$$k_1\int_0^\infty\left(a^4 r^2 e^{2ar} + 4a^3re^{2ar} + 4a^2e^{2ar} \right) dr$$

$$ =k_1\int_0^\infty\left(\frac{a^4}{(2a)^3} \bar r^2 e^{\bar r} + \frac{4a^3}{(2a)^2}\bar r e^{\bar r} + \frac{4a^2}{(2a)}e^{\bar r} \right) d\bar r$$

$$= k_1(-a + 2a - 2a)$$

$$= -k_1 a $$

Therefore an improved testable $E \approx E_0 + E_1$ estimate is obtainable via $ E_1 \approx \int \int \int \Psi_{111} H_1 \Psi_{111} d\tau$.

---

Eg.

Suppose that further proposed fine structure corrections for hydrogen are of the form $H_1 = \frac{\kappa}{r^3}$.

Then $$E_1 \approx \int \int \int \Psi_{111} H_1 \Psi_{111} d\tau$$ $$=\int_{r=0}^\infty ke^{ar}\frac{\kappa}{r^3}ke^{ar}(r^2dr)$$ $$=\kappa \int_{r=0}^\infty e^{2ar}\frac{1}{r}dr$$

may no longer use integration by parts since $\int_{r=0}^\infty \frac{e^r}{r} dr = \log r + r + ...$ is unbounded as $r \rightarrow \infty$.

However, higher approximate solutions of the form $\Psi_{nlm} = r^l(1+...+r^{n-l-1})e^{\frac{r}{n}}\gamma_{ln}(\theta,\phi)$ can yield ever higher powers of $r^l$ which may be used to cancel the $\frac{1}{r}$ term. For instance, in the case of

$$\Psi_{210} = kre^{ar}$$

the constraint equation becomes

$$\frac{1}{r^2}\frac{d}{dr}(r^2\frac{d}{dr}\Psi_{210}) + \frac{1}{r} + \frac{1}{r^2} + \frac{1}{r^3} + \frac{1}{r^4} = E\Psi_{210} $$

$$\color{orange}{k} + \color{red}{\frac{e^{ar}}{r} + \frac{1}{r^2} + \frac{1}{r^3} + \frac{1}{r^4}} = E\Psi_{210} $$

Then

$$E_1 \approx \int_{r=0}^\infty \Psi_{210}H_1 \Psi_{210} (r^2 dr)$$ $$\int_{r=0}^\infty kre^{ar}\left( \frac{e^{ar}}{r} + \frac{1}{r^2} + \frac{1}{r^3} + \frac{1}{r^4} \right) kre^{ar} (r^2 dr)$$ $$\int_{r=0}^\infty ke^{ar}\left( r^3e^{ar} + r^2 + r + 1 \right) ke^{ar} dr$$

which may be solved using integration by parts.

---

Eg. (Coefficient extremization/variation):

$$ \Psi = ke^{-bx^2}$$

is an exact solution to the constraint equation

$$ (-\frac{d^2}{dx^2} + x^2 )\Psi = E \Psi$$

since

$$ -\frac{d^2}{dx^2} ke^{-bx^2} + x^2 ke^{-bx^2}$$ $$ = (2b-4b^2x^2)ke^{-bx^2} + x^2 ke^{-bx^2}$$ $$ = \left[ 2b + \color{red}{(1 - 4b^2)}x^2 \right]\Psi$$ $$ = E\Psi$$

when $$\color{red}{1 - 4b^2} = 0$$ $$ b = \pm \frac{1}{2}$$

---

4-Dimensional Solutions

$$\Psi = e^{k_0w}e^{k_1x}e^{k_2y}e^{k_3z}$$

is an exact solution to the constraint equation

$$(-\frac{d^2}{dw^2} + \frac{d^2}{dx^2} + \frac{d^2}{dy^2} + \frac{d^2}{dz^2})\Psi = E\Psi$$

since $$(-\frac{d^2}{dw^2} + \frac{d^2}{dx^2} + \frac{d^2}{dy^2} + \frac{d^2}{dz^2})e^{k_0w}e^{k_1x}e^{k_2y}e^{k_3z}$$ $$= (-k_0^2+k_1^2+k_2^2+k_3^2)e^{k_0w}e^{k_1x}e^{k_2y}e^{k_3z}$$ $$=E\Psi$$

for various compositions of $E=-k_0^2+k_1^2+k_2^2+k_3^2$.

---

Vector functions of the form

$$\Psi = \begin{bmatrix} (1)ke^{aw}e^{bx}e^{cy}e^{dz} \\ (0)ke^{aw}e^{bx}e^{cy}e^{dz} \\ \frac{d}{a+E}ke^{iaw}e^{bx}e^{cy}e^{dz} \\ \frac{b + ic}{a+E}ke^{aw}e^{bx}e^{cy}e^{dz} \end{bmatrix} $$

are exact solutions to the constraint equation (where the signs $+,-,-,-$ have been absorbed into the matrices)

$$\begin{bmatrix} 1&0&0&0 \\ 0&1&0&0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1\end{bmatrix} \frac{d\Psi}{dw} + \begin{bmatrix} 0&0&0&-1 \\ 0&0&-1&0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0\end{bmatrix}\frac{d\Psi}{dx} + \begin{bmatrix} 0&0&0&i \\ 0&0&-i&0 \\ 0 & -i & 0 & 0 \\ i & 0 & 0 & 0\end{bmatrix}\frac{d\Psi}{dy} + \begin{bmatrix} 0&0&-1&0 \\ 0&0&0&1 \\ 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0\end{bmatrix}\frac{d\Psi}{dz} = E\Psi$$

since

$$a\begin{bmatrix} 1&0&0&0 \\ 0&1&0&0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1\end{bmatrix}\begin{bmatrix} (1)ke^{aw}e^{bx}e^{cy}e^{dz} \\ (0)ke^{aw}e^{bx}e^{cy}e^{dz} \\ \frac{d}{a+E}ke^{iaw}e^{bx}e^{cy}e^{dz} \\ \frac{b + ic}{a+E}ke^{aw}e^{bx}e^{cy}e^{dz} \end{bmatrix} + b\begin{bmatrix} 0&0&0&-1 \\ 0&0&-1&0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0\end{bmatrix}\begin{bmatrix} (1)ke^{aw}e^{bx}e^{cy}e^{dz} \\ (0)ke^{aw}e^{bx}e^{cy}e^{dz} \\ \frac{d}{a+E}ke^{iaw}e^{bx}e^{cy}e^{dz} \\ \frac{b + ic}{a+E}ke^{aw}e^{bx}e^{cy}e^{dz} \end{bmatrix} + c\begin{bmatrix} 0&0&0&i \\ 0&0&-i&0 \\ 0 & -i & 0 & 0 \\ i & 0 & 0 & 0\end{bmatrix}\begin{bmatrix} (1)ke^{aw}e^{bx}e^{cy}e^{dz} \\ (0)ke^{aw}e^{bx}e^{cy}e^{dz} \\ \frac{d}{a+E}ke^{iaw}e^{bx}e^{cy}e^{dz} \\ \frac{b + ic}{a+E}ke^{aw}e^{bx}e^{cy}e^{dz} \end{bmatrix} + d\begin{bmatrix} 0&0&-1&0 \\ 0&0&0&1 \\ 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0\end{bmatrix}\begin{bmatrix} (1)ke^{aw}e^{bx}e^{cy}e^{dz} \\ (0)ke^{aw}e^{bx}e^{cy}e^{dz} \\ \frac{d}{a+E}ke^{iaw}e^{bx}e^{cy}e^{dz} \\ \frac{b + ic}{a+E}ke^{aw}e^{bx}e^{cy}e^{dz} \end{bmatrix} = m\Psi$$

$$ =\begin{pmatrix} \frac{a^2+aE}{a+E} - \frac{b^2+ibc}{a+E} - \frac{c^2-ibc}{a+E} - \frac{d^2}{a+E} \\ -\frac{bd}{a+E} - \frac{icd}{a+E} + \frac{bd+icd}{a+E} \\ -\frac{ad}{a+E} + \frac{ad + Ed}{a+E} \\ -\frac{ab + iac}{a+E} + \frac{ab + Eb}{a+E} + \frac{iac + iEc}{a+E} \end{pmatrix}\begin{bmatrix} ke^{aw}e^{bx}e^{cy}e^{dz} \\ ke^{aw}e^{bx}e^{cy}e^{dz} \\ ke^{iaw}e^{bx}e^{cy}e^{dz} \\ ke^{aw}e^{bx}e^{cy}e^{dz} \end{bmatrix} $$

$$ =\begin{pmatrix} \frac{\color{red}{a^2}+Ea-b^2-c^2-d^2}{a+E} \\ 0 \\ \frac{Ed}{a+E} \\ \frac{Eb + iEc}{a+E}\end{pmatrix}\begin{bmatrix} ke^{aw}e^{bx}e^{cy}e^{dz} \\ ke^{aw}e^{bx}e^{cy}e^{dz} \\ ke^{iaw}e^{bx}e^{cy}e^{dz} \\ ke^{aw}e^{bx}e^{cy}e^{dz} \end{bmatrix}$$

Assuming

$$\boxed{E^2 = a^2 - b^2 - c^2 - d^2}$$

$$ =\begin{pmatrix} \frac{\color{red}{E^2 + b^2 + c^2 + d^2}+Ea-b^2-c^2-d^2}{a+E} \\ 0 \\ \frac{Ed}{a+E} \\ \frac{Eb + iEc}{a+E}\end{pmatrix}\begin{bmatrix} ke^{aw}e^{bx}e^{cy}e^{dz} \\ ke^{aw}e^{bx}e^{cy}e^{dz} \\ ke^{iaw}e^{bx}e^{cy}e^{dz} \\ ke^{aw}e^{bx}e^{cy}e^{dz} \end{bmatrix}$$

$$ =\begin{pmatrix} \frac{E(a+E)}{a+E} \\ 0 \\ \frac{Ed}{a+E} \\ \frac{Eb + iEc}{a+E}\end{pmatrix}\begin{bmatrix} ke^{aw}e^{bx}e^{cy}e^{dz} \\ ke^{aw}e^{bx}e^{cy}e^{dz} \\ ke^{iaw}e^{bx}e^{cy}e^{dz} \\ ke^{aw}e^{bx}e^{cy}e^{dz} \end{bmatrix}$$

$$ = E \begin{pmatrix} 1 \\ 0 \\ \frac{d}{a+E} \\ \frac{b + ic}{a+E}\end{pmatrix}\begin{bmatrix} ke^{aw}e^{bx}e^{cy}e^{dz} \\ ke^{aw}e^{bx}e^{cy}e^{dz} \\ ke^{iaw}e^{bx}e^{cy}e^{dz} \\ ke^{aw}e^{bx}e^{cy}e^{dz} \end{bmatrix}$$

$$=E\Psi$$