What does $|\Psi\rangle_S, |\Psi\rangle_H$ mean in bra-ket notation?

Question

Could anyone familiar with bra-ket notation help explain

the generalized Schrodinger equation: $$ H |\Psi(t)\rangle_S = i \frac{d}{dt}|\Psi(t)\rangle_S$$ One can therefore write: $$|\Psi(t)\rangle_S = e^{-iHt}|\Psi\rangle_H$$

1. What do the subscripts $S, H$ mean in $|\Psi(t)\rangle_S, |\Psi(t)\rangle_H$?

2. How does one get from equation 1 to equation 2?

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Attempt:

The $|\Psi\rangle$ seems related to $\Psi$ according to (in spherical coordinates):

$$ |\Psi\rangle = \int_{r=0}^\infty \int_{\theta=0}^\pi \int_{\phi=0}^{2\pi} \Psi dr d\theta d\phi$$

I am unable to see how the subscripts $|\Psi\rangle_S, |\Psi\rangle_H$ can come into play in the above integral?

Answer 1

I guess S means in Schrodinger's picture as opposed to H which is Heisernberg's picture. And as for how equation 2 is derived from 1, just plug equation 2 into 1 and solve.

Answer 2

I think $|\Psi\rangle_S$ and $|\Psi\rangle_I$ are used as follows. First $H$ is split into the two parts:

$$H = H_0 + H_1$$

where $H_1$ contains the interesting interacting parts and $H_0$ are to be eliminated wherever possible.

Then we have

$$ \boxed{\color{blue}{i\frac{d\Psi}{dt}} = H\Psi = (H_0 + H_1)\Psi}$$

Now consider the quantity $|\Psi\rangle_I = e^{iH_0t}|\Psi\rangle_S=e^{iH_0t}\Psi$:

$$ i\frac{d}{dt}\left(\color{red}{e^{iH_0t}\Psi}\right) = i \frac{d}{dt}(e^{iH_0t}\Psi) = \color{orange}{-H_0e^{iH_0t}\Psi} + \color{orange}{e^{iH_0t}H_0\Psi} + e^{iH_0t}H_1\Psi = H_1 \left(\color{red}{e^{iH_0t} \Psi }\right)$$

Solving this first order differential equation for $e^{iH_0t}\Psi$,

$$ i\frac{d}{dt}\color{red}{e^{iH_0t}\Psi} = H_1\color{red}{e^{iH_0t}\Psi}$$ $$ e^{iH_0t}\Psi = -i\int_{t=0}^{t1} H_1e^{iH_0t}\Psi dt + c$$

By physical considerations,

$$\boxed{c = 1}$$

$W$ on the right hand side can be integrated similarly up to some $t_2$ less than $t_1$, and again ad infinitum:

$$ = 1 - i\int_{t=0}^{t_1} H_1 \left(\color{purple}{1 - i\int_{t_2=0}^{t_2} H_1e^{iH_0t}\Psi dt_2}\right) dt_1 $$ $$= 1 - i\int_{t_1=0}^{t_1} H_1(t_1)dt_1 - H_1(t_1) i\int_{t_1=0}^{t1}\int_{t_2=0}^{t_2} H_1(t_2) e^{iH_0t}\Psi dt_2 dt_1$$ $$= 1 - i\int_{t_1=0}^{t_1} H_1(t_1)dt_1 - H_1(t_1) i\int_{t_1=0}^{t_1}\int_{t_2=0}^{t_2} H_1(t_2) \left(\color{purple}{1 - i\int_{t_3=0}^{t_2} H_1e^{iH_0t}\Psi dt_3}\right) dt_2 dt_1$$ $$= 1 + (- i)\int_{t_1=0}^{t_1} H_1(t_1)dt_1 + (-i)^2\int_{t_1=0}^{t_1}\int_{t_2=0}^{t_2}H_1(t_1) H_1(t_2) dt_2 dt_1 + (-i)^3\int_{t_1=0}^{t_1}\int_{t_2=0}^{t_2}\int_{t_3=0}^{t_3} H_1(t_1) H_1(t_2) H_1(t_3) e^{iH_0t}\Psi dt_3 dt_2 dt_1$$

In general,

$$ e^{iH_0t}\Psi = \sum_{n=0}^\infty (-i)^n \int_{t_1=0}^{t_1}\int_{t_2=0}^{t_2}\int_{t_3=0}^{t_3}...\int_{t_n=0}^{t_n}H_1(t_1)H_2(t_2)H_3(t_3)...H_n(t_n)$$

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Term $(-i)^2$:

$$ T_2 = \int_{t_1=0}^{t_1}\int_{t_2=0}^{t_2}H_1(t_1) H_1(t_2) dt_2 dt_1 $$ $$ = \int_{t_1=0}^{t}\int_{t_2=0}^{t}\theta(t_1-t_2) H_1(t_1) H_1(t_2) dt_2 dt_1 $$

where $ \theta(t_1-t_2) = 1$ only if $t_1-t_2 > 0$.

Term $(-i)^3$:

$$ T_3 = \int_{t_1=0}^{t_1}\int_{t_2=0}^{t_2}\int_{t_3=0}^{t_3} H_1(t_1) H_1(t_2) H_1(t_3) dt_3 dt_2 dt_1 $$ $$ \int_{t_1=0}^{t}\int_{t_2=0}^{t}\int_{t_3=0}^{t} \theta(t_1-t_2)\theta(t_2-t_3) H_1(t_1) H_1(t_2) H_1(t_3) dt_3 dt_2 dt_1 $$

since $\theta(t_1 - t_2)\theta(t_2 - t_3) = 1$ only if $t_1 > t_2 $ and $ t_2 > t_3$.

In general,

$$ \small{e^{iH_0t}\Psi = \sum_{n=0}^\infty (-i)^n \int_{t_1=0}^{t}\int_{t_2=0}^{t}...\int_{t_n=0}^{t}\theta(t_1-t_2)\theta(t_2-t_3)...\theta(t_{n-1}-t_n)H_1(t_1)H_2(t_2)...H_n(t_n)}$$

is numerically solvable to high summation orders.

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$$\boxed{\Psi_{x<0} = ae^{ikx} + be^{-ikx}}$$ $$\boxed{\Psi_{x>a} = fe^{ikx} + ge^{-ikx}}$$

since

$$ -\frac{d^2\Psi}{dx^2} + (0)\Psi$$ $$ = -\frac{d^2}{dx^2}(ae^{ikx} + be^{-ikx})$$ $$ = -\frac{d}{dx}(ikae^{ikx} + (-ik)be^{-ikx})$$ $$ = -(ik)^2ae^{ikx} - (-ik)^2be^{-ikx}$$ $$ = k^2ae^{ikx} + k^2be^{-ikx}$$ $$ = k^2\Psi = E\Psi$$

for any mixtures of $a,b,f,g$.

$$\boxed{\Psi_{0<x<a} = ce^{\kappa x} + de^{-\kappa x}}$$

since

$$ -\frac{d^2\Psi}{dx^2} + (V_0)\Psi$$ $$ = -\frac{d^2}{dx^2}(ce^{\kappa x} + de^{-\kappa x}) + V_0\Psi$$ $$ = -\frac{d}{dx}(\kappa ce^{\kappa x} + (-\kappa)de^{-\kappa x}) + V_0\Psi$$ $$ = -(\kappa)^2ce^{\kappa x} - (-\kappa)^2de^{-\kappa x} + V_0\Psi$$ $$ = (-\kappa ^2+V_0)\Psi = E\Psi$$

since the lack of $i$ reduces the $E =-\kappa^2+V_0$ requirement.

Ensuring continuity at boundary $x=0$:

$$\Psi_{x<0} = \Psi_{0<x<a}$$ $$ae^{ik(0)} + be^{-ik(0)} = ce^{\kappa(0)} + de^{-\kappa(0)}$$ $$\boxed{a+b = c+d}$$

and

$$ \frac{d}{dx}\Psi_{x<0} = \frac{d}{dx}\Psi_{0<x<a}$$ $$ \frac{d}{dx}(ae^{ikx} + be^{-ikx}) = \frac{d}{dx}(ce^{\kappa x} + de^{-\kappa x})$$ $$ ikae^{ik(0)} -ikbe^{-ik(0)} = \kappa ce^{\kappa(0)} - \kappa de^{-\kappa(0)}$$ $$ \boxed{ik(a - b) = \kappa (c - d)}$$

By inspection,

$$ \boxed{c = \frac{a+b}{2} + i\frac{k(a-b)}{2\kappa}\\d = \frac{a+b}{2} - i\frac{k(a-b)}{2\kappa}}$$

Ensuring continuity at boundary $x=a$:

$$ \Psi_{0<x<a} = \Psi_{x>a}$$ $$ \boxed{ce^{\kappa a} + d e^{-\kappa a} = fe^{ika} + ge^{-ika}} $$

and

$$ \frac{d}{dx}\Psi_{0<x<a} = \frac{d}{dx}\Psi_{x>a}$$ $$ \frac{d}{dx}(ce^{\kappa x} + de^{-\kappa x}) = \frac{d}{dx}(fe^{ikx} + ge^{-ikx})$$ $$ \kappa ce^{\kappa x} -\kappa de^{-\kappa x} = ikfe^{ikx} - ikge^{-ikx}$$

$$ \boxed{-i\frac{\kappa}{k} ce^{\kappa a} + i\frac{\kappa}{k} de^{-\kappa a} = fe^{ika} -ge^{-ika}} $$

By inspection,

$$ \boxed{f = \frac{e^{-ika}}{2}\left[ce^{\kappa a}\left(1 -i\frac{\kappa}{k}\right) + de^{-\kappa a}\left(1 + i\frac{\kappa}{k}\right)\right] \\g = \frac{e^{ika}}{2}\left[ce^{\kappa a}\left(1 + i\frac{\kappa}{k}\right) + d e^{-\kappa a}\left(1 - i\frac{\kappa}{k}\right)\right]}$$

Thus, there exist $a,b$ coefficients of $\Psi_{x<0}$ for which there remains non-vanishing $f,g$ coefficients of $\Psi_{x>a}$.

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Two-particle wavefunction over 1D barrier:

$$\boxed{\Psi_2(x,y) = e^{px} e^{qy} - e^{qx} e^{py}}$$

is a solution to the constraint equation with barrier potential $V_0$

$$-\frac{1}{2}\left(\frac{d^2}{dx^2}\Psi + \frac{d^2}{dy^2}\Psi \right) + V_0 \Psi = E\Psi$$

since

$$ -\left(\frac{d^2}{dx^2}(e^{px} e^{qy} - e^{qx} e^{py}) + \frac{d^2}{dy^2}(e^{px} e^{qy} - e^{qx} e^{py}) \right) + V_0 (e^{px} e^{qy} - e^{qx} e^{py})$$

$$ =-\left(\frac{d}{dx}(pe^{px} e^{qy} - qe^{qx} e^{py}) + \frac{d}{dy}(qe^{px} e^{qy} - p e^{qx} e^{py}) \right) + V_0 (e^{px} e^{qy} - e^{qx} e^{py})$$

$$ =-\left(p^2e^{px} e^{qy} - q^2e^{qx} e^{py} + q^2e^{px} e^{qy} - p^2 e^{qx} e^{py} \right) + V_0 (e^{px} e^{qy} - e^{-qx} e^{-py})$$

$$=-(p^2+q^2)(e^{px} e^{qy} - e^{qx} e^{py}) + V_0 (e^{px} e^{qy} - e^{qx} e^{py}) $$

$$ (-p^2-q^2+V_0)\Psi = E\Psi$$

allows for tunnelling solution $E = -p^2-q^2+V_0$.

Similarly,

$$\boxed{\Psi_1(x,y) = e^{ipx} e^{iqy} - e^{iqx} e^{ipy} }$$

is a solution to the constraint equation outside the barrier

$$-\left(\frac{d^2}{d{x}^2}\Psi + \frac{d^2}{d{y}^2}\Psi \right) + (0)\Psi = E\Psi$$

since

$$ -\left(\frac{d^2}{d{x}^2}(e^{ipx} e^{iqy} - e^{iqx} e^{ipy}) + \frac{d^2}{d{y}^2}(e^{ipx} e^{iqy} - e^{iqx} e^{ipy}) \right)$$ $$ (p^2 + q^2)\Psi = E\Psi$$

Thus, there exist suitable $p,q$ that will satisfy the $E$ requirement throughout for tunneling through a barrier of any large $V_0$.

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The wave function of entangled multi-particles cannot be represented as a product of the wave functions of each particle.

Eg.

Using 4-component ground state wavefunction of hydrogen:

$$ \Psi_1 = \begin{bmatrix} e^{-r} \\ 0 \\ i\cos\theta e^{-r} \\ i\sin\theta e^{i\phi}e^{-r} \end{bmatrix}, \Psi_2 = \begin{bmatrix} 0 \\ e^{-r} \\ i\sin\theta e^{i\phi}e^{-r} \\ -i\cos\theta e^{-r} \end{bmatrix}$$

Define

$$\Psi_{slater}(a, b) = \Psi_1(a)\Psi_2(b) - \Psi_2(a)\Psi_1(b)$$

and any two coordinates

$$ q_1 = [r_1,\theta_1,\phi_1]$$ $$ q_2 = [r_2,\theta_2,\phi_2]$$

Then

$$\Psi_{slater} (q_1, q_2) = - \Psi_{slater} (q_2, q_1)$$

and

$$\Psi_{slater} (q_1, q_1) = 0$$ $$\Psi_{slater} (q_2, q_2) = 0$$

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Single slit screen probability:

Double slit controlled initial phase:

Double slit uncontrolled (decoherent) initial phase 30 trials:

Measurement decoherence at slit 1 only:

Decoherence in 1 slit is enough to wash out the entire interference.