Transition probability density between two transmitted states

[amjad-sh]

suppose we are working on a step potential problem, and two transmitted wave functions,corresponding to one particle, are obtained. Let's name them $|1>$ and $|2>$. How can we interpret physically the case where $<1|2>$=$-<2|1>$? or in position representation,$\psi_1^*(x)\psi_2(x)=-\psi_2^*(x)\psi_1(x)$?
I know this means that the probability of transition from state 1 to state 2 is the same as from state 2 to state 1, but what is the role of the negative sign?

 

[vanhees71]

It's simply telling you that the scalar product is purely imaginary. There's no specific physical meaning of that.

It's important to note that the position representation of the scalar product doesn't provide a new function but a complex number! It's given by inserting an identity operator (a great deal of QT is understood quite well when you know which decomposition of the identity opreator to use at the right place ;-))).
$$\langle 1|2 \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 x \langle 1|\vec{x} \rangle \langle \vec{x}|2 \rangle = \int_{\mathbb{R}^3} \mathrm{d}^3 x \psi_1^*(x) \psi_2(x).$$

 

[amjad-sh]

It's simply telling you that the scalar product is purely imaginary. There's no specific physical meaning of that.

It's important to note that the position representation of the scalar product doesn't provide a new function but a complex number! It's given by inserting an identity operator (a great deal of QT is understood quite well when you know which decomposition of the identity opreator to use at the right place ;-))).
$$\langle 1|2 \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 x \langle 1|\vec{x} \rangle \langle \vec{x}|2 \rangle = \int_{\mathbb{R}^3} \mathrm{d}^3 x \psi_1^*(x) \psi_2(x).$$

suppose we are dealing with a step potential problem, where there is SOC present at the interface.
the wave function for z>0 is represented by:
$\phi(z>0)=(e^{ikz}+(t_0+t_x)e^{-ikz})\chi_{+}$
$\phi^{\dagger}(z>0)=\chi_{+}^{\dagger}(e^{-ikz}+(t_0^{*}+t_x^{*})e^{ikz})$
where $\chi_+^{\dagger}=\begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix}$
$t_x$ is reflection coefficient that arises from the spin orbit coupling.
$t_0$ is the transmission coefficient that arise from the difference of potential between region(z<0) and the region (z>0).
Can we consider $t_0\chi_+e^{-ikz}$ and $t_x\chi_{+}e^{-ikz}$ as two different quantum states, where $t_0$ and $t_x$ are general complex numbers?
If we let $|1 \rangle=t_0\chi_+e^{-ikz}$ and $|2 \rangle=t_x\chi_{+}e^{-ikz}$, would $\chi_+^{\dagger}t_0^*t_x\chi_+$ be defined as the transition amplitude from $|2\rangle$ to$|1\rangle$?

 

[vanhees71]

I don't understand which setup you have in mind. What's SOC?

Also the states are strange. Usually you assume that there's an incoming wave starting far from the step potential at $z \rightarrow -\infty$. This implies for the energy modes you have to make the ansatz
$$\phi_<(z)=A \exp(\mathrm{i} k z) + B \exp(-\mathrm{i} k z),\\
\phi_>(z)=C \exp(\mathrm{i} k z),$$
where $\phi_<$ is for $z<0$ and $\phi_>$ for $z>0$ (the potential step is assumed to be at $z=0$).

If you have something with spin, the $A$, $B$, $C$ are spinor valued. For the details, I'd need the specific problem properly stated!

 

[amjad-sh]

the wave functions must be like this :
$\phi_{+}(z>0)=(t_0+t_x\sigma_x+t_y\sigma_y)e^{ikz}\chi_+$
$\phi^{\dagger}_+(z>0)=\chi_+^{\dagger}(t_0^*+t_x^*\sigma_x+t_y^*\sigma_y)e^{-ikz}$
SOC is the abbreviation of spin orbit coupling.
$\chi_+$ is the eigenvector of $\sigma_x$
the Hamiltonian of the system is $\hat{H}=\dfrac{p^{2}}{2m} -\dfrac{\partial_z^2}{2m} +V(z) +V'(z)(\vec{z} \times \vec{p})\cdot \vec{\sigma}$
where $V'(z)(\vec{z} \times \vec{p})\cdot \vec{\sigma}$ is the spin orbit coupling term, and $p^{2}/2m -\dfrac{\partial_z^2}{2m}$ is the kinetic energy term.
$t_x=ap_y$ and $t_y=bp_x$ where a and b are constants.
The wave functions is incident with spin polarized in x-direction and it is in the spin up state.
Particle moving in the x-direction will spin-flip to the spin-down state,when transmitted to z>0, as $\sigma_y\chi_+=-i\chi_-$, which is the spin down state, and -i is just a phase factor.
particle moving in the y-direction will remain polarized in the x direction and in the spin up state, when transmitted to region z>0, as $\sigma_x\chi_+=\chi_+$.
I'm just interested in $(t_0+t_x)e^{ikz}\chi_+$, and my question is that can we consider $t_0e^{ikz}\chi_+$ and $t_xe^{ikz}\chi_+$ as different physical states, since the first stems from the potential difference between region (z<0) and region (z>0) and the second stems from the spin orbit coupling term in the Hamiltonian, or they are physically the same?
and how can we interpret $\chi_+^{\dagger}t_0^*t_x\chi_+$

 

[vanhees71]

It's not an insult. I simply really didn't understand the setup you want to describe. Now I have a better idea, but still don't understand the question. Maybe somebody else can help.

 

[amjad-sh]

It's not an insult. I simply really didn't understand the setup you want to describe. Now I have a better idea, but still don't understand the question. Maybe somebody else can help.

ok, sorry for the misunderstanding.