Probability for a non-hermitian hamiltonian

[shinobi20]

Homework Statement

Given a non-hermitian hamiltonian with V = (Re)V -i(Im)V. By deriving the conservation of probability, it can be shown that the total probability of finding a system/particle decreases exponentially as e^{(-2*ImV*t)/ħ}

Homework Equations

Schrödinger Eqn, conservation of probability (∂P/∂t = -∇⋅j), P is the probability and j is the probability current density, Re V is the real part of V while I am V is the imaginary part of V

The Attempt at a Solution

It is known that ∫∫∫ P d³r = constant, that is ,the total probability for finding the particle should be constant throughout the universe.

iħ ∂ψ/∂t = -ħ²/2m ∇²ψ + Vψ = -ħ²/2m ∇²ψ + (Re)Vψ -i(Im)Vψ
By taking the complex conjugate, we have
-iħ ∂ψ*/∂t = -ħ²/2m ∇²ψ* +V*ψ* = -ħ²/2m ∇²ψ* + (Re)Vψ* +i(Im)Vψ*
By multiplying by ψ* the first equation and by ψ the second equation then subtracting the second from the first, we have
iħ ∂(ψψ*)/∂t = -ħ²/2m (ψ*∇²ψ-ψ∇²ψ*) - 2i (Im)V ψψ* = -∇⋅j - 2i (Im)V ψψ*

I don't know where to go already from here. Any suggestions or hints?

 

[blue_leaf77]

Just use the definition of total probability in quantum mechanics: $\langle \psi(t) | \psi(t) \rangle$. Now $|\psi(t)\rangle = e^{-iVt/\hbar} |\psi(0)\rangle$. What will you get then when calculating $\langle \psi(t) | \psi(t) \rangle$?

 

[shinobi20]

Just use the definition of total probability in quantum mechanics: $\langle \psi(t) | \psi(t) \rangle$. Now $|\psi(t)\rangle = e^{-iVt} |\psi(0)\rangle$. What will you get then when calculating $\langle \psi(t) | \psi(t) \rangle$?

I will get ⟨ψ(t)|ψ(t)⟩ = ⟨ψ(0)|ψ(0)⟩ but what does this tell me? Also, I need to use the conservation of probability equation to derive the question.

 

[blue_leaf77]

I will get ⟨ψ(t)|ψ(t)⟩ = ⟨ψ(0)|ψ(0)⟩ but what does this tell me?

No, you won't because in this problem $V$ is not Hermitian. Try plugging in $V = \textrm{Re}[V] - i \textrm{Im}[V]$.

 

[shinobi20]

No, you won't because in this problem $V$ is not Hermitian. Try plugging in $V = \textrm{Re}[V] - i \textrm{Im}[V]$.

Sorry, I forgot that it is not hermitian, so we will get e^2i(Im)Vt

 

[blue_leaf77]

It will be $\langle \psi(0) |e^{-i\textrm{Im}[V]t/\hbar}| \psi(0) \rangle$, at this point we should know how $\textrm{Im}[V]$ acts on $| \psi(0) \rangle$, i.e. whether it is a function of space at all. But since the question asks you to prove that "decreases exponentially as e^{(-2*ImV*t)/ħ}", I think it is implied that $\textrm{Im}[V]$ is a constant and thus the sandwiched exponential operator may be taken out.

 

[shinobi20]

It will be $\langle \psi(0) |e^{-i\textrm{Im}[V]t/\hbar}| \psi(0) \rangle$, at this point we should how $\textrm{Im}[V]$ acts on $| \psi(0) \rangle$, i.e. whether it is a function of space at all. But since the question asks you to prove that "decreases exponentially as e^{(-2*ImV*t)/ħ}", I think it is implied that $\textrm{Im}[V]$ is a constant and thus the sandwiched exponential operator may be taken out.

Oh! I did something wrong when taking the complex conjugate of V, but anyways, you really cleared everything. Thanks!