Is the Hamiltonian with a Complex Potential Hermitian?

[ehrenfest]

Homework Statement

Let $$V = V_r - iV_i$$, where V_i is a constant. Determine whether the Hamiltonian is Hermitian.

Homework Equations

$$H = \frac{-\hbar^2}{2m}*\Delta^2+V_r - iV_i$$

The Attempt at a Solution

I think you can distribute the Hamiltonian operator as follows:

$$H^{\dag} = \frac{-\hbar^2}{2m}*\left(\Delta^2\right)^{\dag}+V_r^{\dag}-iV_i$$

It doesn't say whether V_r is a constant or not, so we don't need to know that??
How do you take the adjoint of a derivative operator??
And yes those triangles should be upside down.

 

[Gokul43201]

$V_r$ and $-V_i$ refer to the real and imaginary parts of a complex potential, $V$. Assume $V_i \neq 0$, and find the Hermitian adjoint of V. What you've written above is not correct.

 

[olgranpappy]

Just because V_i is "constant" doesn't mean that you don't dagger it... it's a constant *operator*. The hamiltonian is
$$T+V_R-iV_I$$
and you know that $$T=T^{\dagger}$$ and $$V_R=V_R^\dagger$$ and $$V_I=V_I^\dagger$$ and finally, you know that for any operator $$O$$ you have $$(iO)^\dagger=-iO^\dagger$$... so...

 

[olgranpappy]

P.S. you should have mentioned much of this information implicitly in the statement of the problem which is obviously lacking a lot of information...

 

[ehrenfest]

P.S. you should have mentioned much of this information implicitly in the statement of the problem which is obviously lacking a lot of information...

Here the entire problem: Let $$V = V_r - iV_i$$ where the imaginary part V_i is a constant. Determine whether the Hamiltonian is Hermitian? Go through the derivaiton of the continuity equation and show that the total probability for finding the particles decreases with a rate of $$exp\left(-2V_it/\hbar\right)$$ (page 166 Shankar).

I did not think the second part was at all necessary information to do the first part.

Anyway,

Just because V_i is "constant" doesn't mean that you don't dagger it... it's a constant *operator*. The hamiltonian is
$$T+V_R-iV_I$$

You're right about the constant operator. Are you using T to replace -h-bar^2/2m *d^2x/dx^2?

and you know that $$T=T^{\dagger}$$ and $$V_R=V_R^\dagger$$ and $$V_I=V_I^\dagger$$ and finally, you know that for any operator $$O$$ you have $$(iO)^\dagger=-iO^\dagger$$... so...

How do you know that T, V_i, and V_r are all Hermitian?

 

[olgranpappy]

You're right about the constant operator. Are you using T to replace -h-bar^2/2m *d^2x/dx^2?

Yup. The kinetic energy term "T".

How do you know that T, V_i, and V_r are all Hermitian?

Because I know that r and p are both hermitian and thus

$$T=T(p)$$ (a real function of p, i.e. p^2/2m) is hermitian.

And apparently $$V_r=V_r(x)$$ is a real function of x which thus gives a Hermitian operator. For example, if
$$V=\frac{g}{24}x^4$$
where g is a real number then
$$V^{\dagger}=\frac{g}{24}{x^{\dagger}}^4=V$$

We are told that V_i is just a real number times the identity operator and the identity operator is Hermitian, so V_i is.

 

[Dick]

The free particle (derivative) part of the hamiltonian is self-adjoint. Use integration by parts to show this. A non-real potential (v_i!=0) is not. Isn't this bordering on the obvious?

 

[olgranpappy]

Isn't this bordering on the obvious?

That is insulting and inane.

 

[olgranpappy]

dick...

 

[Dick]

That is insulting and inane.

Yes, it is. Sorry. Perhaps we are simply confused. I know I am. It looks like you treating the 'i' in V_i*i (the second one) as an 'identity operator'? It's the imaginary unit 'i'. V_i may be hermitian but i*V_i is not.

 

[Gokul43201]

It looks like you treating the 'i' in V_i*i (the second one) as an 'identity operator'?

I believe you're still misunderstanding what olgramps has written. Vi is a constant - in matrix form, it is just a real number times the identity. i*Vi is not Hermitian because the elements along the main diagonal are no longer real.

Please let's avoid unnecessary bickering. If you think a post is insulting, use the "report post" button, rather than swing back in the thread.

 

[olgranpappy]

Please let's avoid unnecessary bickering. If you think a post is insulting, use the "report post" button, rather than swing back in the thread.

Yes, sorry about that.

Anyways, this thread really has been dragging on, hasn't it? Here's my one last attempt at a good explanation:

All of the information in H is contained in the matrix elements $$<\chi|H|\psi>$$ where chi and psi are arbitrary states.
For the case of single particle quantum mechanics in one-dimension let's write this explicitly:

$$<\chi|H\psi> =\int dx \chi(x)^* \left(\frac{-\nabla^2\hbar^2}{2m}\psi(x) +v_R(x)\psi(x)-iv_I(x)\psi(x) \right)$$
where the potential functions satisfy $$v_R^*=v_R$$ and
$$v_I^*=v_I$$. The K.E. term can be integrated by parts twice (we pick up two minus signs which is equivalent to no sign change) to act on the \chi instead of the \psi. Thus we have
$$\int dx \left((\frac{-\nabla^2\hbar^2}{2m}\chi^*(x))+v_R(x)\chi^*(x)-iv_I(x)\chi^*(x)\right) \psi(x) =\int dx \left((\frac{-\nabla^2\hbar^2}{2m}\chi(x))^*+(v_R(x)\chi(x))^*+(iv_I(x)\chi(x))^*\right) \psi(x)$$
where the i "switched sign" because I moved it under
the * symbol in the last term which really means I did nothing. So, this is
$$\int dx \left((\frac{-\nabla^2\hbar^2}{2m}\chi(x))+(v_R(x)\chi(x))+(iv_I(x)\chi(x))\right)^* \psi(x) =<H^\dagger\chi|\psi>$$
so...