Can a complex potential have real roots?

[eljose]

Let be the Quantum Potential V(x)=A(x)+iB(x) with A and B real functions then my question is if this potential will have real roots if we take the expected value:

$$E_{n}=<\phi|H|\phi>$$ then the complex part of the energies will come from the expected value <B> so for real energies B should be Zero,but using Ehrenfrest,s theorem:

$$i\hbar\frac{d<B>}{dt}=<[B,H]>$$ B and H should commute, but we have that x and p do not commute so <B> can not be zero and there won,t be any real roots...

sorry: with "roots" i meant energies...

 

[eljose79]

Thank you for raising this interesting question about the Quantum Potential V(x)=A(x)+iB(x) and its relation to real roots and energies. I would like to offer my perspective on this issue.

Firstly, let me clarify that the concept of "roots" is usually associated with equations or functions, not potentials. Therefore, it may be more appropriate to refer to "real energies" instead of "real roots" in this context.

Now, let's address your question about whether the potential V(x) will have real roots (energies) if we take the expected value E_n=<\phi|H|\phi>. The answer to this question depends on the specific properties of the potential V(x) and the Hamiltonian H.

In general, the expected value E_n=<\phi|H|\phi> is a complex number, as it is a linear combination of the eigenvalues of the Hamiltonian H. However, if the potential V(x) is a real function, then the complex part of E_n will come from the expected value <B>. This means that for real energies, the expected value <B> should be zero.

However, as you correctly pointed out, the Ehrenfest's theorem i\hbar\frac{d<B>}{dt}=<[B,H]> requires that B and H commute. In other words, B and H should have the same set of eigenfunctions. This condition is not satisfied if the operators x and p do not commute, as in the case of the Quantum Potential V(x)=A(x)+iB(x).

Therefore, it is not possible for the expected value <B> to be zero and for there to be real energies in this scenario. This is because the potential V(x) is not Hermitian and does not have real eigenvalues. Instead, the energies will have a complex component due to the non-commutativity of x and p.

In conclusion, the Quantum Potential V(x)=A(x)+iB(x) may not have real energies if we take the expected value E_n=<\phi|H|\phi>. However, this does not mean that the potential is not physically meaningful. It simply reflects the inherent complexity of quantum systems and the non-commutativity of certain operators. I hope this explanation helps to clarify your question.

 

[denian]

I can confirm that a complex potential can indeed have real roots. The concept of complex numbers allows for the existence of both real and imaginary components in a potential, and the roots of a complex potential can be either real or complex numbers.

In the context of quantum mechanics, the energy levels of a system are related to the roots of the complex potential. The expected value of the potential, which includes both the real and imaginary components, can be used to calculate the energies of the system. However, as you have mentioned, the imaginary component of the energy comes from the expected value of the imaginary part of the potential.

It is true that in order for the energies to be real, the expected value of the imaginary part of the potential must be zero. However, this does not mean that the potential itself cannot have real roots. The expected value of the potential is just one way to calculate the energies, and it is possible for the potential to have both real and imaginary roots.

In conclusion, a complex potential can have real roots, and the expected value of the potential is just one way to calculate the energies of the system. Other methods, such as using the Schrödinger equation, can also be used to determine the energy levels, and these methods may yield different results. It is important to consider all possible approaches when studying complex potentials and their roots.