Ground state wavefunction real?

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The discussion centers on whether the ground-state wavefunction can always be chosen to be real. It is established that this is true if the Hamiltonian has real coefficients and lacks first-order spatial derivatives, particularly for nondegenerate ground states. The conversation highlights that while absolute phases in quantum mechanics are irrelevant, relative phases matter, and choosing a real wavefunction may limit phase choices for other wavefunctions. The broader assertion is that any solution to the time-independent Schrödinger equation can be expressed as a real function under these conditions. Ultimately, the choice of a real ground state wavefunction simplifies the analysis of quantum systems.
krishna mohan
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Hi..

In a textbook, the ground-state wavefunction for any general Hamiltonian was under consideration. Then, a statement was made that this wave function is real since it is the ground state.

Is it true that one can always choose the ground state wave function to be real?

I understand that absolute phases don't matter in quantum physics, but relative phases do. Is it that one can choose the ground state wavefunction to be real and, as a consequence, lose the freedom of choosing the phases for the other wavefunctions?
 
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krishna mohan said:
Hi..

In a textbook, the ground-state wavefunction for any general Hamiltonian was under consideration. Then, a statement was made that this wave function is real since it is the ground state.

Is it true that one can always choose the ground state wave function to be real?

This is always the case if the Hamiltonian has real coefficients and no first order spatial derivatives. In this case, for any eigenfunction psi of H to the eigenvalue E, Re psi and
Im psi have the same property and are real. In particular, if the ground state is nondegenerate, Re psi and I am psi must be proportional, which means that the ground state is a constant multiple of a real function.
 
I understand the rest of it...

But why should there be no first order derivatives?
 
krishna mohan said:
why should there be no first order derivatives?

For simplicity, and since this holds for many concrete Hamiltonians.

The more general statement is that everything holds when the Hamiltonian, expressed in terms of the quantum position and momentum operators, has only real coefficients.
 
I take it you mean the solution to the time independent schordinger equation(TISE). If that is the case, then not just the ground state, any solution can be taken to be real. If psi is a solution of the TISE, then so is its complex conjugate psi* (do it and check it, this is so because the potential is taken to bereal). Since the differential equation is linear, you can now take any linear combination of psi and psi* as it will also be a solution to TISE.

So you make your life easier by choosing psi+psi* and choosing a real wavefunction.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA

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