Undergrad Different expressions about time reversal and confusion

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Sakurai states that if ##\psi(x,t)## is a solution to the Schrödinger equation, then ##\psi^*(x,-t)## is also a solution. In contrast, Bruus and Flensberg specify that this holds only under the condition of time reversal symmetry, where ##H=H^*##. The discussion highlights that Sakurai's assertion assumes a general property of the Hamiltonian. It is clarified that for the Hamiltonian in Sakurai's example, the Hermitian condition ensures that ##H=H^*## is satisfied. Ultimately, the correctness of the statements hinges on the specific conditions of the Hamiltonian in question.
hokhani
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TL;DR
Discrepancy between Sakurai and Bruus
In Sakurai, section 4.4, says that if ##\psi(x,t)## is a solution of Schrodinger equation then ##\psi^*(x,-t)## is always another solution. However, in the corrected version of "Many-body Quantum Theory in Condensed Matter Physics, Bruus and Flensberg, 2016", section 7.1.4, restricts this condition and says that if we have time reversal symmetry ##H=H^*## then ##\psi^*(x,-t)## would be another solution. I would like to know which of these statements is correct.
 
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hokhani said:
TL;DR Summary: Discrepancy between Sakurai and Bruus

In Sakurai, section 4.4, says that if ##\psi(x,t)## is a solution of Schrodinger equation then ##\psi^*(x,-t)## is always another solution. However, in the corrected version of "
Many-body Quantum Theory in Condensed Matter Physics, Bruus and Flensberg, 2016", section 7.1.4, restricts this condition and says that if we have time reversal symmetry ##H=H^*## then ##\psi^*(x,-t)## would be another solution. I would like to know which of these statements is correct.
Sakurai must assume that the Hamiltonian has that property in general. It's not hard too see from the SDE that ##H = H^*## is required for this to make sense.
 
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Thanks, it seems that for the Hamiltonian in Sakurai, ##H=\frac{P^2}{2m}+V(x)##, which is in the position space, the Hermitian condition ##H=H^{\dagger}## fulfils automatically ##H=H^*## since ##V^\dagger=V## demands ##V=V^*##.
 
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