- #1
LagrangeEuler
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- 20
Is there some mathematical prove that for ground state in QM problems wave function doesn't have any zeros?
Avodyne said:Suppose you have a wave function with a simple zero at x=a. Thus, near x=a we can write ψ(x)=c*(x-a), where c is a constant.
Let ε be a small length, and consider the following alternative wave function:
χ(x) = -ψ(x) for x < a-ε
χ(x) = cε (that is, constant) for a-ε < x < a+ε
χ(x) = +ψ(x) for x > a+ε
Compute the expectation value of the hamiltonian in ψ and in χ, and show that the expectation value in χ is smaller than it is in ψ. This implies that ψ cannot be the ground state, because the ground state minimizes the expectation value of H. Therefore, ψ cannot have a zero.
The ground state in 1d problems refers to the lowest possible energy state that a system can have. In other words, it is the state in which the system is most stable and has the least amount of energy.
In 1d problems, the ground state is determined by solving the Schrödinger equation for the system. This equation describes the behavior of quantum particles and allows us to calculate the energy of the system at different states.
The ground state in 1d problems has several properties, including a well-defined position and momentum, as well as a zero-point energy. It also serves as the starting point for understanding the excited states of the system.
The ground state in 1d problems follows the uncertainty principle, which states that there is a fundamental limit to how precisely we can know the position and momentum of a particle. The ground state has the lowest possible uncertainty in these quantities.
Yes, the ground state in 1d problems can be changed by altering the parameters of the system, such as the potential well or the mass of the particles. This can result in a different energy and different properties for the ground state.