Probability density from Wave Function

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The probability density of finding a particle in space is given by the product of the wave function and its complex conjugate, expressed as |\Psi|^2. This is distinct from simply squaring the wave function, \Psi^2, which does not yield the correct probability density due to the presence of imaginary components. While certain eigenstates of atomic or molecular Hamiltonians can be real, making \Psi^2 valid in those specific cases, the general rule remains that |\Psi|^2 is the correct formulation. The distinction is crucial for accurately interpreting quantum mechanics. Thus, the consensus is that the probability density is represented by |\Psi|^2.
Mandelbroth
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A friend of mine recently tried to tell me that the square of the wave function for a particle (that is, \Psi^2) gives the probability density of finding a particle in space.

I disagree. I always thought that the wave function multiplied by its complex conjugate (that is, \Psi \Psi^*) yielded the probability density for the particle. They are definitely not the same, because \forall a,b \neq 0, \ (a+bi)^2 = a^2 + 2abi + b^2 \neq a^2 + b^2.

So, is the probability density given by \Psi^2 or \Psi \Psi^* = |\Psi|^2?
 
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Mandelbroth said:
A friend of mine recently tried to tell me that the square of the wave function for a particle (that is, \Psi^2) gives the probability density of finding a particle in space.

I disagree. I always thought that the wave function multiplied by its complex conjugate (that is, \Psi \Psi^*) yielded the probability density for the particle. They are definitely not the same, because \forall a,b \neq 0, \ (a+bi)^2 = a^2 + 2abi + b^2 \neq a^2 + b^2.

So, is the probability density given by \Psi^2 or \Psi \Psi^* = |\Psi|^2?

It's the |\Psi|^2, a real number
 
Perhaps your friend referred to wave function that is eigenstate of some atomic or molecular Hamiltonian. These can be chosen to be real, so then ##\Psi^2## gives the correct density as well as ##|\Psi|^2##. But the latter is the general expression.
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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