[Physics] Non-complex wave function

complex numbersquantum mechanicsschroedinger equationwavefunction

I am taking an introductory course on QM. I came across the following question:

A particle is described by the wave function $$ \psi(x) = Ae^{-ax^2} $$
where $A$ and $a$ are positive, real constants. If the value of $a$ is increased what effect does this have on the particle's uncertainty in position and particle's uncertainty in momentum.

I thought that the wave function must be a complex and include imaginary variables, but this wave function is not. Also I don't understand how I can relate $a$ to uncertainty.

Best Answer

You are looking at a solution of the time-independent Schrodinger equation as your $\psi(x)$ does not have any time dependence, and the basic solutions of the time-independent equation can often be real. Linear combinations of these basic solutions can be complex.

The solutions to the time-dependent Schrodinger equation are always linear combinations of the form $$ \Psi(x,t)=\sum_n c_n e^{-iE_nt/\hbar} \psi_n(x) $$ and will be complex even if the time-independent functions $\psi_n(x)$ are real.

To relate $a$ to the uncertainty relation you would need to compute $\Delta x^2$ and $\Delta p^2$ using your $\psi(x)$ (which you will have to normalize) and to find how $a$ enters into the product $\Delta x\Delta p$.

To give you a hint I'm including the plot of $\psi(x)^2$ for $a=1$ (black), $a=2$ (blue) and $a=1/2$ (red).

enter image description here

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