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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.

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  • $\begingroup$ The uncertainty is related to how 'spread out' the wavefunction is. So how does changing $a$ affect the spread of the wavefunction in terms of position (try plotting a graph). For the momentum, do you know how to take a Fourier transform? You can then see how $a$ affects how spread out the wavefunction is in momentum. $\endgroup$ – gautampk Apr 10 '17 at 23:48
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    $\begingroup$ The reals are a subset of complex numbers. $\endgroup$ – user121330 Apr 11 '17 at 1:56
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    $\begingroup$ "the wave function must be a complex" - true "and include imaginary variables" - false. $\endgroup$ – tparker Apr 11 '17 at 4:04
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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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Wavefunctions are in general complex, but there is nothing preventing a specific wavefunction from being real. In fact, there are certain cases for which you can show that there is always a real wavefunction that describes the system (it's called a Schmidt decomposition, and applies when your system is made up of an even number of subsystems).

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You are right, the "real" wave function for a particle is a complex function $\Psi(\vec{x}, t)$, which follows the time-dependent Schrödinger equation: $$i\hbar\frac{\partial\Psi(\vec{x}, t)}{\partial t} = \hat{H}\Psi(\vec{x}, t)$$ Where $\hat{H}$ is the hamiltonian operator which gives the total energy of the system, consisting of the kinetic energy and the potential energy, which is given by the potential energy function $V(\vec{x}, t)$: $$\hat{H}\Psi(\vec{x}, t) := -\frac{\hbar^2}{2m}\nabla^2\Psi(\vec{x}, t) + V(\vec{x}, t)\Psi(\vec{x}, t)$$ If the potential energy function is truly a function of time as well as position, this equation becomes very difficult to solve. Luckily, however, the potential energy is almost always only dependent on the position and therefore becomes a simpler function $V(\vec{x})$. In that case, the Schrödinger equation can be solved by assuming the wave function $\Psi(\vec{x}, t)$ to be a simple product of a function $\psi(\vec{x})$ solely dependent on the position and a function $f(t)$ that only depends on the time: $$\Psi(\vec{x}, t) = \psi(\vec{x})f(t)$$ Then we are able to seperate the position and time dependent parts of the equation and from that find that $f(t)$ has to be equal to $e^{-iEt/\hbar}$, where $E$ is some real constant (which can be proven to be the energy of the particle). Furthermore, $\psi(\vec{x})$ must follow the so-called time-independent Schrödinger equation: $$E\psi(\vec{x}) = \hat{H}\psi(\vec{x})$$ where $E$ is the same constant. It can be proven that $\psi(\vec{x})$ can always taken to be a real function (That is, if you do have a solution to the time-independent Schrödinger equation that isn't always realy, you can always express it as a linear combination of ones that are. It does not mean that every solution necessarily has to be real.). Then we have: $$\Psi(\vec{x}, t) = \psi(\vec{x})e^{-iEt/\hbar}$$ Remember, that this is only under the assumption, that $\Psi(\vec{x}, t)$ CAN be written as such a product. However, all the other solutions, can then be expressed as linear combinations of these simple time-independent solutions, as Zero already pointed out.

Your question supposes a time-independent wave function that is equal to $$\psi(x) = Ae^{-\alpha x^2}$$ (This is a one-dimensional function, so I'm leaving the arrow off) Therefore, the time-dependent wave-function you know would be $$\Psi(x, t) = Ae^{-\alpha x^2}e^{-iEt/\hbar}$$ To find $E$, you can apply the Schrödinger equation: $$EAe^{-\alpha x^2} = \hat{H}(Ae^{-\alpha x^2})$$ (However, you would need to know the potential energy function you used to derive that wave function)

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