# Imaginary numbers and the real part of the Schrodinger Equation

1. Apr 15, 2012

### Hypatio

At the moment I am studying the Schrodinger equation using this resource.

In a 1D solution (sec 3.1 in the paper) they show that a wave function can be expressed as

$$\Psi(x,t)=\sqrt{2}e^{-iE_{n_x}t}\sin (n_x\pi x)$$

where n_x is the quantum number. And they show the real part of the solution in Figure 2a for t=0 and for over time in Figure 3a. I do not understand the structure seen in the time-dependent solution. In particular, in my solution I can show exactly what they give in Figure 3 except that I ONLY show the wavefunction being positive at x<0.5 and negative in x>0.5. I can only get all their curves if I assume that the wave function is both positive and negative over time.

I think this might be due to the fact that I do not understand the use of the imaginary number in the equation and solutions. For instance, apparently when the above equation is squared you arrive at

$$\Psi^2=2\sin^2 (n_x\pi x)$$

But I don't see how that operates on the exponential. So what is the function of the imaginary number in the schrodinger equation? Do you just assume that i=1 sometimes and i=-1 othertimes?

2. Apr 15, 2012

### SeannyBoi71

You're not just squaring the wave function, you do the square modulus. This means you multiply the wave function by its complex conjugate, here it is
$$\Psi = \sqrt(2) exp(iEt) sin(n\pi x) [\tex] When you work it out, the i's should cancel out. i2= -1 ALWAYS, from my knowledge at least. Hopefully there is someone with more experience here that may be able to correct me. EDIT: My LaTeX text isn't working, no idea what I'm doing wrong. my apologies Last edited: Apr 15, 2012 3. Apr 15, 2012 ### Bob S The probability density in fig. 2(b) is [tex] P(x,t)=<\overline{\Psi}(x,t)\vert\Psi(x,t)>$$
where the amplitude and its conjugate are
$$\Psi(x,t)=e^{-iE_n t}A_{n}\sin(n\pi x)$$ and
$$\overline{\Psi}(x,t)=e^{+iE_n t}A_{n}\sin(n\pi x)$$

4. Apr 15, 2012

### SeannyBoi71

That's what I was trying to say, except this is much nicer. Thanks Bob!