Imaginary numbers and the real part of the Schrodinger Equation

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SUMMARY

The discussion focuses on the interpretation of imaginary numbers in the context of the Schrödinger equation, specifically the wave function expressed as \Psi(x,t)=\sqrt{2}e^{-iE_{n_x}t}\sin(n_x\pi x). The user struggles with understanding the time-dependent solution and the role of the imaginary unit in the equation. It is clarified that the probability density is calculated using the square modulus of the wave function, which involves multiplying the wave function by its complex conjugate, resulting in the cancellation of the imaginary unit. The discussion emphasizes the importance of correctly applying these concepts to derive the expected results.

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Hypatio
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At the moment I am studying the Schrödinger 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 Schrödinger equation? Do you just assume that i=1 sometimes and i=-1 othertimes?
 
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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]<br /> When you work it out, the i&#039;s should cancel out.<br /> <br /> i<sup>2</sup>= -1 ALWAYS, from my knowledge at least. Hopefully there is someone with more experience here that may be able to correct me.<br /> <br /> EDIT: My LaTeX text isn&#039;t working, no idea what I&#039;m doing wrong. my apologies
 
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The probability density in fig. 2(b) is
P(x,t)=&lt;\overline{\Psi}(x,t)\vert\Psi(x,t)&gt;
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)
 
Bob S said:
The probability density in fig. 2(b) is
P(x,t)=&lt;\overline{\Psi}(x,t)\vert\Psi(x,t)&gt;
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)

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

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