Undergrad Complex Numbers in Wave Function: QM Explained

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The inclusion of the complex number i=√–1 in wave functions is crucial for representing wave-like behavior in quantum mechanics, as it allows for normalization and avoids the limitations of real-valued exponential functions. The Schrödinger equation operates as a complex diffusion equation, enabling phenomena such as wavepackets reflecting off boundaries. Without the complex component, solutions would not exhibit the expected periodic nature of waves. While parts of the wave function can resemble exp(kx-wt), a complete wave function must incorporate complex elements to maintain physical validity. This complexity is essential for accurately describing bound states and various decay processes in quantum systems.
Zaya Bell
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I just need to know. Why exactly what's the complex number i=√–1 put in the wave function for matter. Couldn't it have just been exp(kx–wt)?
 
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Zaya Bell said:
I just need to know. Why exactly what's the complex number i=√–1 put in the wave function for matter. Couldn't it have just been exp(kx–wt)?

It's not possible to normalize that kind of a real-valued exponential function. The Schroedinger equation is a complex diffusion equation, and the complex numbers make it possible to obtain wave-like behavior (wavepackets reflecting from walls, etc.) instead of the simple dissipative dynamics of ordinary diffusion where the solution is a real function.
 
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Is it a periodic function without the ##i## which we would expect from a wave? You may want to look up the connection between the sine (cosine) function and the exponential function.
 
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Zaya Bell said:
I just need to know. Why exactly what's the complex number i=√–1 put in the wave function for matter. Couldn't it have just been exp(kx–wt)?
Sure, you can have parts of the wavefunction look like exp(kx-wt), but not the whole thing. If you consider a WKB approximation for the time independent Schrodinger equation, you get solutions that look like
##\exp(\sqrt{2m(V-E)})##
which looks like an exponential where V > E and an oscillation where E > V. You aren't allowed to have V > E for the whole wavefunction, because, as hilbert2 noted, it is not possible to normalize. But, for a bound state of a delta potential (V < 0), you have a double exponential solution https://en.wikipedia.org/wiki/Delta_potential.

As far as the time part, you get exponential decay for various decay processes.
 
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Thank you all.
 
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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