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In quantum mechanics when we have simple potentials (infinite square well, oscillator harmonic, free particle, delta-Dirac, and finite square barrier) the solution eigenfunctions are different. In infinite potential well the eigenfunctions are sine and cosine. Then, the solutions in finite potential barrier are exponential functions.

When solving the Schrodinger equation, how do we know what functions to use (exponential and sine or cosine) for the solution (wavefunction)?

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  • $\begingroup$ Firstly, please try to express yourself in a clear way :) And regarding the question itself, it depends on what kind of potential you have (as you already said). Usually exponential wave-functions are OK because they tend very quickly to zero, which more or less would be a physical solution. $\endgroup$ – Robert Poenaru Jul 22 '17 at 12:54
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The solutions are like sines and cosines (oscillating) when the energy of the particle is greater than the energy of the potential. Those regions are regions where a classical particle can exist. The solutions are like exponentials when the the energy of the particle is lower than the potential, regions where a classical particle cannot exist.

For example, if a particle with energy $E$ is in a region of no potential $V(x)=0$, the wave function will be sine-like. Suppose that particle encounters a barrier of height $V>E$. A classical particle does not have enough energy to pass by that barrier. The region of higher potential is classically forbidden. The particle will bounce off the barrier. But in quantum mechanics, the wave function extends into the barrier $\ldots$ into the forbidden region. In the forbidden region the wave function will fall exponentially.

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The question is the same as asking when are the solutions to the Schrodinger equation of the form $\psi(x)=A\sin(kx)$ or $\psi(x)=Ae^{\kappa x}$.

Assume $\psi(x)=A\sin(kx)$ is your solution with $k$ constant but an unknown $V(x)$ and plug into the Schrodinger equation \begin{align} -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}A\sin(kx) + V(x)A\sin(kx)&=EA\sin(kx)\, , \\ \frac{\hbar^2k^2}{2m}A\sin(kx)+V(x)A\sin(kx)&=EA\sin(kx) \tag{1}\, . \end{align} Since (1) must hold for every $x$, and assuming $A\sin(kx)\ne 0$ for some $x$, we can cancel the common factor $A\sin(kx)$ for this $x$ and are left with: $$ V(x)=E-\frac{\hbar^2k^2}{2m}\, . $$ In other words, the potential is constant: $V(x)=V_0$. Reorganizing we get $$ \frac{\hbar^2k^2}{2m}=E-V_0\, . $$ Since the left hand side is necessarily positive, the $\psi(x)=A\sin(kx)$ can only occur when $E-V_0>0.$

The same manipulations repeated with $\psi(x)=Ae^{\kappa x}$ yields $$ V(x)=E+\frac{\hbar^2\kappa ^2}{2m}\, , $$ also showing the potential is constant. This time however one concludes that $$ \frac{\hbar^2\kappa ^2}{2m}=V_0-E $$ so that this solution can only occur then the potential is greater than the energy.

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