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I learn quantum physics from Alonso-Finn's book (Amazon link), there's one step of Schrödinger equation for a free particle that I couldn't understand. $$ \frac{\mathrm{d^{2}\Psi } }{\mathrm{d} x^{_{2}}}+k^{2}\Psi =0 $$ How can it become $\Psi \left ( x \right )=Ae^{ikx}$ for a particle moving in $+x$ direction and $\Psi \left ( x \right )=Be^{-ikx} $ for a particle moving in $-x$ direction?

Here is what I get with my own pen:

$$ \ln\Psi=-k^{2}x^{2} $$

$$ \Psi \left ( x \right )=e^{-k^{2}x^{2}} $$

Can anybody explain me what's wrong with my answer?

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    $\begingroup$ There's an implied time dependence of $\exp -i E t / \hbar$. $\endgroup$ – DanielSank Mar 17 '15 at 16:08
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    $\begingroup$ Apply the comment by @DanielSank then see physics.stackexchange.com/q/56338 $\endgroup$ – Bill N Mar 17 '15 at 16:12
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    $\begingroup$ It's a general solution to this Ordinary Differential Equation (ODE). Or perhaps I am not understanding your question? $\endgroup$ – Jiminion Mar 17 '15 at 16:19
  • $\begingroup$ That's what I mean Jiminion, but I couldn't get the same result. Can you explain me, please? $\endgroup$ – Agus Putra Dana Mar 17 '15 at 16:29
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Probably one of the first things one should do after arriving at a solution to an ODE is stick it back in to see if it satisfies the original equation. In your case, you obtain $$ \left(4k^4x^2\Psi-2k^2\Psi\right)+k^2\Psi\neq0 $$ where the term in the parenthesis is the result of $\frac{d^2}{dx^2}\Psi$. So clearly you went wrong somewhere. I'd wager that it was in getting $$ \ln\Psi=-k^2x^2 $$ which I suspect you found via attempting separation of variables: $$ \int\frac{d^2\Psi}{\Psi}=-k^2\int dx^2 $$ which doesn't really make much sense to me.

With 2nd order ODEs, one typically guesses at the function (this method is called trial functions). Since we want the second derivative to be the initial function times some negative constant ($\Psi''=-k^2\Psi$), this suggests a complex exponential: $\Psi\sim\exp\left[-ikf(x)\right]$. Taking the 2nd derivative of this, \begin{align} \frac{d^2}{dx^2}e^{-ikf(x)}&=-\left(\frac{df}{dx}\right)^2k^2e^{-ikf(x)}-i\left(\frac{d^2f}{dx^2}\right)ke^{ikf(x)}\\ &=-\left(\frac{df}{dx}\right)^2k^2\Psi-i\left(\frac{d^2f}{dx^2}\right)k\Psi \end{align} Since we want $-k^2\Psi$ on the right side then it must be that $f(x)=x$ such that $f'=1$ and $f''=0$. Thus our solution is $$ \Psi(x)=\exp\left[-ikx\right] $$

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The general solution to the differential equation you have given is $Ae^{ikx}+Be^{-ikx}$, which you can verify by substituting back in. (This is a common method, putting in some parametric form of a potential solution and determining the values of the parameters after substituting the trial solution). A plane wave has the form $\Psi(x,t)=Ae^{i(kx-\omega t)}$, so that, once the time dependence from that part of the Schrodinger equation is factored in, we can think of the solution to your equation as broken down into two parts, one representing a right moving plane wave, and the other a left moving plane wave.

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When solving differential equations we often us a technique called ansatz. The word ansatz is German for guess. In other words we guess at a possible solution then feed it back into the differential equation to see if it works.

In this case we guess that the solution is $\Psi(x) = e^{ikx}$. If we take the second differential of this we get:

$$ \frac{d^2\Psi}{dx^2} = -k^2 e^{ikx} = -k^2\Psi $$

Feed this back into the Schrodinger equation and we get:

$$ -k^2\Psi + k^2\Psi = 0 $$

which is true. So we guessed right.

It would be nice to claim that the equation was solved with some sophisticated mathematical technique, but when d’Alembert first solved the wave equation in 1746 it was just a guess - sorry, ansatz!

As for the direction of motion, some of the comments have given one way to see this but let me point out a different way. In quantum mechanics the momentum operator is:

$$ \hat{p}_x = -i\hbar\frac{\partial}{\partial x} $$

As it happens, the wavefunctions you have are momentum eigenstates so the momentum, $p_x$, is given by:

$$ p_x\Psi = \hat{p}_x\Psi \tag{1} $$

let's take the state $\Psi = e^{ikx}$ that, as you say, represents a particle moving in the positive $x$ direction. If we use equation (1) we get:

$$ p_x\Psi = -i\hbar\frac{\partial}{\partial x}(e^{ikx}) = \hbar k e^{ikx} = \hbar k \Psi $$

and we find that $p_x = \hbar k$ and this is positive. If the momentum is positive that means the particle must be moving in the positive $x$ direction. If instead you use $\Psi = e^{-ikx}$ you'll get $p_x = -\hbar k$ so the particle is moving in the negative $x$ direction.

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Assuming $\Psi$ is the wave function. The Schrodinger equation biing,

$$ \frac{\mathrm{d^{2}\Psi } }{\mathrm{d} x^{_{2}}}+k^{2}\Psi =0 $$

This is a standard solution of differential equations. If fou have done simple harmonic oscillators, it appears there. In particilar,

$$\frac{d^2f}{dx^2} = -k^2f$$ $$\frac{d}{dx}\left(\frac{df}{dx} \right) = -k^2f$$ Multiplfing bf $\frac{df}{dx}$ on both sides, we get

$$\left(\frac{df}{dx}\right)\frac{d}{dx}\left(\frac{df}{dx} \right) = -k^2f \frac{df}{dx}$$. Assuming, the integration constant is 0.

$$\left(\frac{df}{dx}\right)^2 = k^2f^2 \implies \left(\frac{df}{dx}\right) = kf$$

I hope from this point onwards its clearly to the see the solution to this is $\boxed{e^{ikx}}$. Similarly, for $-x$.

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As pointed out above, this is a standard ODE in which u are normally to assume a solution to the DE as Psi(x)=Aexp(ax) [where A and a are both constants]

If u then substitute for the assumed solution in the Schrodinger equation, u have; a^2+k^2=0 or a^2=-(k^2) or a=+/-(ik) since u now have two values for a (ik and -ik), it means you should have two independent solutions to Psi(x). The sum of these solutions is also a solution as well (check standard ODE textbooks for reasons). Then Psi(x)=(A)exp(ikx)+(B)exp(-ikx) if A=B=1 (B is a constant for the reflected wave exp(-ikx)) Psi(x)=exp(ikx)+exp(-ikx)

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    $\begingroup$ Welcome to Physics! Note that this site has MathJax enabled, which means you can use Latex-like syntax to add in equations for readability. $\endgroup$ – Kyle Kanos Nov 22 '15 at 22:04

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