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http://uqu.edu.sa/files2/tiny_mce/plugins/filemanager/files/4190016/Quantum_Mechanics_1/ch4-virtual-book.pdf

On page 2 of the above pdf they describe how they select their wavefunctions. Finding the general solution is easy but they then go on to say that D= 0 because 'this is not physically meaningful' leaving $\psi=Ce^{ik_2x}$.

Fine, I can see that if the wave is transmitted through the barrier it is best represented by such a travelling wave moving in the +x direction.

Now, my lecturer also supports and uses the above explanation.

However, there are a few university websites and a textbook (Introduction to Quantum Mechanics by Bransden) that state that the wavefunction for the section $V(x) = V_0$ cannot be represented by $\psi=Ce^{ik_2x}$ as this tends to infinity at x= infinity. Also, by the postulates of QM the wavefunction must have a real quantifiable integral which such a function is in violation of.

What do I follow ? Who is right ?

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  • $\begingroup$ "as this tends to infinity at x= infinity" - do you mean that $\int_{-a}^a |\psi|^2 dx \to \infty$ as $a\to\infty$? $\endgroup$ – innisfree Apr 5 '14 at 11:37
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    $\begingroup$ Well according to my textbook $\psi$ will not be finite as we take the limit x tending to infinity since $exp(k_2x)$ will diverge meaning that this term must be set to zero... This is in contradiction with what my lecturer says. $\endgroup$ – Ari Ben Canaan Apr 5 '14 at 11:55
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    $\begingroup$ Comment to the post (v1): It would be good if OP (or somebody else?) could try to make the question formulation self-contained, so one doesn't have to open the link to understand the question. Also please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. $\endgroup$ – Qmechanic Apr 7 '14 at 14:07
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Note that the problem in your link considers a particle with $E>V_0$. This means that $k_2$ is real, and thus $e^{ik_2x}$ is bounded.

On the other hand, if you speak in general, then for $E<V_0$ $k_2$ must be imaginary since kinetic energy is negative in the right side, and thus $e^{ik_2x}$ is real and unbounded behaving like real exponential. In this case you need to take the correct version of the exponent — the one which decays — it's a legal solution, while its opposite, which diverges at infinity, is unphysical (as long as the barrier has infinite thickness).

So, your lecturer is correct only in the constraints of the problem considered, while the websites and textbooks you mention are likely talking about either general case, or low-energy states.

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  • $\begingroup$ Yes. Are you missing a negative sign in the second case ? I think I know what you mean. If E > Vo then $exp(ik_2x)$ is a solution. If E<Vo then $exp(-ik_2x)$ is a solution as a result of reflection. Please note the negative sign. $\endgroup$ – Ari Ben Canaan Apr 5 '14 at 15:02
  • $\begingroup$ No, for any $E$, reflection is given by $\exp(-ik_1x)$ — note the index 1 for $k_1$ — because it's in the left side. But for $E<V_0$, we have $k_2=\pm i|k_2|$, and thus in the right side, $\exp(ik_2x)=\exp(\mp|k_2|x)$, from which you should take "$-$" sign to have decaying solution. Then diverging solution should be dropped as unphysical. $\endgroup$ – Ruslan Apr 5 '14 at 16:08
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I believe you must have misunderstood something in the other textbooks but there is no way to know since I do not have a copy of the book mentioned. The wave comes from -$\infty$ to x and the solution should be normalisable this way.

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  • $\begingroup$ What ? Can you clarify what exactly is the 'correct' solution ? Is the case as described by the link and my lecturer or the textbook ? $\endgroup$ – Ari Ben Canaan Apr 5 '14 at 11:26
  • $\begingroup$ I believe your lecturer is correct. What I am saying is, D=0 because as you said its not a meaningful solution. The reason for this is that the other equation contains is $e^{-ik_2x}$ and cannot be normalised between $-\infty$ and x, which is essentially where you wave begins its journey and ends its journey $\endgroup$ – Constandinos Damalas Apr 5 '14 at 11:29

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