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I just have one question.

I'm doing a problem where I'm told to normalize a wave function, which is split up into two regions, namely where $r \leq r_0$ and $r > r_0$. My question is, why am I doing this? I'm not using any of the math I get out of it later on. The only thing I do after it is applying the continuous condition. So what I was thinking was, that by normalizing my wave function, I also showed that my wave function is in fact continuous. So is that the case, or am I just mistaken all together?

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    $\begingroup$ Well, we can't tell you why you are doing the normalization, but normalization isn't necessary for anything. $\endgroup$
    – ACuriousMind
    Commented Mar 1, 2015 at 21:31
  • $\begingroup$ @DenverDang Why are you doing what? I just guess that the two wave-functions are given as $Af_1(\vec r, t)$ and $Bf_2(\vec r, t)$. A wave-function (w.f.) has to be continuous and with a continuous 1st derivative, so you get some relation between $A$ and $B$, and maybe fix some additional constant. The normalization is the physical condition emerging from the requirement that the abs. square of the w.f. represent density of prob. By normalizing the resulting w.f. i.e. after fixing its continuity and of its 1st derivative, you'll get finally the value of the leading constant. $\endgroup$
    – Sofia
    Commented Mar 1, 2015 at 23:22
  • $\begingroup$ So normalization doesn't really tell me anything, except the fact, that it has to equal one, and from that it may be possible to find some constants if necessary ? $\endgroup$ Commented Mar 1, 2015 at 23:58

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I think what you are asking whether the relationship

$$ \mathrm{normalizable} \iff \mathrm{continuous}$$

holds, which is utterly wrong! The wave function has to be continuous*. Notwithstanding take $\psi(x)=H(x-1/2)-H(x+1/2)$, where $H(x)$ is the Heaviside step function.

$$ \implies \int_{-\infty}^\infty \mathrm{d}x \,\, ||{\psi(x)}||^2 = 1 $$

(Area of a square with sides 1) Thus, although the function isn't continuous, it is normalizable.

Edit: *As ACuriousMind points out the wave function, in general, need not be continuous, although in the physical world it has to be so.

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  • $\begingroup$ Who says the wavefunction has to be continuous? The usual space of states is that of the complex-valued square-integrable functions $L^2(\mathbb{R})$, which contains exactly such non-continuous functions as the Heaviside function. $\endgroup$
    – ACuriousMind
    Commented Mar 2, 2015 at 0:43
  • $\begingroup$ Am I missing something because I was always thought that the wave function has to be continuous. Wikipedia also indicates that this must be the case, so that "the calculations and physical interpretation make sense". I haven't studied QM in such high level that $V(x)=\delta'(x)$ is a legitame potential so I cannot really say, what I have said is completely correct. However they must be $L^2(\mathbb{R})$ that is for sure! $\endgroup$
    – Gonenc
    Commented Mar 2, 2015 at 0:49
  • $\begingroup$ In the usual setting, you are correct that the form of the Hamiltonian forces the actual wavefunctions of physical states to be continuous, because otherwise the differentiation operator could not act upon them, for example. But this is rather a requirement that arises from the specific physical situation than a general constraint of quantum mechanics, so I am reclutant to make the general statement "The wave function has to be continuous". (BTW, the RHS of your first TeX block has a typo.) $\endgroup$
    – ACuriousMind
    Commented Mar 2, 2015 at 0:59
  • $\begingroup$ @ACuriousMind I understand your concern. But given the question, I think this does not play so much role. continious vs. continuous always gets me! $\endgroup$
    – Gonenc
    Commented Mar 2, 2015 at 1:02
  • $\begingroup$ No, it is certainly not really relevant to this question. $\endgroup$
    – ACuriousMind
    Commented Mar 2, 2015 at 1:06
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My question is, why am I doing this?

Becase, by convention, we want the probability account for all possible outcomes to sum to unity. The fact that we will get some outcome at a measurement, is guaranteed, and with this normalization it reflects a total probability of 1.

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Like gonenc pointed out your assumption that normalizing your wave function does not imply continuity. And yes you'll probably won't need the normalization factor in your further calculations. The reason for you doing this could be consistency with the Interpretation of the wave function squared as a probability amplitude:
$$ P = \int|\psi(r,t)|^2 dr $$

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