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Before I started studying quantum mechanics, I thought I knew what normalization was. Just pulling off Google, here's a definition that matches what I've understood normalization to mean:

Normalization -- to multiply (a series, function, or item of data) by a factor that makes the norm or some associated quantity such as an integral equal to a desired value (usually 1).

Most often I have seen normalization that normalizes to 1 or 100% or something like that. For instance, isn't putting things in percentages a kind of normalization? If I take a quiz and get 24 / 25 points, then I "normalize" this by saying I got 96%. That's what I understood normalization to be.

Why I am Confused Now

Ever since I started studying quantum mechanics, I have felt confused by the term normalization. Let me quote this portion from Griffiths to illustrate an example of how he uses the term:

We return now to the statistical interpretation of the wave function, which says taht $|\Psi(x,t)|^2$ is the probability density for finding the particle at point $x$, at time $t$. It follows that the integral of $|\Psi|^2$ must be 1 (the particle's got to be somewhere. \begin{equation} \int^{+\infty}_{-\infty} |\Psi(x,t)|^2 dx = 1 \end{equation} Without this, the statistical interpretation would be nonsense.

However, this requirement should disturb you: After all, the wave function is supposed to be determined by the Schrödinger equation --- we can't go imposing an extraneous condition on $\Psi$ without checking that the two are consistent. Well, a glance at [the time-dependent Schrödinger equation] reveals that if $\Psi(x,t)$ is a solution, so too is $A\Psi(x,t)$, where $A$ is any (complex) constant. What we must do, then, is pick this undetermined multiplicative factor so as to ensure $\int^{+\infty}_{-\infty} |\Psi(x,t)|^2 dx = 1 $ is satisfied. This process is called normalizing the wave function.

I get the idea we need the probability distribution $\rho$ to be 1 over the whole position space. That makes sense and is obvious. So the integral makes sense. But I don't understand a couple things:

  1. What was the wave function like prior to normalization? Why did it need to be normalized in the first place? To use my quiz analogy, why wasn't the test out of 100 points to begin which in which case no normalization would be needed. 96% would be 96 points.
  2. Why if $\Psi(x,t)$ is a solution, so too is $A\Psi(x,t)$?

Perhaps an answer could comment on how my initial definition of normalization relates to normalizing the wave function. Also, if you like to write, adding a comment or two about Dirac normalization would be awesome.

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    $\begingroup$ possible duplicate of Who is doing the normalization of wave function in the time evolution of wave function? $\endgroup$
    – ACuriousMind
    Feb 26, 2015 at 0:22
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    $\begingroup$ Not an exact duplicate, but the answer is the same - normalisation of states is convenience, not necessity. $\endgroup$
    – ACuriousMind
    Feb 26, 2015 at 0:22
  • $\begingroup$ How can it be for convenience though? I thought that was Griffiths point: without normalizing, the notion of probability wouldn't make any sense. $\endgroup$ Feb 26, 2015 at 0:24
  • $\begingroup$ Look at the ugly second Born rule in my answer. You can still get probability, it just becomes uglier, so you normalize for convenience. Griffith, as so many, is talking about normalization because he wants to have an easy rule to get probability. If you didn't normalize the function, every time you want to get the probability of finding the particle in an interval, you'd have to divide the integral of the square of the wavefunction by the total integral over the entire space every time, so you just set it so that that integral is 1. $\endgroup$
    – ACuriousMind
    Feb 26, 2015 at 0:27
  • $\begingroup$ In your answer, I think you touched on a point I am not understanding: "The basic principle says that states are rays in the Hilbert space, so that $|\psi\rangle$ and $c|\psi\rangle$ represent the same state for all $c\in\Bbb{C}$, and are, for all purposes, fully equivalent representants of the same state." Why is this? I think that's essentially what my question 2 is trying to ask. But I didn't know how to put it into the bra-ket formalism like you did. $\endgroup$ Feb 26, 2015 at 0:39

2 Answers 2

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Let us take a canonical coin toss to examine probability normalization. The set of states here is $\{|H\rangle,|T\rangle\}$. We want them to occur in equal amounts on average, so we suggest a simple sum with unit coefficients: $$\phi=|H\rangle+|T\rangle$$ When looking at probabilities, we fundamentally care about ratios. Since the ratio of the coefficients is one, we get a 1:1 distribution. We simply define the unnormalized probability as $$P(\xi)=|\langle\xi|\phi\rangle|^2$$ Plugging the above state in, we see we get a probability of 1 for both states. The probability (as we normally think of it), is the unnormalized probability divided by the total probability: $$P(\xi)=\frac{|\langle\xi|\phi\rangle|^2}{\langle\phi|\phi\rangle}$$ If we make the conscious choice of $\langle\phi|\phi\rangle$ every time, we don't have to worry about this normalized definition.

For your 2., note that the SE is linear. Thus $A\Psi$ is also a solution.

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  • $\begingroup$ Let me try to understand your comments. So in this example, we have two states obviously. I take it this is a regular coin, so probability 1/2 for each side. I don't understand what $\phi$ here represents. The 1:1 distribution makes sense obviously. Is $\xi$ the state that the coin ends up in after flipping? Is $\langle \phi | \phi \rangle$ just any complex number? I have seen again and again this $$\int |p\rangle \langle p |dp$$ Is that relate? $\endgroup$ Feb 26, 2015 at 1:07
  • $\begingroup$ @StanShunpike That integral is the identity operator for a continuous normalized basis. Yes, $\xi$ is the state the coin ends up in. $\langle\phi|\phi\rangle$ is the squared norm of the coin wave function. $\endgroup$
    – Ryan Unger
    Feb 26, 2015 at 1:14
  • $\begingroup$ for that $P(\xi)$ fraction. In the top we have the probability of ending up in state $\xi$ right? For the bottom, as you say, its the squared norm of the coin wave function. Why are these two things needed to define $P(\xi)$? Isn't the top alone sufficient? $\endgroup$ Feb 26, 2015 at 4:37
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You are the one who comes up with the normalization in the first place.

Suppose I ask you to find a solution to the particle-in-a-box problem, where the box is of length L. You will of course immediately say $$ \sin(\pi x/L) $$ is a solution. And it is! It is because $\sin(\pi x/L)$ is zero at 0 and L.

The function $\sin(\pi x/L)$ is a solution to the Schrodinger equation. Where did it come from? Well, you just told me it was a solution!

Why didn't you tell me the normalized solution? Because you don't care, you only care about coming up with some (any) solution to the Schrodinger equation, which you did. Now that you've got a solution, you can clearly go ahead and normalize it as suggested in Griffiths.

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