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We all know what is a wave function. There is a technique to convert non normalized wave function to normalized wave function.

In case of non normalized wave function integration of square of wave function with respect to 'dt' in a length 'dx' will be any constant other than 1. In case of normalized wave equation the probability of finding will be maximum i.e. 1.

The process of converting a non-normalized wave function to a normalized wave function is known as Normalization technique. In this technique we multiply the wave function with some constant and makes it equal to 1 (condition for a normalized wave function). That means after conversion the probability of finding electron is becoming maximum.

If we multiply a wave function with any constant it's nature doesn't change. So, my question is after normalization of the wave function, the nature of the wave doesn't changes and how come the probability of finding electron becomes maximum i.e. 1???

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I swear I read the question many times, but I still not quite sure what it is about. So I will try to clarify some things hoping that they are relevant.


First off, any complex multiple of a wave function indeed describes exactly the same state of a system. This directly follows from the postulate of quantum mechanics which states that the expectation value of an observable $A$ represented by a self-adjoint operator $\hat{A}$ is given as follows, $$ \langle A \rangle = \frac{\langle \Psi | \hat{A} | \Psi \rangle}{\langle \Psi | \Psi \rangle} \, , $$ where $\Psi$ is the wave function. It is indeed easy to show that given $\Psi$ we can multiply it by any complex number $c$ and this would not change the expectation value of any observable, $$ \langle A \rangle = \frac{\langle c \Psi | \hat{A} | c \Psi \rangle}{\langle c \Psi | c \Psi \rangle} = \frac{c^{*} c \langle \Psi | \hat{A} | \Psi \rangle}{c^{*} c \langle \Psi | \Psi \rangle} = \frac{\langle \Psi | \hat{A} | \Psi \rangle}{\langle \Psi | \Psi \rangle} \, , $$ since $c^{*} c$ in numerator and denominator perfectly cancels each other. Also note that the following definition of the expectation value $$ \langle A \rangle = \langle \Psi | \hat{A} | \Psi \rangle \, , $$ holds true only for a normalized wave function $\Psi$ and trivially follows from the above mentioned more general definition.


Secondly, for the simplest case of one particle in one spatial dimension, if one would like to interpret the wave function $\Psi(x,t)$ as a probability amplitude that the particle is at $x$, and consequently, it square modulus, $ \left|\Psi(x, t)\right|^2 = {\Psi(x, t)}^{*}\Psi(x, t)$ as the probability density that the particle is at $x$, he/she have to normalize it as follows, $$ \int\limits_{-\infty}^\infty d x \, |\Psi(x,t)|^2 = 1 \, . $$ This is required because one of the axioms of the probability theory requires that

the probability that some elementary event in the entire sample space will occur is 1.

The sample space is just a set of all possible outcomes in an experiment, which in our case is the set of events of finding a particle at any possible position, i.e anywhere between $-\infty$ and $+\infty$. Thus, in our case the probability of finding the particle anywhere between $-\infty$ and $+\infty$ have to be 1, and consequently, the wave function has to be normalized. Otherwise its interpretation as a probability amplitude would be inconsistent with the probability theory.


Hope it helps.

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Normalization means that the integral of the wave function times its complex conjugate over all space is 1. It has nothing to do with maximizing anything. It is only a way of saying "there is one electron (or other particle)".

In other words, if I say "there is an 80% chance Vamsi is at school, 40% chance Vamsi is at home, 50% chance Vamsi it at work, and a 30% chance Vamsi is shopping" it doesn't make any sense.

But if I say "there is an 40% chance Vamsi is at school, 20% chance Vamsi is at home, 25% chance Vamsi it at work, and a 15% chance Vamsi is shopping" it does make sense.

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  • $\begingroup$ Then What is the difference between a normalized wave function and non normalized wave function?? $\endgroup$ – vamsi Jan 27 '15 at 15:23
  • $\begingroup$ @vamsi, in short, only normalized wave function can be used to calculate probabilities. See my answer for details. $\endgroup$ – Wildcat Jan 27 '15 at 15:39

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