Normalisation and normalising wavefunctions

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Normalisation of wavefunctions in quantum mechanics ensures that the total probability of finding a particle in infinite space equals one. This is achieved by evaluating the integral of the wavefunction's squared magnitude over all space and setting it to one. The process involves determining a multiplicative constant that adjusts the wavefunction accordingly. While normalisation aids in probability calculations, it does not carry physical significance in the wave equation itself. Understanding this concept is crucial for accurate expectation value calculations, although interpretations may vary among instructors.
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In our physics class of quantum mechanics, we constantly talk about normalisation and normalising wavefunctions such that the total probability of finding the particle in infinite space is one. I don't get why do we normalise and how do we normalise(I have not taken up statistics course). It would also be benificial if you can provide me with external links.
 
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indeterminate said:
why do we normalise

If the wave function ψ is normalized, then we can calculate the probability that the particle lies between x=a and x=b (in the one-dimensional case) simply by evaluating the integral $$P(a \leq x \leq b) = \int^b_a {\psi^*\psi dx}$$

and how do we normalise

An un-normalized ψ has an arbitrary constant overall multiplicative factor, call it A. For example, we might have $$\psi(x) = Ae^{-x^2}$$ To normalize ψ, we find the value of A that makes this equation true: $$\int^{+\infty}_{-\infty} {\psi^*\psi dx} = 1$$ That is, we evaluate the integral, whose result must include a factor of A2, solve the equation for A, and finally substitute our newly-found value of A back into the original formula for ψ.
 
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so, basically normalisation is a technique which helps in mathematical calculations of probability. It has no physical significance in the wave equation.
 
indeterminate said:
so, basically normalisation is a technique which helps in mathematical calculations of probability. It has no physical significance in the wave equation.
As long as you always evaluate expectation values as \langle A\rangle = \frac{\langle \Psi|\hat A|\Psi\rangle}{\langle\Psi|\Psi|\rangle},
then yes, that is a valid perspective---but not one shared by all instructors, so be careful in examinations. But in any case, note that the wave function property of being normalizable (not normalized) can have important consequences.
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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