# How Do You Normalize the Wavefunction ψ=Ae^(-λχ)e^(-iδt)?

• baldywaldy
In summary, the given wavefunction is not valid for negative x and may need to be corrected to account for this. The '2' in front of the A could be due to using symmetry in the integration process.
baldywaldy

## Homework Statement

Normalise ψ=Ae^(-λχ)e^(-iδt)

## Homework Equations

I know you have to intergrate ψ^2 i.e (ψxψ*)

## The Attempt at a Solution

Im literally just stuck at the first bit , i can do the rest. I have the solutions manual and I don't understand how they get 2|A|^2 e^(-2λχ) from ψxψ*well only the 2 in front of A^2 really when i do it i get |A|^2 e^(-2λχ)

thanks for the help

baldywaldy said:

## Homework Statement

Normalise ψ=Ae^(-λχ)e^(-iδt)

## Homework Equations

I know you have to intergrate ψ^2 i.e (ψxψ*)

## The Attempt at a Solution

Im literally just stuck at the first bit , i can do the rest. I have the solutions manual and I don't understand how they get 2|A|^2 e^(-2λχ) from ψxψ*well only the 2 in front of A^2 really when i do it i get |A|^2 e^(-2λχ)

thanks for the help
I'm not sure, but it might have something to do with the fact that the wavefunction you specified, as it is specified,

$$\Psi = Ae^{- \lambda x}e^{-i\delta t}$$

is not a valid wavefunction for negative x. It blows up toward infinity as x becomes more negative.

Are you sure it's not something like,
$$\Psi = Ae^{- \lambda |x|}e^{-i\delta t}$$?

Maybe the '2' in front of the A comes taking symmetry into account. Check the limits of integration used in the solutions manual. Are the limits from negative infinity to infinity, taking the entire wavefunction into account? Or is only half the wavefunction integrated (taking advantage of symmetry), and integrated from 0 to infinity?

Last edited:

## 1. What is "squaring the wavefunction"?

Squaring the wavefunction refers to the mathematical operation of multiplying a wavefunction by its complex conjugate. This is often done in quantum mechanics to calculate the probability density of finding a particle in a certain position or state.

## 2. Why is "squaring the wavefunction" important in quantum mechanics?

By squaring the wavefunction, we can determine the probability of finding a particle in a specific state or position. This is crucial in understanding the behavior of particles at the quantum level, as it helps us make predictions and calculations about their behavior.

## 3. How is "squaring the wavefunction" related to the uncertainty principle?

The uncertainty principle states that it is impossible to know both the exact position and momentum of a particle at the same time. Squaring the wavefunction allows us to calculate the probability of finding a particle in a certain position, but it also increases the uncertainty in its momentum.

## 4. Can the wavefunction be squared for all particles?

Yes, the wavefunction can be squared for all particles, regardless of their mass or energy. However, it may not always accurately predict the behavior of particles in certain situations, such as in the presence of strong magnetic fields or at high energies.

## 5. Is "squaring the wavefunction" the same as taking the absolute value?

No, squaring the wavefunction involves multiplying the complex conjugate of the wavefunction, while taking the absolute value simply involves removing any negative signs. These operations may produce different results and have different interpretations in quantum mechanics.

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