# Does this wavefunction make sense?

• I
• Isaac0427
In summary, the conversation discussed the wavefunction ##\left| \psi (x) \right|^2 = e^{-ax^2+1}+e^{-bx^2-1}## and its normalization. It was noted that even if the function is normalized, there are still two spots where ##\left| \psi \right|^2 > 1##, which may seem odd. However, it was clarified that ##|\psi^2|## is a probability density and the integral of this function must be ##\le 1##, not equal to 1. It was also mentioned that despite being normalized, the function can still be made arbitrarily large, but this is acceptable as it is a probability density.

#### Isaac0427

Hi all!
Consider a wavefunction, where ##\left| \psi (x) \right|^2 = e^{-ax^2+1}+e^{-bx^2-1}## where a and b are real, positive numbers that satisfy normalization (they are purpously inside the exponent). Even if it is normalized, there are still 2 spots that ##\left| \psi \right|^2 > 1## which makes no sense. What is going on here?
Thanks!

Recall that ##|\psi^2|## is a probability density. The probability for finding a particle in an interval ##[a,b]## is ##\int\limits_a^b|\psi^2|\mathrm{d}x##. This integral is what must be ##\le 1## (where it equals one when you integrate over all of space). To convince yourself, take one of the Gaussian functions you wrote and normalize it, evaluate at ##x=0## and start increasing ##a##. Despite still being normalized ##|\psi^2|## can be made arbitrarily large. Why is that alright?

Isaac0427
Haborix said:
Recall that ##|\psi^2|## is a probability density. The probability for finding a particle in an interval ##[a,b]## is ##\int\limits_a^b|\psi^2|\mathrm{d}x##. This integral is what must be ##\le 1## (where it equals one when you integrate over all of space). To convince yourself, take one of the Gaussian functions you wrote and normalize it, evaluate at ##x=0## and start increasing ##a##. Despite still being normalized ##|\psi^2|## can be made arbitrarily large. Why is that alright?
Ah, I get that. I was confused as I thought of it as a probability and not a probability density.

## 1. Does the wavefunction follow the laws of quantum mechanics?

Yes, the wavefunction is a fundamental concept in quantum mechanics and it must obey the laws of quantum mechanics, such as the Schrödinger equation.

## 2. Is the wavefunction normalized?

Normalization is a fundamental requirement for wavefunctions in quantum mechanics, and it ensures that the probability of finding the particle in any location is equal to 1. Therefore, a valid wavefunction must be normalized.

## 3. Why does the wavefunction have complex values?

In quantum mechanics, the wavefunction is a complex-valued function because it allows for the interference of different states, which is a key aspect of quantum behavior.

## 4. How can I interpret the wavefunction?

The wavefunction is a mathematical representation of a physical system, and it provides information about the probability of finding a particle in a particular state. It cannot be directly observed, but it is a useful tool for calculating observable quantities in quantum mechanics.

## 5. Can the wavefunction be used to predict exact outcomes?

No, the wavefunction does not give precise predictions for individual measurements. Instead, it describes the probability of obtaining different outcomes when a measurement is made on a quantum system.