Does this wavefunction make sense?

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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.
  • #1
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!
 
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  • #2
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?
 
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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.

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