Calculating the probability density of a superposition

In summary, the author is arguing that the probability density for a superposition of two wavefunctions is not always the sum of the squared factors for the individual wavefunctions, but is instead the product of the squared factors.
  • #1
Archeon
7
0
This ought to be some simple gap in my knowledge, but it bugs me nonetheless. Let me present the argument as I see it, I'm fairly certain that there is just some tiny part that I didn't learn correctly.
Let us assume a wavefunction $$\Psi$$ is defined as a superposition of two wavefunctions:
\begin{equation}
\Psi(x) = \alpha \psi_1 + \beta \psi_2
\end{equation}
with both subwavefunctions being normalized.

Now (and correct me if I'm wrong here) even if the wavefunctions arent orthogonal, the norm squared and thus the probability density can be expressed like so:
\begin{equation}
|\Psi(x) |^2 = |\alpha \psi_1 + \beta \psi_2|^2 = |\alpha|^2 + |\beta|^2 + \lambda = 1
\end{equation}
with $$\lambda$$ being some real number as a result of the multiplication.
And here is the problem I face. Does this mean that $$|\alpha|^2$$ and $$|\beta|^2$$ only represent probabilities if the wavefunctions are orthogonal? Or am I missing something else here?

Thanks
 
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  • #2
Archeon said:
even if the wavefunctions arent orthogonal, the norm squared and thus the probability density can be expressed like so:
$$
\begin{equation} |\Psi(x) |^2 = |\alpha \psi_1 + \beta \psi_2|^2 = |\alpha|^2 + |\beta|^2 + \lambda = 1 \end{equation}
$$
with
$$
\lambda
$$
being some real number as a result of the multiplication.

Where does ##\lambda## come from? And why must it be real?
 
  • #3
It would be the double product of ψ1 with ψ2: $$\lambda = \alpha^* \psi_1^* \beta \psi_2 + \beta^* \psi_2^* \alpha \psi_1 = 2 Re(\alpha^* \psi_1^* \beta \psi_2)$$
which can be seen as some complex number plus its conjugate, thus twice its real part and a real number.
 
  • #4
Archeon said:
It would be the double product of ψ1 with ψ2: $$\lambda = \alpha^* \psi_1^* \beta \psi_2 + \beta^* \psi_2^* \alpha \psi_1 = 2 Re(\alpha^* \psi_1^* \beta \psi_2)$$
which can be seen as some complex number plus its conjugate, thus twice its real part and a real number.
You need to take the integral over all ##x## instead of only using ##|\psi(x)|^2## in order to come to the probabilistic interpretation.
 
  • #5
Archeon said:
And here is the problem I face. Does this mean that ##|\alpha|^2## and ##|\beta|^2## only represent probabilities if the wavefunctions are orthogonal? Or am I missing something else here?
Probabilities of what? If they are not orthogonal, they cannot be two distinct measurement results of the same type of measurement. There is no measurement that would give either the first or the second state, therefore there is no reason why the squared prefactors should sum to 1.
 
  • #6
Ah, I think I see the error in my thinking. Also, what I meant to write was that the integral of the Real part would have to add up with the factors squared to be equal to one.
I see now, of course the two wavefunctions that make up the superposition would have to be eigenstates of some observable that gets measured, but can I assume that such eigenstates are always orthogonal for every measurable quantity?
 
  • #8
Enough already...!The normalisation factor of the superposition will be different. You cannot add two normalised wavefunctions together and say the result is normalised!
 

1. What is a superposition in probability density?

A superposition in probability density refers to the combination of two or more probability distributions to create a new distribution. This is commonly used in quantum mechanics to describe the state of a system.

2. How do you calculate the probability density of a superposition?

To calculate the probability density of a superposition, you must first determine the individual probability distributions of the components. Then, you can use the formula for combining probability distributions to calculate the overall probability density of the superposition.

3. Can the probability density of a superposition be greater than 1?

No, the probability density of a superposition cannot be greater than 1. This value represents the total probability of an event occurring, and it cannot exceed 100%.

4. What is the difference between a superposition and a mixture in probability density?

A superposition refers to the combination of probability distributions, while a mixture refers to the combination of actual events. In a superposition, the individual components retain their original probabilities, while in a mixture, the probabilities are adjusted based on the combination of events.

5. How does calculating the probability density of a superposition relate to real-life applications?

Calculating the probability density of a superposition has applications in various fields, such as finance, biology, and physics. It allows for the analysis of complex systems and the prediction of outcomes based on multiple variables. For example, it can be used in risk assessment or in understanding the behavior of particles in quantum mechanics.

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