# Why is the probability of measuring an eigenvalue its coefficient squared?

• timn
In summary, the conversation discusses a question about the probability of measuring the eigenvalue -2 in a quantum physics example. The solution involves taking the absolute square of the eigenfunction and using orthogonality to simplify the integral, ultimately showing that the squares of the coefficients are the correct measure of probabilities for different eigenstates. The conversation ends with the asker having an epiphany about the concept of projection.
timn

## Homework Statement

This is an example from Gasiorowicz's Quantum Physics. "Example 3-1" is a particle in an infinite potential-well, but that should not matter.

## The Attempt at a Solution

Why is P(-2) (which I suppose is the probability that the eigenvalue -2 is measured) the coefficient squared?

The sum of the squares of the coefficient should be normalised, so it makes sense, but I don't understand why.

To figure out how much an eigenfunction contributes to the probability function -- psi^2 -- I'd square psi as follows:

$$\left( \frac{N}{4}\sqrt{2\pi}(u_ {-2}+2u_0+u_2) \right)^2 = \frac{N^2\pi^2}{8}(u_ {-2}^2+2u_0^2+u_2^2+2u_{-2}u_2+4u_{-2}u_0+4u_0u_2)$$

followed by being completely lost.

Could anyone explain this or make it seem plausible for me?

Edit: For reference, the answer is P(-2)=1/6.

timn said:

## Homework Statement

This is an example from Gasiorowicz's Quantum Physics. "Example 3-1" is a particle in an infinite potential-well, but that should not matter.

## The Attempt at a Solution

Why is P(-2) (which I suppose is the probability that the eigenvalue -2 is measured) the coefficient squared?

The sum of the squares of the coefficient should be normalised, so it makes sense, but I don't understand why.

To figure out how much an eigenfunction contributes to the probability function -- psi^2 -- I'd square psi as follows:

$$\left( \frac{N}{4}\sqrt{2\pi}(u_ {-2}+2u_0+u_2) \right)^2 = \frac{N^2\pi^2}{8}(u_ {-2}^2+2u_0^2+u_2^2+2u_{-2}u_2+4u_{-2}u_0+4u_0u_2)$$

followed by being completely lost.

Could anyone explain this or make it seem plausible for me?

Edit: For reference, the answer is P(-2)=1/6.

You can start this way if you want. Once you take the (absolute) square of psi, then remember that psi is normalized so let:

$$\int\left( \frac{N}{4}\sqrt{2\pi}(u_ {-2}+2u_0+u_2) \right)^2 d\phi = \frac{N^2\pi^2}{8}\int(u_ {-2}^2+2u_0^2+u_2^2+2u_{-2}u_2+4u_{-2}u_0+4u_0u_2)d\phi=1$$

Use the othogonality of the eigenfunctions to simplify the integral. Notice that terms like $u_n^*u_m$ with $m\not= n$ will integrate to zero. Thus after all the integrations are done, you should be left with:

$$\frac{2\pi N^2}{16}\int(|u_2|^2+4|u_0|^2+|u_{-2}|^2)d\phi=1$$

Simplify this by integrating. You will see that the sum of the absolute squares of the coefficients will be equal to one, and thus it also follows from your line of reasoning that the squares of the coefficients are the correct measure of probabilities for different eigenstates.

Last edited:
Aha! I had forgotten about the orthogonality. Also, I had an epiphany when I realized that $$A_n = \int_0^a u_n^*(x) \psi(x) dx$$ is just a projection.

Thank you!

## 1. Why is the probability of measuring an eigenvalue its coefficient squared?

The probability of measuring an eigenvalue is equal to its coefficient squared because of the fundamental principles of quantum mechanics. In quantum mechanics, the amplitude of a wave function represents the probability of finding a particle in a particular state. The square of the amplitude gives the probability of finding the particle in that state. In the case of eigenvalues, the coefficients of the wave function represent the amplitude, so squaring the coefficient gives the probability of measuring that eigenvalue.

## 2. Can the probability of measuring an eigenvalue ever be greater than 1?

No, the probability of measuring an eigenvalue can never be greater than 1. This is because the probability is calculated by squaring the coefficient of the eigenvalue, which is always a number between 0 and 1. Therefore, the probability will always be between 0 and 1, with 1 representing a 100% chance of measuring the eigenvalue.

## 3. How does the probability of measuring an eigenvalue change if the system is in a superposition of states?

If the system is in a superposition of states, the probability of measuring an eigenvalue will be the sum of the probabilities for each individual state. This is because in quantum mechanics, the principle of superposition states that a system can exist in multiple states simultaneously. Therefore, the probability of measuring an eigenvalue will be the combined probabilities of each state that makes up the superposition.

## 4. Is the probability of measuring an eigenvalue affected by the measurement process?

No, the probability of measuring an eigenvalue is not affected by the measurement process. In quantum mechanics, the measurement process does not change the state of the system, but rather reveals the state of the system at the time of measurement. Therefore, the probability of measuring an eigenvalue remains the same regardless of the measurement process.

## 5. Can the probability of measuring an eigenvalue be negative?

No, the probability of measuring an eigenvalue cannot be negative. As mentioned before, the probability is calculated by squaring the coefficient of the eigenvalue, which will always result in a positive number. Negative probabilities do not make physical sense, and therefore, the probability of measuring an eigenvalue must always be positive.

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