Probability integral, Expectation Value and Square of Psi

In summary: However, what this integral actually does is to give the probability of finding the particle in a given interval, or set of points. So in this context, it's connecting the calculation to an observable.My hypothesis about the difference between the two integrals is that the first one gives you the probability of finding the particle in a certain region while the expectation value tells you something about a specific point. If my hypothesis is correct, what exactly does it tell you about said point?In a thread I made once before, someone stated that it was not the probability of finding the particle at said point.That's not a correct statement. The expectation value is basically the average:
  • #1
space-time
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I have come across a bit of conflict in wording of some physics and chemistry textbooks about the probability of finding particles in certain places. To be more specific, I have come across 3 different statements:

1. $$\int_a^b {| \psi(x) |^2 dx}$$

The above integral is said to give the probability of finding a particle between x=a and x=b

2. $$\int_{-\infty}^{+\infty} {x | \psi(x) |^2 dx}$$

The above integral is said to yield the expectation value for finding the particle at point x. The problem is: My quantum physics book and other sources say that this expectation value and expectation values in general are what tie the calculations to actual physical observables. That is a problem because, if the first integral that I listed gives the probability of finding the particle in a certain region, then wouldn't that integral be the one that connects the calculations to the physical observable of position (and not the expectation value integral)? In other words, what exactly does this expectation value tell you? My hypothesis about the difference between the two integrals is that the first one gives you the probability of finding the particle in set region while the expectation value tells you something about a specific point. If my hypothesis is correct, what exactly does it tell you about said point? In a thread I made once before, someone stated that it was not the probability of finding the particle at said point.

3. | \psi(x) |^2

My chemistry book says that the square of the magnitude of the wave function evaluated at x gives you the probability of finding the particle at point x. I, having already known about the first integral that I mentioned, felt that simply squaring the wave function is way too simple. If you can get precision to a single point by just squaring the wave function, then integrating over a region seems almost useless. Does the square of the magnitude of the wave function alone really give you the probability of finding the probability at a point, or do you think that my AP chemistry book was simply simplifying it for students who have not taken calculus? Also, if I am right in saying that this is an over simplification, then what exactly does plugging a point directly into a wave function tell you anyway?
 
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  • #2
space-time said:
1. $$\int_a^b {| \psi(x) |^2 dx}$$

The above integral is said to give the probability of finding a particle between x=a and x=b
That is correct.

space-time said:
2. $$\int_{-\infty}^{+\infty} {x | \psi(x) |^2 dx}$$

The above integral is said to yield the expectation value for finding the particle at point x.
The statement is not correct, particularly as x is not a point but a variable in the equation. The expectation value is basically the average:
$$
\langle \hat{A} \rangle \equiv \int_{-\infty}^{\infty} \psi^*(x) \hat{A} \psi(x) dx
$$
Given an ensemble of systems all prepared in the same state ##\psi##, then the average over all systems of the value of the observable ##\hat{A}## will be ##\langle \hat{A} \rangle##.

space-time said:
3. | \psi(x) |^2

My chemistry book says that the square of the magnitude of the wave function evaluated at x gives you the probability of finding the particle at point x.
That's sloppy language. You can check easily that ##| \psi(x) |^2## has units, thus is not a probability. It is a probability distribution function, and must be integrated as in 1. or 2. in order to yield probabilities. By itself it instructs on how the probability varies in space, and is vary useful for visualization.
 
  • #3
space-time said:
I have come across a bit of conflict in wording of some physics and chemistry textbooks about the probability of finding particles in certain places. To be more specific, I have come across 3 different statements:

1. $$\int_a^b {| \psi(x) |^2 dx}$$

The above integral is said to give the probability of finding a particle between x=a and x=b

2. $$\int_{-\infty}^{+\infty} {x | \psi(x) |^2 dx}$$

The above integral is said to yield the expectation value for finding the particle at point x. The problem is: My quantum physics book and other sources say that this expectation value and expectation values in general are what tie the calculations to actual physical observables. That is a problem because, if the first integral that I listed gives the probability of finding the particle in a certain region, then wouldn't that integral be the one that connects the calculations to the physical observable of position (and not the expectation value integral)? In other words, what exactly does this expectation value tell you?

That's really a question of probability theory, independent of quantum mechanics.

If you have a probability density [itex]P(x)[/itex], then that means that you will have (approximately, in the limit as [itex]\delta x \Rightarrow 0[/itex])probability [itex]P(x) \delta x[/itex] of finding a particle in the region between [itex]x[/itex] and [itex]x+\delta x[/itex].

Now, if you rerun the same experiment many, many times, starting with the same probability distribution each time, and you measure the position where the object is found each time, then you will get a sequence of values: [itex]x_1[/itex], [itex]x_2[/itex]... Let [itex]N[/itex] be the number of times that you perform the measurement. Then the average value of [itex]x[/itex] for your measurements is

[itex]\langle x\rangle_N = \frac{1}{N} \sum_i x_i[/itex]

The prediction of probability theory is that in the limit as [itex]N \Rightarrow \infty[/itex],

[itex]\langle x \rangle_N \Rightarrow \int x P(x) dx[/itex]

Expectation values is a way to compute averages of measurables.
 
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  • #4
space-time said:
I have come across a bit of conflict in wording of some physics and chemistry textbooks about the probability of finding particles in certain places. To be more specific, I have come across 3 different statements:

1. $$\int_a^b {| \psi(x) |^2 dx}$$

The above integral is said to give the probability of finding a particle between x=a and x=b

2. $$\int_{-\infty}^{+\infty} {x | \psi(x) |^2 dx}$$

The above integral is said to yield the expectation value for finding the particle at point x. The problem is: My quantum physics book and other sources say that this expectation value and expectation values in general are what tie the calculations to actual physical observables. That is a problem because, if the first integral that I listed gives the probability of finding the particle in a certain region, then wouldn't that integral be the one that connects the calculations to the physical observable of position (and not the expectation value integral)? In other words, what exactly does this expectation value tell you? My hypothesis about the difference between the two integrals is that the first one gives you the probability of finding the particle in set region while the expectation value tells you something about a specific point. If my hypothesis is correct, what exactly does it tell you about said point? In a thread I made once before, someone stated that it was not the probability of finding the particle at said point.

3. | \psi(x) |^2

My chemistry book says that the square of the magnitude of the wave function evaluated at x gives you the probability of finding the particle at point x. I, having already known about the first integral that I mentioned, felt that simply squaring the wave function is way too simple. If you can get precision to a single point by just squaring the wave function, then integrating over a region seems almost useless. Does the square of the magnitude of the wave function alone really give you the probability of finding the probability at a point, or do you think that my AP chemistry book was simply simplifying it for students who have not taken calculus? Also, if I am right in saying that this is an over simplification, then what exactly does plugging a point directly into a wave function tell you anyway?
1. This is the correct expression when the eigenvalues are continuous.
2. The expectation value only coincides with the probability distribution in the idealized case of a pure state(only one value in the diagonal of the density matrix) so 1. and 2. are giving you quite different information in general.
3. This is a simplification for a pure state probability distribution, in practice only mixed states are measured, so you either use 2. or if you are considering pure states probabilities you use the integral as in 1. for continuous eigenvalues or the sum Σi for discrete ones.
 
  • #5
TrickyDicky said:
1. This is the correct expression when the eigenvalues are continuous.
2. The expectation value only coincides with the probability distribution in the idealized case of a pure state(only one value in the diagonal of the density matrix) so 1. and 2. are giving you quite different information in general.
3. This is a simplification for a pure state probability distribution, in practice only mixed states are measured, so you either use 2. or if you are considering pure states probabilities you use the integral as in 1. for continuous eigenvalues or the sum Σi for discrete ones.

What do you mean by pure and mixed states? Also, how would I know whether an eigenvalue was continuous or not before calculating the integral (integral 1.)?
 
  • #6
space-time said:
What do you mean by pure and mixed states?
It is quite unfortunate that most most undergraduate QM books only deal with pure states and mention mixed states only in passing or as a footnote. A pure state is one that is in a superposition, a mixed state is a mixture of pure states. Both can be represented with a statistical operator.
Also, how would I know whether an eigenvalue was continuous or not before calculating the integral (integral 1.)?
You know what observable you are going to measure right? Then you know if their iegenvalues are discrete or continuous. i.e. position is continuous, spin discrete...
 
  • #7
General mathematics statement. Let f(x) be the probability density of some random variable X.

[itex]P(a<X<b) = \int_a^b f(x)dx [/itex]
[itex]E(X)=\int_{-\infty}^{\infty} xf(x)dx [/itex], where E(X) is average
 

1. What is the probability integral in quantum mechanics?

The probability integral, also known as the normalization condition, is a fundamental concept in quantum mechanics that states the total probability of finding a particle in all possible states must equal to 1.

2. How is the expectation value calculated in quantum mechanics?

The expectation value, also called the mean value or average, is calculated by taking the integral of the product of the probability density function and the variable of interest. In quantum mechanics, this variable is usually the position or momentum of a particle.

3. What is the significance of the square of Psi in quantum mechanics?

The square of Psi, also known as the wave function squared, represents the probability density of finding a particle in a specific location or state. It is a crucial concept in quantum mechanics as it allows us to determine the likelihood of a particle being in a particular state or position.

4. Can the probability integral, expectation value, and square of Psi be directly measured in experiments?

No, these values cannot be directly measured in experiments. However, they can be calculated theoretically using mathematical equations and then compared to experimental results to validate the predictions of quantum mechanics.

5. How do the probability integral, expectation value, and square of Psi relate to the uncertainty principle?

The uncertainty principle states that it is impossible to know both the precise position and momentum of a particle simultaneously. The probability integral, expectation value, and square of Psi provide a way to calculate the likelihood of a particle being in a particular state, but they cannot predict the exact position and momentum of the particle due to the uncertainty principle.

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