What is the Relationship Between Probability Amplitudes and the Sum of Terms?

[isochore]

Homework Statement

The probability of an event is $\vert c_1 \psi_1 + c_2 \psi_2 \vert ^2$. If $\vert c_1 \vert ^2 = \vert c_2 \vert ^2 = \frac{1}{2}$, what can you say about the total probability?

Relevant Equations

See above

I am not sure what I can do with the equation. I realize that $\vert c_1 \vert ^2 = \vert c_2 \vert ^2 = \frac{1}{2}$ does not mean that $c_1 ^2 = c_2 ^2 = \frac{1}{2}$ or that $c_1 = c_2$, so I don't know how to use it. I think ideally I might have something like $P = \vert c_1 \vert ^2 \vert \psi_1 \vert ^2 + \vert c_2 \vert ^2 \vert \psi_2 \vert ^2 +$ (more terms involving $\psi_1 \psi_2$), but is there a formula or something?

 

[PeroK]

Homework Statement:: The probability of an event is $\vert c_1 \psi_1 + c_2 \psi_2 \vert ^2$. If $\vert c_1 \vert ^2 = \vert c_2 \vert ^2 = \frac{1}{2}$, what can you say about the total probability?
Relevant Equations:: See above

I am not sure what I can do with the equation. I realize that $\vert c_1 \vert ^2 = \vert c_2 \vert ^2 = \frac{1}{2}$ does not mean that $c_1 ^2 = c_2 ^2 = \frac{1}{2}$ or that $c_1 = c_2$, so I don't know how to use it. I think ideally I might have something like $P = \vert c_1 \vert ^2 \vert \psi_1 \vert ^2 + \vert c_2 \vert ^2 \vert \psi_2 \vert ^2 +$ (more terms involving $\psi_1 \psi_2$), but is there a formula or something?

What are $\psi_1, \psi_2$ and what can you assume about them?

 

[isochore]

What are ψ1,ψ2ψ1,ψ2\psi_1, \psi_2 and what can you assume about them?

I think they're just supposed to be wave functions corresponding to some experiment, and $c_1$ and $c_2$ are supposed to be scalars.

 

[PeroK]

I think they're just supposed to be wave functions corresponding to some experiment, and $c_1$ and $c_2$ are supposed to be scalars.

What can you assume about $\psi_1, \psi_2$ mathematically?

 

[isochore]

What can you assume about ψ1,ψ2ψ1,ψ2\psi_1, \psi_2 mathematically?

I guess I'm not following here?

 

[PeroK]

I guess I'm not following here?

Have you ever heard the terms "normalised" and "orthogonal"?

 

[isochore]

Yeah, I have, and it makes sense that it would matter. This is all the info I got with the problem (my professor just asked us to consider it, rather than it being a formal problem), so I guess they're normalised/orthogonal?

 

[PeroK]

Yeah, I have, and it makes sense that it would matter. This is all the info I got with the problem (my professor just asked us to consider it, rather than it being a formal problem), so I guess they're normalised/orthogonal?

It's the same if that expression involved ordinary vectors. The magnitude would depend on the magnitude of the vectors and their inner product.

I think it's safe to assume $\psi_1, \psi_2$ are orthogonal and normalised.

 

[isochore]

Makes sense. My initial thought for getting something like $P = \vert c_1 \vert ^2 \vert \psi_1 \vert ^2 + \vert c_2 \vert ^2 \vert \psi_2 \vert ^2 +$ (more terms involving $\psi_1 \psi_2$) was by using the triangle identity, but I don't see a way to get the mixed terms. Is there a different direction to go in?

 

[PeroK]

Makes sense. My initial thought for getting something like $P = \vert c_1 \vert ^2 \vert \psi_1 \vert ^2 + \vert c_2 \vert ^2 \vert \psi_2 \vert ^2 +$ (more terms involving $\psi_1 \psi_2$) was by using the triangle identity, but I don't see a way to get the mixed terms. Is there a different direction to go in?

You need to find out what normalised (magnitude of $1$) and orthogonal (inner product of $0$) mean.

Note that the wavefunctions are functions/vectors not numbers.

You must have course notes on this.

 

[isochore]

I guess I can expand it, like
$P = (c_1^* \psi_1^* + c_2^* \psi_2^*) (c_1 \psi_1 + c_2 \psi_2) = c_1^* c_1 \psi_1^* \psi_1 + c_2^* c_2 \psi_2^* \psi_2 + c_1^* c_2 \psi_1^* \psi_2 + c_2^* c_1 \psi_2^* \psi_1 = \vert c_1 \vert^2 \vert \psi_1 \vert^2 + \vert c_2 \vert^2 \vert \psi_2 \vert^2 + c_1^* c_2 \psi_1^* \psi_2 + c_2^* c_1 \psi_2^* \psi_1$
So if they are normalised/orthogonal, $\vert \psi_1 \vert^2 = \vert \psi_2 \vert^2 = 1$ and $\psi_1^* \psi_2 = \psi_2^* \psi_1 = 0$, so I would be left with $\vert c_1 \vert^2 + \vert c_2 \vert^2$?
This seems to imply that $\vert c_1 \psi_1 + c_2 \psi_2 \vert^2 = \vert c_1 \psi_1 \vert ^2 + \vert c_2 \psi_2 \vert ^2$, but my professor talked about the importance of differentiating between these two in experiments, specifically that you had to sum the probability amplitudes and not the probabilities because of interference effects? Is there some distinction I don't understand?

 

[PeroK]

You are treating the wavefunctions like complex numbers. They are functions. I'm on my phone so it's difficult to type too much.

$c_1\psi_1 + c_2\psi_2$ is a function. It's magnitude is actually an integral. But the integral has the properties of an inner product, so you don't always have to write out an integral. And the functions have properties of vectors.

The magnitude of a function is defined by

$|\psi|^2 = \langle \psi |\psi \rangle = \int \psi^*(x)\psi(x)dx$

However, when you have an orthonormal basis of wavefunctions, then this is the same as you have for ordinary vectors. The magnitude of a function/vector is simply the square root of the sum of the squares (modulus squared for complex coefficients) of the components. Which is just a generalisation of Pythagoras.

 

[PeroK]

So if they are normalised/orthogonal, $\vert \psi_1 \vert^2 = \vert \psi_2 \vert^2 = 1$ and $\psi_1^* \psi_2 = \psi_2^* \psi_1 = 0$, so I would be left with $\vert c_1 \vert^2 + \vert c_2 \vert^2$?

This is not right. Orthogonal means:

$\langle \psi_1 |\psi_2 \rangle = 0$

 

[isochore]

I was using $\psi^*$ to be $\langle \psi \vert$, but that's probably not good notation.
So $P = (c_1^* \langle \psi_1 \vert + c_2^* \langle \psi_2 \vert) (c_1 \vert \psi_1\rangle + c_2 \vert \psi_2\rangle) = c_1^* c_1 \langle \psi_1 \vert \psi_1\rangle + c_2^* c_2 \langle \psi_2 \vert \psi_2\rangle + c_1^* c_2 \langle \psi_1 \vert \psi_2\rangle+ c_2^* c_1 \langle \psi_2 \vert \psi_1\rangle = \vert c_1 \vert^2 \vert \psi_1 \vert^2 + \vert c_2 \vert^2 \vert \psi_2 \vert^2 + c_1^* c_2 \langle \psi_1 \vert \psi_2\rangle + c_2^* c_1 \langle \psi_2 \vert \psi_1\rangle$.
So if they are normalised/orthogonal, $\vert \psi_1 \vert^2 = \vert \psi_2 \vert^2 = 1$ and $\langle \psi_1 \vert \psi_2 \rangle = \langle \psi_2 \vert \psi_1 \rangle = 0$, so I would be left with $\vert c_1 \vert^2 + \vert c_2 \vert^2$?

 

[PeroK]

Generally $*$ denotes the complex conjugate.

Note that because we are dealing with probability amplitudes we have $|c_1|^2 + |c_2|^2 = 1$.

If we were dealing with classical probabilities we would have $p_1 + p_2 = 1$. And in this case it's not $c_1 + c_2$ we are interested in.

And, although for an orthonormal basis the cross-terms cancel, they do not in general. What we have done here is not valid for all wavefunctions $\psi_1, \psi_2$.

 

[isochore]

Makes sense - thanks!