Discussion on the probabilistic interpretation of quantum mechanics

[alebruna]

TL;DR

Considering a possible (not sure) intepretation of the probability by the means of product of two wavefunctions.

I'm an undergraduate trying to understand quantum physics.
So, I'm trying to understant the plane wave basis for a general 3D box.
I understood that the plane wave basis is used to define a period boundary to model real-world situations where the space in which the experiment is performed is limited.
Then for the plane wave basis my function has the form:

$$
\phi(\vec{x})=\frac{e^{i \vec{k}\cdot\vec{x}}}{\sqrt{L^3}}
$$

If we consider one of the boundaries of our 3D box we have that

$$
\phi(x_1 +L ,x_2 ,x_3 )=\frac{e^{i k_1 L}}{\sqrt{L^3}} e^{i \vec{k}\cdot\vec{x}}
$$

Now considering another state $\vec{q}=(q_1,q_2,q_3)$ independent of the first one $\vec{k}=(k_1,k_2,k_3)$ we can find the probability $(\phi_{\vec{k}},\phi_{\vec{q}})$ that SHOULD (I'm not sure) be the probability of finding the states $\phi_{\vec{k}}$ and $\phi_{\vec{q}}$ in our box. By definition the bracket is:

$$
(\phi_{\vec{k}},\phi_{\vec{q}})=\int_V d^3 x \frac{e^{i \vec{x}( \vec{q}-\vec{k})}}{\sqrt{L^3}}
$$

Now this is my interpretation.
Since we're talking about two independent states, in terms of statistics the bracket should be the probability of finding both states, that equates to the intersection of state k and state q. Since uncorrelated the probability of the intersection is the probability of k times the probability of q.
This means also that since k is independent from q, if k and q are the same the probability (by Parseval formula) is 1, otherwise the probability is 0.

Is my interpretation correct? Am I missing something? This type of reasoning can only be used in this case or can be applied anywhere in quantum mechanics?

[Moderator's note: attachment deleted, see post #7.]

 

[PeterDonis]

It's my first thread I don't know how to put math scripts that show the actual math formula and not latex code

LaTeX is how we post math here. There is a "LaTeX Guide" link at the bottom left of each post window. Please use it.

 

[PeterDonis]

@alebruna I have used magic moderator powers to edit your OP to add the markers that make the LaTeX show up the way it should. For separate equations you use double dollar signs $$at the start and end; for inline LaTeX you use double pound signs ##at the start and end.

 

[PeterDonis]

@alebruna where is the attachment in your OP from? Can you provide a link?

 

[PeterDonis]

Since we're talking about two independent states, in terms of statistics the bracket should be the probability of finding both states

No, that's not what the bracket (inner product) means. It means the probability that if you prepare the system in one state, $\phi_{\vec{k}}$, that when you measure it using a device for which another state, $\phi_{\vec{q}}$, is an eigenstate, the measurement will give the eigenvalue $\vec{q}$ corresponding to $\phi_{\vec{q}}$.

In other words: the bracket $(\phi_{\vec{k}}, \phi_{\vec{q}})$ gives the probability that, if you prepare the particle in the plane wave state $\phi_{\vec{k}}$, and then you measure whether or not its momentum points along the direction $\vec{q}$, you will get a "yes" result. (Actually, in this case it will be a probability density, since we are talking about continuous variables. It's actually better, IMO, to try to understand states and inner products with a discrete state space, like spin, first, so the inner products represent actual probabilities in finite sample spaces. But unfortunately many textbooks don't take that approach.)

 

[alebruna]

@alebruna where is the attachment in your OP from? Can you provide a link?

My attachment comes from study material that our professor published on the site of university but it's not publicly available so I only have the pdf not a link. I'll put below the complete file.

[Moderator's note: attachment deleted, see post #7.]

 

[PeterDonis]

My attachment comes from study material that our professor published on the site of university but it's not publicly available

If it's not, then we can't have it publicly visible here, unfortunately.

 

[alebruna]

No, that's not what the bracket (inner product) means. It means the probability that if you prepare the system in one state, $\phi_{\vec{k}}$, that when you measure it using a device for which another state, $\phi_{\vec{q}}$, is an eigenstate, the measurement will give the eigenvalue $\vec{q}$ corresponding to $\phi_{\vec{q}}$.

For example: suppose $\phi_{\vec{k}}$ is a momentum eigenstate (i.e., a plane wave), and you are measuring position. Then the bracket $(\phi_{\vec{k}}, \phi_{\vec{q}})$ gives the probability that, if you prepare the particle in the plane wave state $\phi_{\vec{k}}$, you will measure its position to be $\vec{q}$. (Actually, in this case it will be a probability density, since we are talking about continuous variables. It's actually better, IMO, to try to understand states and inner products with a discrete state space, like spin, first, so the inner products represent actual probabilities in finite sample spaces. But unfortunately many textbooks don't take that approach.)

Thank you for your explanation, I think I understood better what we're talking about

 

[PeterDonis]

Thank you for your explanation

Please read the edited version of that post. I had misread your OP; I thought $\vec{q}$ was intended to refer to a position, but it's not, it's another momentum. (The letter $q$ is usually used to denote position in QM, which is why I was confused.)

 

[alebruna]

No, that's not what the bracket (inner product) means. It means the probability that if you prepare the system in one state, $\phi_{\vec{k}}$, that when you measure it using a device for which another state, $\phi_{\vec{q}}$, is an eigenstate, the measurement will give the eigenvalue $\vec{q}$ corresponding to $\phi_{\vec{q}}$.

In other words: the bracket $(\phi_{\vec{k}}, \phi_{\vec{q}})$ gives the probability that, if you prepare the particle in the plane wave state $\phi_{\vec{k}}$, and then you measure whether or not its momentum points along the direction $\vec{q}$, you will get a "yes" result. (Actually, in this case it will be a probability density, since we are talking about continuous variables. It's actually better, IMO, to try to understand states and inner products with a discrete state space, like spin, first, so the inner products represent actual probabilities in finite sample spaces. But unfortunately many textbooks don't take that approach.)

So in this case $(\psi_{\vec{k}},\psi_{\vec{q}})$ indicates the projection of $\psi_{\vec{k}}$ on $\psi_{\vec{q}}$ and since in this case they are orthogonal the only possibility for which that projection exists (and it's also 1) is that \vec{k} and \vec{q} are equal.
But since in this case the brackets acts as a projection i wonder if it works as a projection also for a general wavefunction $\psi(\vec{x})=\sum_j C_j \psi_j (\vec{x})$. In this case for a bracket like $(\psi, \phi_i)$ we have that this is the projection of $\psi$ over $\phi_i$ whatever $i$ is, but for the bracket $(\psi,\psi)$ why would it be the probability density $\rho$? Am I completely wrong on the approach of the bracket as the "projection"?

 

[PeterDonis]

the only possibility for which that projection exists (and it's also 1) is that \vec{k} and \vec{q} are equal.

Yes.

for the bracket $(\psi,\psi)$ why would it be the probability density $\rho$? Am I completely wrong on the approach of the bracket as the "projection"?

The projection of a state you prepare onto an eigenstate of the measurement you're doing is the probability density. That's what a projection means in this case.

In the case of plane wave momentum states, what you said at the top of your post is equivalent to saying that if you prepare a particle in momentum state $\phi_{\vec{k}}$, then measure its momentum, you have probability $1$ of getting the result $\vec{k}$ and probability $0$ of getting any other result $\vec{q}$.

 

[bhobba]

Once you gain more experience in QM and Born's rule, the following (Gleason's Theorem) will likely be of interest:
https://arxiv.org/abs/quant-ph/9909073

Thanks
Bill