Calculating Probabilities in a Quantum System

[phyzmatix]

Homework Statement

The components of the initial state $$|\psi_i>$$ of a quantum system are given in a complete and orthonormal basis of three states $$|\phi_1>, |\phi_2>, |\phi_3>$$ by

$$<\phi_1|\psi_i>=\frac{i}{\sqrt{3}}$$
$$<\phi_2|\psi_i>=\sqrt{\frac{2}{3}}$$
$$<\phi_3|\psi_i>=0$$

Calculate the probability of finding the system in a state $$|\psi_f>$$ whose components are given in the same basis by

$$<\phi_1|\psi_f>=\frac{1+i}{\sqrt{3}}$$

$$<\phi_2|\psi_f>=\frac{1}{\sqrt{6}}$$

$$<\phi_3|\psi_f>=\frac{1}{\sqrt{6}}$$

The Attempt at a Solution

Actually, I must admit that I don't really know what I have to do to answer this question. However, while experimenting with possible approaches to a solution I got to

$$P_1=|<\phi_1|\psi_i>|^2=|\frac{i}{\sqrt{3}}|^2=-\frac{1}{3}$$

$$P_2=|<\phi_2|\psi_i>|^2=|\sqrt{\frac{2}{3}}|^2=\frac{2}{3}$$

$$P_3=|<\phi_3|\psi_i>|^2=|0|^2=0$$

But is it possible to get a negative probability?

Also, since the three states are orthonormal, shouldn't they automatically be normalized, and the total probability $$\sum{P_i}=1$$?

Any help here will be greatly appreciated.
phyz

 

[nickjer]

First, what is the equation that describes the probability of finding the initial state in the final state?

In your work, you make no mention of the final state. Also, it is impossible to get a negative probability. The $||^2$ mean to multiply the value by its complex conjugate, so you shouldn't get a negative value.

 

[phyzmatix]

Hi there nickjer! Cheers for your reply.

First, what is the equation that describes the probability of finding the initial state in the final state?

I'm not sure actually...could it be

$$P|<\psi_f|\psi_i>|^2$$

?

But how would I find the states from the given info? (I'm sorry if these questions are trivial and elementary, but I'm studying these subjects without the aid of lectures and often struggle to figure these things out by myself).

In your work, you make no mention of the final state. Also, it is impossible to get a negative probability. The $||^2$ mean to multiply the value by its complex conjugate, so you shouldn't get a negative value.

Ah, you see, I didn't realize this from the available examples (they obviously skip steps assumed as known).

 

[Matterwave]

Your basis is complete and orthnormal. This means you can express any state as:

$$|\psi>=x_1 |\phi_1>+x_2 |\phi_2>+x_3 |\phi_3>$$

The conditions given to you tell you what the x's are for the different states. Now, you just need to take the inner product.

On a side note, the wording of a problem is bad imo. One never finds a particle in some random "state" since the wave-function is not observable. One can only ever observe the observables of that state. Anyways, this is just pedantic...but still...

 

[vela]

I'm not sure actually...could it be

$$P|<\psi_f|\psi_i>|^2$$

?

Yup. (I assume you meant P equals the modulus squared, not P times that quantity.)

But how would I find the states from the given info? (I'm sorry if these questions are trivial and elementary, but I'm studying these subjects without the aid of lectures and often struggle to figure these things out by myself).

You have a complete set, so you can say

$$\sum_{j=1}^3 |\phi_j\rangle\langle\phi_j| = 1$$

If you apply this to $|\psi_i\rangle$, you get

$$|\psi_i\rangle = 1|\psi_i\rangle = \sum_{j=1}^3 |\phi_j\rangle\langle\phi_j|\psi_i\rangle$$

To evaluate $|\langle\psi_f|\psi_i\rangle|^2$, try inserting a 1 between the bra and ket.

 

[nickjer]

Exactly as vela said :)

 

[kidsmoker]

sorry, wrong topic!

 

[phyzmatix]

Your basis is complete and orthnormal. This means you can express any state as:

$$|\psi>=x_1 |\phi_1>+x_2 |\phi_2>+x_3 |\phi_3>$$

The conditions given to you tell you what the x's are for the different states. Now, you just need to take the inner product.

On a side note, the wording of a problem is bad imo. One never finds a particle in some random "state" since the wave-function is not observable. One can only ever observe the observables of that state. Anyways, this is just pedantic...but still...

Thanks for the input!

Yup. (I assume you meant P equals the modulus squared, not P times that quantity.)

Typo

You have a complete set, so you can say

$$\sum_{j=1}^3 |\phi_j\rangle\langle\phi_j| = 1$$

If you apply this to $|\psi_i\rangle$, you get

$$|\psi_i\rangle = 1|\psi_i\rangle = \sum_{j=1}^3 |\phi_j\rangle\langle\phi_j|\psi_i\rangle$$

To evaluate $|\langle\psi_f|\psi_i\rangle|^2$, try inserting a 1 between the bra and ket.

Thanks for this, it's brought to light a whole new angle on my understanding of these "bras", "kets" and bloomin wave functions!

Exactly as vela said :)