Probability of quantum transition

I have a question about a task:

We have a particle, which is in a linear combination of the first two states of the harmonic oscillator, which we can parametrise as $|\psi\rangle=\cos(\frac{\theta}{2})\space |0\rangle +e^{i\phi}\sin(\frac{\theta}{2})|1\rangle$, where $0\leq\theta \leq \pi$ and $\leq \phi<2\pi$, what is the probability of finding the particle in the state of $|a\rangle=\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$.

I know that the right answer is the projection of $|a\rangle$ onto $|\psi\rangle$ squared, so $P=|\langle \psi,a\rangle|^2$, but why is that so ? I understand that we need to project $|\psi\rangle$ onto a eigenvector ($|0\rangle$, $|1\rangle$), if we want to find the probability of measuring/finding the particle in a eigenstate.

What about finding the particle in a non eigenstate, what is the interpretation of this, since we can only measure eigenstates?

• Why is that so? - It is called the Born rule, and it is a postulate of QM. What is the interpretation of finding the particle in a non eigenstate? - How do you expect to do this given that, as you've said yourself, we can only measure eigenstates of operators? – Prof. Legolasov Nov 7 '16 at 1:39
• "The transition probability $P(\psi,\theta)$ from a state $\psi$ to a state $\theta$, or, in other words the probability of a "quantum jump" from $\psi$ to $\theta$ is $P(\psi,\theta)=|\langle \psi,\theta\rangle|^2$ " This is what I have found in one of the pdfs, and I think this is what I am looking for. I just dont understand why that is so, why would a projection like that give a probability... – Luka8281 Nov 7 '16 at 9:21
• I am terribly sorry I have misread, I have now edited the question . Sorry ! – Luka8281 Nov 7 '16 at 9:47

In the specific case you mention, the particle either is in the state you are looking for or not, so that you have to set $\theta$ and $\phi$ so that they fit the state sought. You want $cos(\theta)=sin(\theta)$ and $e^{i\phi}=0$ so $\theta=\pi/4$ and $\phi=0$.

Yet in general a state can evolve over time and you may wonder "when will it reach a given state $|\chi\rangle$ during its evolution".

You then impose $\langle \chi |\psi\rangle=1$ which is not a jump probability (the state is not jumping, it's evolving) but rather a projection (it is one when the states are the same).

In your case, even if it is not evolving, you have to project. You then get: $${1\over \sqrt{2}}(\langle 0 | +\langle 1 |)|(\cos(\theta) |0 \rangle+\sin(\theta)e^{i\phi}|1\rangle)=1$$

You are then imposing that the states are the same one by imposing that their scalar product is unitary (assuming everything is normalised).

Then: $${1\over \sqrt{2}}\cos(\theta)+{1\over \sqrt{2}}\sin(\theta)e^{i\phi}=1$$ Which you can solve, but it is solved by the condition mentioned above: $\cos(\theta)=\sin(\theta)$ and $e^{i\phi}=0$ so $\theta=\pi/4$ and $\phi=0$.

(The complex part is zero only if $\phi=0$ or $\theta=0$. If $\theta=0$ you would get for the real part $\cos(0)=\sqrt{2}$ so it has to be $\phi=0$. And then $\theta=\pi/4$. Plus periodicity, of course.)

In general though, the probability that a state $|\psi\rangle$ when measured gives you a given result $|\chi\rangle$ is $\langle \chi|\psi\rangle^2$ as you mentioned. Think of it has: "I am using a magical operator which has $|\chi\rangle$ as eigenvector. Then of course is such operator does not exist you will never measure it, yet $\langle \chi |\psi\rangle$ still gives you an indication about the "amount" of $|\chi\rangle$ inside $|\psi\rangle$ (or rather, their superposition in vector space). In your case:

$${1\over \sqrt{2}}\cos(\theta)+{1\over \sqrt{2}}\sin(\theta)e^{i\phi}$$ gives you, as $\theta$ and $\phi$ vary, of how similar your state is to ${1\over 2}(|0\rangle+|1\rangle)$.

• But the projection depends on $\theta$ and $\phi$, so for a given $\theta$ and $\phi$ I will get a state, and a value of the projection/probability of measruing the state $|a\rangle$, but if my system is now in a certain state, shouldn't I allways measure that state, that the system is in, not the state $|a\rangle$ ?? $|a\rangle$ actually means measuring the $|1\rangle$ or $|0\rangle$ with the same probabilty = 1/2 . – Luka8281 Nov 7 '16 at 11:06
• That is what I was trying to say: you are mixing up two things: 1-if the system is in state a and you measure it with an operator O you can only get its eigenvectors as new states. 2-in your case, your state is fixed (depending on the values of $\theta$ and $\phi$). Thus it will never be in another state, unless $\theta$ and $\phi$ are the right one. The projection only tells you how close you are to the state you are looking for or you can impose perfect projection (1) to find the right values of the parameters. If the state is given you can only measure an operator eigenstates – JalfredP Nov 7 '16 at 11:14
• Given a you can measure 1 or 2 using the harmonic hamiltonian. You may measure other stuff with another operator which does not commute with the Hamiltonian. If you devised an operator capable of measuring 1+2 (meaning its eigenstates include 1+2) you could try to measure it, but will not measure it unless the projection is different from 0. If it is one you will measure it with certainty, otherwise it depends on the sqared projection. – JalfredP Nov 7 '16 at 11:16
• Oh... I see what you did with the magical operator, thank you very much! – Luka8281 Nov 7 '16 at 13:37

You measure eigenstates of the operator you use to measure. In general a state may live as a superposition of states (eigenstates) of another operator.

• I am terribly sorry I have misread, I have now edited the question . Sorry ! – Luka8281 Nov 7 '16 at 9:47