Probability of Measuring Two-State System in State |2>

[logic smogic]

I have a pretty general question about probabilities/expectation values.

Let's say you have a two-state quantum system, with orthogonal states $$|1>,|2>$$.
It's prepared in state $$|\psi>=c_{1}|1>+c_{2}|2>$$, for complex numbers $$c_{1},c_{2}$$.

How do you find the probability of measuring the system to be in state $$|2>$$?

I know that orthogonality means the inner product is zero, i.e., $$<2|1>=0$$, and that there's an expectation value involved here, but from there on I'm a bit lost.

Thanks!

 

[Rach3]

So you know that the probability of measuring a state |psi> to be in the state |2> is |<2|psi>|^2; this is one of the postulates/axioms of QM. So you use the following:

• p(2) = |<2|psi>|^2
  
• |psi>=c1|1> + c2|2>
  
• <2|1> = 0
  
• <2|2> = 1
  
• linearity

Can you figure it out? (Expectation values are not involved, actually.)

 

[logic smogic]

Okay, I think so. I'm just becoming familiar with bra-ket notation, so that's part of the problem (I think).

P(2) = |<2|psi>|^2
|psi>=c1|1> + c2|2>
<2|psi> = <2|c1|1> + <2|c2|2>
= 0 + c2
So, |<2|psi>|^2 = c2*c2 (where c2* is the complex conjugate of c2).

In otherwords, do the bra's commute like that? And does <2|c2|2> = c2, from the identities you listed? Thanks!

 

[Rach3]

Okay, I think so. I'm just becoming familiar with bra-ket notation, so that's part of the problem (I think).

P(2) = |<2|psi>|^2
|psi>=c1|1> + c2|2>
<2|psi> = <2|c1|1> + <2|c2|2>
= 0 + c2
So, |<2|psi>|^2 = c2*c2 (where c2* is the complex conjugate of c2).

In otherwords, do the bra's commute like that? And does <2|c2|2> = c2, from the identities you listed? Thanks!

Yup! <2|c2|2> = c2(<2|2>) (just a scalar factor). Chapter 1 of Sakurai is a very good intrudction to this braket stuff.

In otherwords, do the bra's commute like that?

<2|(|1>+|2>) = <2|1> + <2|2>, if that's what you're asking. It's part of the whole "linearity" business.

 

[logic smogic]

Ah, thanks! I'll have to stop by the physics library and see if it's on reserve (or elsewhere).

 

[dextercioby]

How do you find the probability of measuring the system to be in state $$|2>$$?

Apply the 3-rd axiom

$$\mathcal{P}\left(a_{2},|2\rangle\right)=\frac{\left\langle c_{1}1+c_{2}2\left | \right\hat{P}_{\mathcal{H}_{a}_{2}}\left | \right c_{1}1+c_{2}2\right\rangle}{\Vert |c_{1}1+c_{2}2\rangle \Vert^{2}}$$

Daniel.

 

[cesiumfrog]

Apply the 3-rd axiom

Really? Would you mind enumerating the axioms of QM?

 

[dextercioby]

1. Mathematical description of physical quantum states.
2. Mathematical description of observables and canonical quantization receipt.
3. Probabilistic description of measurement outcomes of observables.
4. Time evolution equation either for states (Schroedinger picture; Interaction picture) or/and observables (Heisenberg picture; Interaction picture)
5. Von Neumann's state vector/density operator reduction postulate (thus applying Copenhagian interpretation)
6. Symmetrization postulate for the description of a quantum system made up of identical subsystems.


Daniel.

 

[masudr]

Apply the 3-rd axiom

$$\mathcal{P}\left(a_{2},|2\rangle\right)=\frac{\left\langle c_{1}1+c_{2}2\left | \right\hat{P}_{\mathcal{H}_{a}_{2}}\left | \right c_{1}1+c_{2}2\right\rangle}{\Vert |c_{1}1+c_{2}2\rangle \Vert^{2}}$$

Daniel.

What is $a_2$ in all this?

 

[dextercioby]

The solution to the spectral equation $A|2\rangle =a_{2}|2\rangle$, where A is the operator <------------measured observable.

Daniel.