Is a Complex Probability Result Valid in a Quantum Measurement?

[Bacat]

Homework Statement

Consider a quantum system described by a basis $$\mid 1 \rangle$$ and $$\mid 2 \rangle$$.

The system is initially in the state: $$\psi_i = \frac{i}{\sqrt3} \mid 1 \rangle + \sqrt{\frac{2}{3}} \mid 2 \rangle$$.

(a) Find the probability that the initial system is measured to be in the state: $$\psi_f = \frac{1 + i}{\sqrt 3} \mid 1 \rangle + \frac{1}{\sqrt{3}} \mid 2 \rangle$$

Homework Equations

The basis is assumed to be orthonormal, hence $$\langle 1 \mid 1 \rangle = \langle 2 \mid 2 \rangle = 1$$

Probability is calculated as $$(\langle \psi_f \mid \psi_i \rangle)^2$$

The Attempt at a Solution

Calculating this, I get a complex answer. I'm not sure but I think a probability (a real observable) should be a real number. Is that right?

The answer I get is $$\frac{2 + 2\sqrt{2}}{9}(1+i)$$.

 

[Bacat]

I just tried to normalize the two states and found that they are both unit-normal already...

This is confusing.

 

[gabbagabbahey]

Probability is calculated as $$(\langle \psi_f \mid \psi_i \rangle)^2$$

Surely you mean:

$$\frac{\langle \psi_i \vert \psi_f \rangle \langle \psi_f \vert \psi_i \rangle}{\langle \psi_i \vert \psi_i \rangle \langle \psi_f \vert \psi_f \rangle}=\vert\langle \psi_i \vert \psi_f \rangle \vert^2$$

(for normalized $\psi_i$ and $\psi_f$)...right?

 

[Bacat]

Thanks for the response ggh.

Since my states are normalized, the denominator drops to 1. The numerator seems correct. I think you are right that I missed the absolute value signs.

...

After a quick calculation, I see that's where I screwed up. I get a real answer if I take the absolute value before squaring.

Thanks!