Can Quantum Mechanics Ket and Bra Methods Solve Scalar Equations?

[Ugnius]

Homework Statement

$\left|m\right\rangle = \frac{1}{\sqrt{A}}\binom{1-2i}{\alpha} , \left|n\right\rangle = \frac{1}{\sqrt{14}}\binom{-3+2i}{\beta}$

Relevant Equations

Find unknown constants A, α and β, we know that β is real positive integer , α has both real and imaginary parts

Not really even sure how to approach this problem , I would guess if we need scalar answer we would need to combine these two given equations together but I'm unfamiliar with such methods, in the book there is methods to make a ket to a bra and then matrix part transposes and multiplies with the original while constant squares like:
$\left\langle m \right|\left|m\right\rangle = ({\frac{1}{\sqrt{A}}}^2)*\binom{1-2i}{\alpha}*(1-2i ,\alpha)$

Would that be an approach?

 

[topsquark]

Homework Statement: $\left|m\right\rangle = \frac{1}{\sqrt{A}}\binom{1-2i}{\alpha} , \left|n\right\rangle = \frac{1}{\sqrt{14}}\binom{-3+2i}{\beta}$
Relevant Equations: Find unknown constants A, α and β, we know that β is real positive integer , α has both real and imaginary parts

Not really even sure how to approach this problem , I would guess if we need scalar answer we would need to combine these two given equations together but I'm unfamiliar with such methods, in the book there is methods to make a ket to a bra and then matrix part transposes and multiplies with the original while constant squares like:
$\left\langle m \right|\left|m\right\rangle = ({\frac{1}{\sqrt{A}}}^2)*\binom{1-2i}{\alpha}*(1-2i ,\alpha)$

Would that be an approach?

If you normalize $\mid n \rangle$, then yes, you can find $\beta$. But you would write this as
$\langle n \mid n \rangle = \dfrac{1}{\sqrt{14}} \begin{pmatrix} -3-2i & \beta^* \end{pmatrix} \dfrac{1}{\sqrt{14}} \begin{pmatrix} -3+2i \\ \beta \end{pmatrix}$
(You wrote your bra and ket in the wrong order for $\langle m \mid m \rangle$ in your OP.)

We know that $\beta$ is real, so $\beta^* = \beta$, and you can go from there.

To find A and $\alpha$ we need to know something about how $\mid m \rangle$ and $\mid n \rangle$ relate to each other. Are they orthogonal? ie. $\langle n \mid m \rangle = 0$?

-Dan

 

[Ugnius]

Thank you.
They are orthogonal yes , I'm brute forcing the solution right now , I solved for \beta. Now I need to solve \langle n \mid m \rangle = 0 and i'll get back to you for confirmation if I did it correctly