Normalization of electron Spin state

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The discussion revolves around normalizing a spin state represented as a complex vector. It highlights the confusion regarding whether to integrate over spatial dimensions or to use inner products for normalization. The correct approach involves calculating the inner product of the spin state with itself rather than performing an integral. The norm of the spin state should be derived using the complex conjugate, leading to a clearer understanding of the normalization constant. The emphasis is on treating the spin state as a two-component vector and applying appropriate quantum mechanics principles for normalization.
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Homework Statement
Given the Spin Operator $$\hat{\vec{S}}=\frac{\hbar}{2}\hat{\vec{\sigma}}$$ with the Pauli matrices $$\hat{\vec{\sigma}}$$ calculate the Normalizationconstant A for the given Spinstate $$\chi$$
Relevant Equations
$$\chi=A\begin{pmatrix}
3i\\
4
\end{pmatrix}$$

$$\sigma_x=\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix}$$

$$\sigma_y=\begin{pmatrix}
0 & -i\\
i & 0
\end{pmatrix}$$

$$\sigma_z=\begin{pmatrix}
1 & 0\\
0 &-1
\end{pmatrix}$$
I don't really know where to begin.
1. idea: For a spatial wave funtion I'd have to calculate the integral over dxdydz for -inf to +inf. But that doesn't seem very reasonable to me here.
$$\int \chi dxdydz=\int A\begin{pmatrix}
3i\\
4
\end{pmatrix} dxdydz$$

Do have to substitute dxdydz with something and get the pauli matrizes involved?

2. idea: If I treat the spinstate like a regular vector the norm would just be $$\sqrt{3i^2+4^2}=\sqrt{16-9}=\sqrt{5}$$. But can I treat a spinstate like this?
 
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Anton02 said:
Homework Statement: Given the Spin Operator $$\hat{\vec{S}}=\frac{\hbar}{2}\hat{\vec{\sigma}}$$ with the Pauli matrices $$\hat{\vec{\sigma}}$$ calculate the Normalizationconstant A for the given Spinstate $$\chi$$
Relevant Equations: $$\chi=A\begin{pmatrix}
3i\\
4
\end{pmatrix}$$

$$\sigma_x=\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix}$$

$$\sigma_y=\begin{pmatrix}
0 & -i\\
i & 0
\end{pmatrix}$$

$$\sigma_z=\begin{pmatrix}
1 & 0\\
0 &-1
\end{pmatrix}$$

I don't really know where to begin.
1. idea: For a spatial wave funtion I'd have to calculate the integral over dxdydz for -inf to +inf. But that doesn't seem very reasonable to me here.
$$\int \chi dxdydz=\int A\begin{pmatrix}
3i\\
4
\end{pmatrix} dxdydz$$

Do have to substitute dxdydz with something and get the pauli matrizes involved?
That is a bizarre idea!
Anton02 said:
2. idea: If I treat the spinstate like a regular vector the norm would just be $$\sqrt{3i^2+4^2}=\sqrt{16-9}=\sqrt{5}$$. But can I treat a spinstate like this?
Why not?
 
You should review how to deal with a complex vector space. The norm of given state isn't ##\sqrt 5##.
 
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Anton02 said:
Homework Statement: Given the Spin Operator $$\hat{\vec{S}}=\frac{\hbar}{2}\hat{\vec{\sigma}}$$ with the Pauli matrices $$\hat{\vec{\sigma}}$$ calculate the Normalizationconstant A for the given Spinstate $$\chi$$
Relevant Equations: $$\chi=A\begin{pmatrix}
3i\\
4
\end{pmatrix}$$

$$\sigma_x=\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix}$$

$$\sigma_y=\begin{pmatrix}
0 & -i\\
i & 0
\end{pmatrix}$$

$$\sigma_z=\begin{pmatrix}
1 & 0\\
0 &-1
\end{pmatrix}$$

I don't really know where to begin.
1. idea: For a spatial wave funtion I'd have to calculate the integral over dxdydz for -inf to +inf. But that doesn't seem very reasonable to me here.
$$\int \chi dxdydz=\int A\begin{pmatrix}
3i\\
4
\end{pmatrix} dxdydz$$

Do have to substitute dxdydz with something and get the pauli matrizes involved?

2. idea: If I treat the spinstate like a regular vector the norm would just be $$\sqrt{3i^2+4^2}=\sqrt{16-9}=\sqrt{5}$$. But can I treat a spinstate like this?
Several problems with this.

One for normalizing you take the inner product it doesnt necessarily have to involve Integration. Since there’s no spatial wavefunction you shouldnt integrate in this case of just a spin state just take the inner product of the spin state χ with itself. Also for the magnitude its necessary to multiply by the complex conjugate to get the normalization constant squared then divide by A.
 
Of course for a two-component vector (or rather a Weyl spinor) the scalar product of course reads
$$\langle \psi|\phi \rangle=\psi_1^* \phi_1 + \psi_2^* \phi_2,$$
and thus
$$\langle \psi|\psi \rangle=|\psi_1|^2 + |\psi_2|^2.$$
Now it should make more sense.
 
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