Normalization of electron Spin state

  • #1
Anton02
1
0
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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  • #2
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?
 
  • #3
You should review how to deal with a complex vector space. The norm of given state isn't ##\sqrt 5##.
 
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  • #4
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.
 
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  • #5
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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Likes DrClaude

FAQ: Normalization of electron Spin state

What is normalization of electron spin state?

Normalization of an electron spin state refers to the process of ensuring that the total probability of finding the electron in any of its possible spin states is equal to one. This is a fundamental requirement in quantum mechanics, ensuring that the wave function describing the electron's spin state is physically meaningful.

Why is normalization important in quantum mechanics?

Normalization is important because it ensures that the probabilities of all possible outcomes of a quantum measurement add up to one. This aligns with the probabilistic interpretation of quantum mechanics, where the square of the wave function's amplitude represents the probability density of finding a particle in a particular state.

How do you normalize an electron spin state?

To normalize an electron spin state, you calculate the norm (or magnitude) of the wave function and then divide the wave function by this norm. For a spin state represented by a two-component spinor, you ensure that the sum of the squares of the absolute values of the components equals one.

What is the mathematical representation of a normalized electron spin state?

A normalized electron spin state is often represented by a two-component spinor, such as \( \chi = \begin{pmatrix} \alpha \\ \beta \end{pmatrix} \), where \( |\alpha|^2 + |\beta|^2 = 1 \). This condition ensures that the spin state is normalized.

Can an electron spin state be unnormalized, and what does that imply?

Yes, an electron spin state can initially be unnormalized, which means that the total probability of finding the electron in any spin state is not equal to one. This is often the case in intermediate steps of calculations. However, for physical interpretation and meaningful results, the spin state must be normalized.

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