**By: Jenard B. Cerbanez, MS Physics I, MSU-IIT**

Show that ħ

Solution:

Given that,

, where

,

Thus,

however, . Now we are left with the terms,

where

Therefore,

ħ, where ħ (Canonical Commutator).

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**By: Jenard B. Cerbanez, MS Physics I, MSU-IIT**

Show that ħ

Solution:

Given that,

, where

,

Thus,

however, . Now we are left with the terms,

where

ħ

Therefore,

ħ, where ħ (Canonical Commutator).

Posted in Quantum Science Philippines **|** No Comments »

**By: Jenard B. Cerbanez, MS Physics I, MSU-IIT**

Show that ħ

Solution:

Given that,

, where

,

Thus,

however, . Now we are left with the terms,

where ħ (Canonical Commutator)

Therefore,

ħ, where .

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By: **Maria Christine L. Lugo**, MS Physics I, MSU-IIT

Show that

a.)

b.)

Solution:

a.) We can expand by its components and S_{2}^{2},

But we know that,

(See link https://www.quantumsciencephilippines.com/?p=5698 )

Therefore,

b.) Using the eigenvalue equation of and substituting the basis

We have,

Therefore, the matrix representation will give us

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### Orthogonality of Two Eigenvectors

*by: Mariel A. Escobal, MS Physics I, MSU-IIT*

### Prove that two eigenvectors of a Hermitian operator corresponding to two different eigenvalues are orthogonal.

Consider two eigenvectors **|ψ⟩** and** |φ⟩ **of the Hermitian operator **Â,**

Â|ψ⟩ = λ|ψ⟩ (1)

Â|φ⟩ = μ|φ⟩ (2)

Since **Â **is Hermitian, we can write (2) as

⟨φ|Â = μ⟨φ| (3)

Multiplying (1) by **⟨φ| **on the left and (3) by **|ψ⟩ **on the right, we get

⟨φ|Â|ψ⟩ = λ⟨φ|ψ⟩ (4)

⟨φ|Â|ψ⟩ = μ⟨φ|ψ⟩ (5)

Subtracting (4) and (5), we get

(**λ-μ)⟨φ|ψ⟩ = 0**

Consequently, the fact that **λ≠μ**, (**λ-μ) ≠ 0** which then implies that **⟨φ|ψ⟩ = 0.**

And as we know, this is the condition for two orthogonal eigenvectors.

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