Chapter 3: Q3P (page 98)

Show that if $⟨ h ∣ \hat{Q} h ⟩ = ⟨ \hat{Q} h ∣ h ⟩$ for all functions $h$ (in Hilbert space), then $⟨ f ∣ \hat{Q} g ⟩ = ⟨ \hat{Q} f ∣ g ⟩$ for allrole="math" localid="1655395250670" $f$ and $g$ (i.e., the two definitions of "Hermitian" -Equations 3.16 and 3.17- are equivalent).

Short Answer

Expert verified

Hermitian equations have two definitions, both of which are equivalent.

$⟨ f ∣ \hat{Q} g ⟩ = ⟨ \hat{Q} f ∣ g ⟩$

$⟨ g ∣ \hat{Q} f ⟩ = ⟨ \hat{Q} g ∣ f ⟩$

Step by step solution

Concept used

For a Hermitian operator (say Q), the following condition must be satisfied:

$\int_{a}^{b} g^{*} Qfdx = \int_{a}^{b} f {(Qg)}^{*} dx$

Given information from question

For a Hermitian operator (say Q), the following condition must be satisfied:

$\int_{a}^{b} g^{*} Qfdx = \int_{a}^{b} f {(Qg)}^{*} dx$

This condition can be written as: Using the more compact bracket notation, this condition can be written as:

$⟨ g ∣ \hat{Q} f ⟩ = ⟨ \hat{Q} g ∣ f ⟩$

Explanation

Start with a less restrictive condition that is:

$⟨ h ∣ Q^{h} ⟩ = ⟨ Q^{h} ∣ h ⟩ . ..... (1)$

Let , $h = f + g$ so:

$\begin{matrix} ⟨ h ∣ \hat{Q} h ⟩ = ⟨ f + g ∣ \hat{Q} f + \hat{Q} g ⟩ . ..... (2) \\ = ⟨ f ∣ \hat{Q} f ⟩ + ⟨ g ∣ \hat{Q} g ⟩ + ⟨ f ∣ \hat{Q} g ⟩ + ⟨ g ∣ \hat{Q} f ⟩ . ..... (3) \end{matrix}$

We can also write this equation as:

$\begin{matrix} ⟨ \hat{Q} h ∣ h ⟩ = ⟨ \hat{Q} f + \hat{Q} g ∣ f + g ⟩ . ..... (4) \\ = ⟨ \hat{Q} f ∣ f ⟩ + ⟨ \hat{Q} g ∣ g ⟩ + ⟨ \hat{Q} f ∣ g ⟩ + ⟨ \hat{Q} g ∣ f ⟩ . ..... (5) \end{matrix}$

Use assumption $⟨ h ∣ \hat{Q} h ⟩ = ⟨ \hat{Q} h ∣ h ⟩$ for all functions, we have $⟨ f ∣ \hat{Q} f ⟩ = ⟨ \hat{Q} f ∣ f ⟩$ and $⟨ g ∣ \hat{Q} g ⟩ =$ $⟨ Q \hat{Q} g ∣ g ⟩$ so, equating (2) and (3) and cancel out the common terms, we get: $⟨ f ∣ \hat{Q} g ⟩ + ⟨ g ∣ \hat{Q} f ⟩ = ⟨ \hat{Q} f ∣ g ⟩ + ⟨ \hat{Q} g ∣ f ⟩$

Now we do the same to above but with $h = f + ig$ , so:

$\begin{matrix} ⟨ h ∣ \hat{Qh} ⟩ = ⟨ f + ig ∣ \hat{Q} f + i \hat{Q} g ⟩ \\ = ⟨ f ∣ \hat{Q} f ⟩ + (- i) (i) ⟨ g ∣ \hat{Q} g ⟩ + i ⟨ f ∣ \hat{Q} g ⟩ - i ⟨ g ∣ \hat{Q} f ⟩ \\ = ⟨ f ∣ \hat{Q} f ⟩ + ⟨ g ∣ \hat{Q} g ⟩ + i ⟨ f ∣ \hat{Q} g ⟩ - i ⟨ g ∣ \hat{Q} f ⟩ . ..... (6) \\ ⟨ Qh ∣ h ⟩ = ⟨ \hat{Q} f ∣ f ⟩ + ⟨ \hat{Q} g ∣ g ⟩ + i ⟨ \hat{Q} f ∣ g ⟩ - i ⟨ Qg ∣ f ⟩ . ..... (7) \end{matrix}$

Using the assumption $⟨ h ∣ \hat{Q} h ⟩ = ⟨ \hat{Q} h ∣ h ⟩$ for all functions, we have $⟨ f ∣ \hat{Q} f ⟩ =$ $⟨ \hat{Q} ∣ f ⟩$ and $⟨ g ∣ \hat{Q} g ⟩ = ⟨ \hat{Q} ∣ g ⟩$ so, equating (2) and (3) and cancel out the common terms, we get: $⟨ f ∣ \hat{Q} g ⟩ - ⟨ g ∣ \hat{Q} f ⟩ = ⟨ \hat{Q} f ∣ g ⟩ - ⟨ \hat{Q} g ∣ f ⟩ . ..... (8)$

Add equation (4) and (8), we get,

$⟨ f ∣ \hat{Q} g ⟩ = ⟨ \hat{Q} f ∣ g ⟩$

If we subtract equation (8) from equation (4), we get,

$⟨ g ∣ \hat{Q} f ⟩ = ⟨ \hat{Q} g ∣ f ⟩$

Thus, the given statement is proved, i.e., $⟨ g ∣ \hat{Q} f ⟩ = ⟨ \hat{Q} g ∣ f ⟩$ .

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Show that if ⟨h∣Q^h⟩=⟨Q^h∣h⟩for all functionsh(in Hilbert space), then⟨f∣Q^g⟩=⟨Q^f∣g⟩for allrole="math" localid="1655395250670" fandg(i.e., the two definitions of "Hermitian" -Equations 3.16 and 3.17- are equivalent).

Short Answer

Step by step solution

Concept used

Given information from question

Explanation

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Wave Optics

Atoms and Radioactivity

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Study anywhere. Anytime. Across all devices.