Metamath Proof Explorer


Theorem polidi

Description: Polarization identity. Recovers inner product from norm. Exercise 4(a) of ReedSimon p. 63. The outermost operation is + instead of - due to our mathematicians' (rather than physicists') version of Axiom ax-his3 . (Contributed by NM, 30-Jun-2005) (New usage is discouraged.)

Ref Expression
Hypotheses polid.1 ⊢ A ∈ ℋ
polid.2 ⊢ B ∈ ℋ
Assertion polidi ⊢ A ⋅ ih B = norm ℎ ⁡ A + ℎ B 2 - norm ℎ ⁡ A - ℎ B 2 + i ⁢ norm ℎ ⁡ A + ℎ i ⋅ ℎ B 2 − norm ℎ ⁡ A - ℎ i ⋅ ℎ B 2 4

Proof

Step Hyp Ref Expression
1 polid.1 ⊢ A ∈ ℋ
2 polid.2 ⊢ B ∈ ℋ
3 1 2 2 1 polid2i ⊢ A ⋅ ih B = A + ℎ B ⋅ ih A + ℎ B - A - ℎ B ⋅ ih A - ℎ B + i ⁢ A + ℎ i ⋅ ℎ B ⋅ ih A + ℎ i ⋅ ℎ B − A - ℎ i ⋅ ℎ B ⋅ ih A - ℎ i ⋅ ℎ B 4
4 1 2 hvaddcli ⊢ A + ℎ B ∈ ℋ
5 4 normsqi ⊢ norm ℎ ⁡ A + ℎ B 2 = A + ℎ B ⋅ ih A + ℎ B
6 1 2 hvsubcli ⊢ A - ℎ B ∈ ℋ
7 6 normsqi ⊢ norm ℎ ⁡ A - ℎ B 2 = A - ℎ B ⋅ ih A - ℎ B
8 5 7 oveq12i ⊢ norm ℎ ⁡ A + ℎ B 2 − norm ℎ ⁡ A - ℎ B 2 = A + ℎ B ⋅ ih A + ℎ B − A - ℎ B ⋅ ih A - ℎ B
9 ax-icn ⊢ i ∈ ℂ
10 9 2 hvmulcli ⊢ i ⋅ ℎ B ∈ ℋ
11 1 10 hvaddcli ⊢ A + ℎ i ⋅ ℎ B ∈ ℋ
12 11 normsqi ⊢ norm ℎ ⁡ A + ℎ i ⋅ ℎ B 2 = A + ℎ i ⋅ ℎ B ⋅ ih A + ℎ i ⋅ ℎ B
13 1 10 hvsubcli ⊢ A - ℎ i ⋅ ℎ B ∈ ℋ
14 13 normsqi ⊢ norm ℎ ⁡ A - ℎ i ⋅ ℎ B 2 = A - ℎ i ⋅ ℎ B ⋅ ih A - ℎ i ⋅ ℎ B
15 12 14 oveq12i ⊢ norm ℎ ⁡ A + ℎ i ⋅ ℎ B 2 − norm ℎ ⁡ A - ℎ i ⋅ ℎ B 2 = A + ℎ i ⋅ ℎ B ⋅ ih A + ℎ i ⋅ ℎ B − A - ℎ i ⋅ ℎ B ⋅ ih A - ℎ i ⋅ ℎ B
16 15 oveq2i ⊢ i ⁢ norm ℎ ⁡ A + ℎ i ⋅ ℎ B 2 − norm ℎ ⁡ A - ℎ i ⋅ ℎ B 2 = i ⁢ A + ℎ i ⋅ ℎ B ⋅ ih A + ℎ i ⋅ ℎ B − A - ℎ i ⋅ ℎ B ⋅ ih A - ℎ i ⋅ ℎ B
17 8 16 oveq12i ⊢ norm ℎ ⁡ A + ℎ B 2 - norm ℎ ⁡ A - ℎ B 2 + i ⁢ norm ℎ ⁡ A + ℎ i ⋅ ℎ B 2 − norm ℎ ⁡ A - ℎ i ⋅ ℎ B 2 = A + ℎ B ⋅ ih A + ℎ B - A - ℎ B ⋅ ih A - ℎ B + i ⁢ A + ℎ i ⋅ ℎ B ⋅ ih A + ℎ i ⋅ ℎ B − A - ℎ i ⋅ ℎ B ⋅ ih A - ℎ i ⋅ ℎ B
18 17 oveq1i ⊢ norm ℎ ⁡ A + ℎ B 2 - norm ℎ ⁡ A - ℎ B 2 + i ⁢ norm ℎ ⁡ A + ℎ i ⋅ ℎ B 2 − norm ℎ ⁡ A - ℎ i ⋅ ℎ B 2 4 = A + ℎ B ⋅ ih A + ℎ B - A - ℎ B ⋅ ih A - ℎ B + i ⁢ A + ℎ i ⋅ ℎ B ⋅ ih A + ℎ i ⋅ ℎ B − A - ℎ i ⋅ ℎ B ⋅ ih A - ℎ i ⋅ ℎ B 4
19 3 18 eqtr4i ⊢ A ⋅ ih B = norm ℎ ⁡ A + ℎ B 2 - norm ℎ ⁡ A - ℎ B 2 + i ⁢ norm ℎ ⁡ A + ℎ i ⋅ ℎ B 2 − norm ℎ ⁡ A - ℎ i ⋅ ℎ B 2 4