In general for two operators to be equal, all their (matrix) elements must be equal

$$A = B \rightarrow \langle \phi_1|A| \phi_2\rangle=\langle \phi_1|B| \phi_2\rangle$$

However, I am asked to show that in complex vector spaces it is enough to just state:

$$A = B \rightarrow \langle \phi_1|A| \phi_1\rangle=\langle \phi_1|B| \phi_1\rangle$$

In my attempt to do show this I did the following:

$$ | \phi_1\rangle = | \psi_1\rangle + i| \psi_2\rangle \\ \langle \phi_1|A| \phi_1\rangle = (\langle \psi_1| + i \langle \psi_2|)A(| \psi_1\rangle + i| \psi_2\rangle) = (\langle \psi_1| + i \langle \psi_2|)B(| \psi_1\rangle + i| \psi_2\rangle) = \langle \phi_1|B| \phi_1\rangle$$

which when expanded out gave me $$ \langle \psi_1|A| \psi_1\rangle + i\langle \psi_1|A| \psi_2\rangle - i\langle \psi_2|A| \psi_1\rangle + \langle \psi_2|A| \psi_2\rangle = \langle \psi_1|B| \psi_1\rangle + i\langle \psi_1|B| \psi_2\rangle - i\langle \psi_2|B| \psi_1\rangle + \langle \psi_2|B| \psi_2\rangle $$

cancelling out terms on either side leaves me with:

$$ i\langle \psi_1|A| \psi_2\rangle - i\langle \psi_2|A| \psi_1\rangle = i\langle \psi_1|B| \psi_2\rangle - i\langle \psi_2|B| \psi_1\rangle $$

In addition to this I constructed another equality by following these steps, but starting from:

$$\langle \phi_1|A^\dagger| \phi_1\rangle=\langle \phi_1|B$\dagger| \phi_1\rangle $$

and in doing so obtained:

$$ i\langle \psi_1|A^\dagger| \psi_2\rangle - i\langle \psi_2|A^\dagger| \psi_1\rangle = i\langle \psi_1|B^\dagger| \psi_2\rangle - i\langle \psi_2|B^\dagger| \psi_1\rangle $$

my plan was to attempt to combine the two equality's in an attempt to produce $$\langle \psi_1|A| \psi_2\rangle=\langle \psi_1|B| \psi_2\rangle $$

as someone in the class mentioned they had luck with this method, but I'm stumped with where to go from here, or if I've made a mistake along the way. Any help would be greatly appreciated, I've been wracking my brain trying to think of something else to try.

  • 1
    $\begingroup$ What you are looking for is called polarization identity in Hilbert spaces (you have to use the complex one). $\endgroup$
    – yuggib
    Oct 12 '15 at 7:50
  • $\begingroup$ I do not think that the polarization identity, though related, can be directly exploited, unless $A-B$ is Hermitian. I wrote a direct proof in my answer. $\endgroup$ Oct 12 '15 at 9:14
  • $\begingroup$ @ValterMoretti In your proof you essentially prove that two sesquilinear forms are equal if the related quadratic forms are equal; and that immediate once the polarization identity is proved. $\endgroup$
    – yuggib
    Oct 12 '15 at 9:25
  • $\begingroup$ $(x,y) = \langle x| A y\rangle$ is not sequilinear because $(y,x) \neq \overline{(x,y)}$ unless $A \subset A^*$. That's my point... $\endgroup$ Oct 12 '15 at 9:29
  • $\begingroup$ We have a different definition of sesquilinear. For me sesquilinear is just linear in the right argument, antilinear in the left one. $\endgroup$
    – yuggib
    Oct 12 '15 at 9:30

Yes, your conjecture is true (there is no need for involving the adjoint operators in the proof). Indeed, in a complex Hilbert space $\cal H$ (more generally a complex vector space equipped with an Hermitian scalar product) we have the following proposition.

Proposition. Let $A,B : D \to \cal H$ be a pair of linear operators defined in the dense subspace $D\subset \cal H$. $A=B$ if and only if $\langle x|Ax \rangle = \langle x|Bx \rangle$ for all $x\in D$.

Proof. It is enough proving that $\langle x|Ax \rangle=0$ for all $x\in D$ implies $A=0$. To this end, first use $x= y\pm z$ and then $x= y\pm iz$ in the identity above observing that $\langle y|Ay \rangle= 0$ and $\langle z|Az \rangle=0$. Linearity in the right entry and anti linearity in the left entry easily produce $\langle y|Az \rangle=0$ for all $y,z \in D$. Since $D$ is dense, there is a sequence $D \ni y_n \to Az$. Continuity of the scalar product immediately yields $||Az||^2 = \lim_{n\to +\infty} \langle y_n | Az\rangle =0$ for all $z\in D$. In other words $A=0$. $\Box$

In real Hilbert spaces the proposition is false. For instance, in $\mathbb R^n$, antisymmetric matrices satisfy $\langle x|Ax \rangle =0$ for every $x \in \mathbb R^n$, but $A\neq 0$ in general.


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