Equality of operators [closed]

$$\def\bra#1{\langle#1|} \def\ket#1{|#1\rangle}$$I'm trying to prove this little result. I'm sure its proof is very simple, but I'm stuck. Could you please give me a hint? Thanks.

Let $$\bra{\psi}A\ket{\psi} = \bra{\psi}B\ket{\psi}$$ for all $$\psi$$. Prove that $$A = B$$, in the sense that $$\bra{\phi_{1}}A\ket{\phi_{2}} = \bra{\phi_{1}}B\ket{\phi_{2}}$$ for all $$\phi_{1}$$ and $$\phi_{2}$$.

By direct inspection, using right linearity and left antilinearity $$4 \langle x| T y\rangle = \langle x+y| T (x+y)\rangle - \langle x-y| T (x-y)\rangle -i\langle x+iy| T (x+iy)\rangle + i \langle x-iy| T (x-iy)\rangle.$$ This identity implies that if $\langle z|T z\rangle =0$ for every $z$, then $\langle x| T y \rangle =0$ for every $x,y$. Thus, assuming $x=T y$, we conclude that $||T y||^2=0$ for every $y$, that means $T =0$. Using linearity of the scalar product, defining $A-B=T$, this result implies the initial thesis.

Hermiticity does not play any role here. What matters is that the scalar product is antilinear in the left entry and linear in the other, and finally that it gives rise to a norm.

REMARK. It is interesting to note that this result does not work for real Hilbert spaces (where the scalar product is real symmetric): think of antisymmetric matrices in $\mathbb R^n$. Instead, it still holds in quaternionic Hilbert spaces.

$$\def\bra#1{\langle#1|} \def\ket#1{|#1\rangle}$$You need to use linearity. Apply the assumption $$\bra{\psi}A\ket{\psi} = \bra{\psi}B\ket{\psi}$$ to the case $$\psi=\phi_1+\phi_2$$ and expand using the linearity of the inner product in the first term and the conjugate symmetry. Then cancel terms you know to be equal from the assumption.

You'll should then arrive at the expression (I'm writing the inner products out in full rather than bras and kets):

$$\langle\phi_1,\,\Delta^\dagger\,\phi_2\rangle +\left(\langle\phi_1,\,\Delta\,\phi_2\rangle\right)^\ast = 0\tag{1}$$

where $$\Delta = A-B$$ and with $$\dagger$$ standing for the adjoint operator and $$\ast$$ standing for the complex conjugate (one can use the same symbol for both, because their domains differentiate them, but I've used the two symbols for easier reading). Now apply first-argument-linearity to the above expression to the case where we use $$i\,\phi_1$$ instead of $$\phi_1$$. You should get

$$\langle\phi_1,\,\Delta^\dagger\,\phi_2\rangle -\left(\langle\phi_1,\,\Delta\,\phi_2\rangle\right)^\ast = 0\tag{2}$$

and (1) and (2) will get you what you need.

• You can't assume the operator is Hermitian- the result is general. – Chris Mar 3 '18 at 2:51
• I'm not sure I follow. Regardless, even if it's not completely general, it is at least much more general than just "Hermitian." With complex fields, it's easy to prove that if the result holds for Hermitian operators, it also holds for anti-Hermitian operators, and from there to any operator. – Chris Mar 3 '18 at 3:56
• @Chris Yes, you are indeed good in the complex case, as I've worked out above, but the stated property is not a general property of billinear forms. In general, you need to assume symmetry of the form to make it work. – Selene Routley Mar 3 '18 at 5:10
• I suspect this is for an elementary quantum mechanics class, so a proof for the complex case is probably all they were looking for. This is a bit more than a "hint" now, though. – Chris Mar 3 '18 at 5:19

I think the answer from WetSavannaAnimal aka Rod Vance is basically correct, although in the statement of the problem there is no assumption about $A$ and $B$ being Hermitian, let alone the inner product is over the field of real numbers.

Nonetheless, one can deal with a complex-valued inner product by successively taking $\psi = \phi_1 + \phi_2$ and $\psi = \phi_1 + i \phi_2$, which yields a condition on the real and imaginary part of $\langle\phi_1|A|\phi_2\rangle$, respectively.

As for the Hermiticity of $A$ and $B$, I presume it is essential in order for the assertion to be proven.

• You can't assume the operator is Hermitian- the result is general. – Chris Mar 3 '18 at 2:51