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I was attending a Quantum Mechanics lecture when the instructor casually mentioned the following theorem:

$\langle \alpha \rvert A \rvert \alpha \rangle = 0 ~\forall \alpha \implies A=0$, where $A$ is an operator and $\rvert\alpha\rangle$ is an arbitrary ket in the complex Hilbert space.

I have always assumed that the above theorem was 'obvious', but on second thought, it doesn't seem to be easy or trivial to prove. I tried looking at various sources for the theorem, but it seems to be surprisingly difficult to find this theorem or proof anywhere.

I would be very glad if someone would point me towards the proof of the theorem, and provide a small outline of it if possible.

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  • $\begingroup$ en.wikipedia.org/wiki/Operator_norm $\endgroup$ Sep 7 '17 at 0:09
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    $\begingroup$ @CountIblis OP did not say $A$ needs to be bounded. $\endgroup$
    – Ryan Unger
    Sep 7 '17 at 0:12
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    $\begingroup$ @CountIblis $\langle \alpha \vert A \lvert \alpha \rangle$ is not the norm of $A$ - that would be $\langle \alpha \vert A^\dagger A \vert \alpha \rangle$. $\endgroup$
    – ACuriousMind
    Sep 7 '17 at 0:21
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    $\begingroup$ For example: if you consider the harmonic oscillator problem, its Fock space and the usual orthonormal basis, then the property you are describing is shared by the creation-anihillation operators: $\langle n \rvert \alpha \rvert n \rangle=\langle n \rvert \alpha^\dagger \rvert n \rangle=0$ $\forall n$. But these are certainly not equal to zero! $\endgroup$
    – QCrypt
    Sep 7 '17 at 2:04
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    $\begingroup$ @QCrypt The property described is supposed to hold for all vectors, not only for a basis. The c/a operators do not fulfill this for all operators, e.g. $\langle \alpha \vert a \vert \alpha \rangle$ is non-zero for $\lvert \alpha\rangle = \lvert 0\rangle + \lvert 1 \rangle$, so they are not a counterexample. $\endgroup$
    – ACuriousMind
    Sep 7 '17 at 9:31
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Pick any orthonormal basis $\lvert \psi_i\rangle$ of our Hilbert space. Then $\langle \psi_i\vert A \vert \psi_i \rangle = 0$ for all $i$ by assumption, and for $\lvert \phi_{ij}(a,b)\rangle := a\lvert \psi_i\rangle + b\lvert \psi_j\rangle$ we find $$ \langle \phi_{ij} \vert A \vert \phi_{ij}\rangle = a^\ast b \langle \psi_i \vert A \vert \psi_j\rangle + ab^\ast \langle \psi_j \vert A \vert \psi_i\rangle = 0,$$ which for $a,b = 1$ implies $$ \langle \psi_i \vert A \vert \psi_j \rangle = - (\langle \psi_j \vert A \vert \psi_i\rangle )^\dagger = - \langle \psi_i \vert A^\dagger \vert \psi_j \rangle $$ which means that $A = -A^\dagger$, i.e. $A$ is anti-Hermitian. Since anti-Hermitian operators are in particular normal, they are diagonalizable by the spectral theorem, and therefore $\langle \alpha \vert A \vert \alpha\rangle = 0$ means that all eigenvalues are 0, i.e. the diagonalization of $A$ is the zero matrix, which also means $A = 0$.

Note that the application of the spectral theorem relies on the space being a complex vector space, and that the assertion would be false over a real vector space - $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ is a counterexample on $\mathbb{R}^2$ (but not on $\mathbb{C}^2$, since its expectations values do not vanish for all vectors there).

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    $\begingroup$ Why $\langle \psi_i \vert A \vert \psi_j \rangle = - \langle \psi_i \vert A^\dagger \vert \psi_j \rangle$ implies that $A = -A^\dagger$ ? $\endgroup$
    – QCrypt
    Sep 7 '17 at 0:07
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    $\begingroup$ @QCrypt If $\langle \psi_i \vert A \vert \psi_j\rangle = \langle \psi_i \vert B \rvert \psi_j\rangle$ for all $i,j$ in a basis $\psi$, then $A=B$. In other words, operators whose matrix elements are the same are the same operators. $\endgroup$
    – ACuriousMind
    Sep 7 '17 at 0:18
  • $\begingroup$ But isn't this what the OP is all about ? $\endgroup$
    – QCrypt
    Sep 7 '17 at 0:20
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    $\begingroup$ This proof implicitly requires $A^*$ to be defined everywhere. This is equivalent to $A$ being closed. Since $A$ is defined everywhere (implicit from OP), the closed graph theorem implies $A$ must be bounded. So this proof requires the additional hypothesis that $A$ is bounded. $\endgroup$
    – Ryan Unger
    Sep 7 '17 at 0:32
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    $\begingroup$ Well, generally $\langle \psi_i \vert A \vert \psi_j\rangle = \langle \psi_i \vert B \rvert \psi_j\rangle$ for all $i,j$ in some basis $\psi$, implies that $A-B$ is contained in the kernel of the specific representation (and thus its action is zero). Not that it is necessarilly equal to $0$ (as an abstract operator). But maybe this is enough for the OP's meaning. $\endgroup$
    – QCrypt
    Sep 7 '17 at 0:34
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I will give a semi mathematical point of view, and a completely intuitive point of view.

Mathematical(ish) POV

We have $$\langle \alpha |A|\alpha\rangle = 0 \ \ \forall |\alpha\rangle$$

Switching to the matrix view of operators we can say

$$A = \sum \limits_{i,j} a_{ij} |a_i\rangle\langle a_j|$$

Thus $$|a_n|^2 ||\langle \alpha|a_n\rangle||^2 = 0 $$

Besides the trivial case the inner product will not be 0, but we also have the condition from the axioms of QM that $|a_n|^2 \geq 0 $. Thus the only way for this to be true for an arbitrary $|\alpha\rangle$ is for the coefficients to be $0$ which is equivalent to saying $A = 0$.

(Another way of seeing this is since $|\alpha \rangle$ is arbitrary, choose it to be one of the eigenvectors, $|a_n\rangle$. Each time you do this the only solution is that the coefficient is 0. Repeat $\forall n$ and you'll see that $A$ must be the $0$ operator.

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  • $\begingroup$ In your second equation - don't we first have to prove that $A$ is diagonalizable? $\endgroup$ Mar 28 at 1:53
  • $\begingroup$ I don't think so. The matrix view just requires any linearly dependent set of vectors that span the space. Although, the fact that A is hermitian by the postulates of QM guarantees that it is diagonilizable. $\endgroup$
    – Jake Rose
    Mar 28 at 23:28
  • $\begingroup$ If $A$ were not diagonalizable, then we could not write it that way :) I was being nice but you do have to show this first. Otherwise in general we can only write $A = \sum_{i,j}a_{ij} |a_i \rangle |a_j \rangle \langle a_i | \langle a_j|$. $\endgroup$ Mar 29 at 14:27
  • $\begingroup$ It depends on the level which you're attacking it from. From an undergrad physics perspective my answer was sufficient. The postulates of QM guarantee the diagonaliszability, and often it is simply stated that you can do this. $\endgroup$
    – Jake Rose
    Mar 31 at 13:15
  • $\begingroup$ Oh apologies. I just realised it says it's summed over n. When it should be a double index. I don't think your equation is right. I'm not satisfied with my answer but I'lld edit it when I get a chance. $\endgroup$
    – Jake Rose
    Mar 31 at 13:17
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I am a little late on this, the correct mathematical theorem (in the sense implied by @Ryan Unger in the comments) states:

Theorem: If $A:D(A)\subsetneq \mathcal{H}\rightarrow \mathcal{R}(A)\subset\mathcal{H}$ is a linear operator in a complex separable Hilbert space, then we have the following equivalence: $$ \forall \psi\in D(A), \langle \psi,A\psi\rangle =0 \Rightarrow A\equiv \hat{0}_{\mathcal{H}} \Leftrightarrow D(A)\text{ is dense everywhere in } \mathcal{H} $$

Ideas for a proof: The vanishing of the expectation values of the domain means that $\left(D(A)\right)^{\perp}= 0$. We actually have that $\overline{D(A)}^{\perp} = 0$, because one can show that $\left(D(A)\right)^{\perp}=\overline{D(A)}^{\perp}$ by continuity of the scalar product, therefore $\overline{D(A)}=\mathcal{H}$, in other words $A$ is densely defined.

As stated, the result in the quote from the OP is valid for operators defined everywhere on the Hilbert space and is nothing but the trivial assertion that follows from $\mathcal{H}^\perp = 0$, namely that $\mathcal{R}(A)=0\Leftrightarrow A \equiv 0$.

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