Let $H$ be a Hermitian operator, so it can be written as $$H=\lambda_1P_1+\lambda_2P_2........\lambda_kP_k,$$ where $\lambda_i$ are eigen values and $P_i$ corresponding projector operators for the corresponding eigenspace where $1 \leq i \leq k$. Also if $i \ne j$, $\;$ $P_iP_j=0$.
I am not sure about the above but if its true then I am struggling with the question that the projector operator for a space can be written in infinitely many ways in bra-ket outer product form (by choosing any set of orthonormal basis for that space). So all the $P_i$ above independently can be written in many ways, thus $H$ can be written in many ways, are all these representations for the same $H$, am I missing something?

Let the eigen space corresponding to eigen value $\lambda_i$ have dimension $d_i$. Also let an orthonormal basis corresponding to it be ${|j_{i1}\rangle,|j_{i2}\rangle,.....|j_{id_i}\rangle}$. Thus $H$ can be written as $$H=\sum_{i=1}^{i=k} \lambda_i \sum_{l=1}^{l=d_i}|j_{il}\rangle \langle j_{il}|$$ Now if I take a different orthonormal basis for each eigen space then also it is the same operator, ie say $$H^{'}=\sum_{i=1}^{i=k} \lambda_i \sum_{l=1}^{l=d_i}|m_{il}\rangle \langle m_{il}|$$ $$H=H^{'}......(1)$$ where ${|m_{i1}\rangle,|m_{i2}\rangle,.....|m_{id_i}\rangle}$ is the new orthonormal basis for space corresponding to eigen value $\lambda_i$. But one orthonormal basis can be obtained from another by some unitary transformation thus $$H^{'}=UHU^{\dagger}.....(2)$$ for some unitary operator $U$. But $(1)$ and $(2)$ are contradictory, because using both we get $H=UHU^{\dagger}$, but $H=H$.Where am I going wrong ?

  • $\begingroup$ Just as 0=1+(-1), 0=2+(-2), 0=2+(-1)+(-1), etc.. Should this be more irritating in this case? -- Otherwise, if that's not what you are up to you should make your question more precise. $\endgroup$ Commented Jan 13, 2015 at 9:51
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    $\begingroup$ My suggestion would be that you use formulas to illustrate how exactly you mean $H$ can be rewritten, i.e., in which "outer product form". There could be different interpretations of that, such as ket-bra outer products, products of an operator basis (e.g. Paulis), etc.. Using formulas is always a good idea to make things more clear. $\endgroup$ Commented Jan 13, 2015 at 10:30
  • $\begingroup$ @NorbertSchuch thanks I tried to make the changes. $\endgroup$ Commented Jan 13, 2015 at 10:49
  • $\begingroup$ Thanks, it is already much more clear now. But why are (1) and (2) contradictory? Can you elaborate on that? (Ideally using formulas!) --- Note for instance that for $H=I$ (the identity matrix), $H=UHU^\dagger$ for any unitary $U$. $\endgroup$ Commented Jan 13, 2015 at 11:05
  • $\begingroup$ yes, but it will not be true in general for any hermitian operator $H \neq I$ $\endgroup$ Commented Jan 13, 2015 at 11:10

3 Answers 3


I) Let us for simplicity assume that the Hilbert space $H$ is finite-dimensional. Let $A:H\to H$ be a normal operator [and hence orthogonally diagonalizable] with eigenvalues

$${\rm Spec}(A)~=~\{\lambda_1, \ldots, \lambda_n\}.$$

We can uniquely decompose the Hilbert space

$$H~=~\oplus_{i=1}^n H_i,$$

in orthogonal eigenspaces

$$H_i~:=~ {\rm ker}(A-\lambda_i{\bf 1}_H) ~\subseteq~H$$

for $A$. There exist unique corresponding projection operators $P_i:H\to H$, so that

$$ P_i^2~=~P_i, \qquad P_i^{\dagger}~=~P_i,\qquad {\rm Im}(P_i)=H_i.$$

Then we may uniquely write $$A~=~\sum_{i=1}^n \lambda_iP_i. $$

II) We see that it is enough to analyze and restrict attention to a single eigenspace $H_i$ for a fixed index $i\in\{1,\ldots,n\}$. Let us therefore drop the index $i$ from the notation, i.e. call $H_i$ for $H$. In other words, we may assume without loss of generality that the operator $A:H\to H$ is proportional to the identity

$$ A~=~\lambda{\bf 1}_H. $$

We can now pick an orthonormal basis $\left(|e_k\rangle\right)_k$ for $H$ in infinitely many ways. However the "outer product representation" of the unit operator

$${\bf 1}_H~=~\sum_k |e_k\rangle\otimes \langle e_k|$$

will be independent of the choice of orthonormal basis $\left(|e_k\rangle\right)_k$. Moreover it will commute with any other operator. In particular, unitary operators.

  • $\begingroup$ so is $H=UHU^{\dagger}$ a contradiction ? I am still not clear $\endgroup$ Commented Jan 13, 2015 at 13:01
  • $\begingroup$ @sasha : No, it is not a contradiction, because $A$ [=what you call $H$] and $U$ are on block form, and within each block $A$ is proportional to the identity, and hence commute with $U$. $\endgroup$
    – Qmechanic
    Commented Jan 13, 2015 at 13:06
  • $\begingroup$ So does it mean that in a hermitian operator has a unique bra-ket representation in diagonal form ? $\endgroup$ Commented Jan 13, 2015 at 14:47
  • $\begingroup$ @sasha : There are infinitely many choices of orthonormal basis within each eigenspace. Even if the eigenspaces are non-degenerate, there is a choice of phases on each eigenvector. $\endgroup$
    – Qmechanic
    Commented Jan 13, 2015 at 14:55
  • $\begingroup$ what do you mean by the last line "Moreover it will commute with any other operator. In particular, unitary operators" ? $\endgroup$ Commented Jan 14, 2015 at 7:31

When you give the form of an operator with respect to a certain basis, it would be good to explicitly state which basis you are using. The fact that you get $H' = UHU^*$ is giving you a way to represent the same operator in a different basis. Let's say you have bases $B$ and $B'$ for a Hilbert space. What you have is the operator $H$ in the basis $B$, and we shall denote this by $[H]_B$. Now the same operator in the basis $B'$ is related to $B$ by a unitary transformation, so $$[H]_{B'} = U[H]_BU^*.$$ So an operator must not be confused with its representation, because operator algebras can be given in the abstract through, say, generators and relations alone, and there is, in principle, no representation Hilbert space.


You are right. Think of the limit case of the Heritian identity operator $I$, it can be written in infinitely many ways, one for each Hilbertian basis of the Hilbert space.

ADDENDUM. Referring to the EDIT, if you explicitly perform computations you see that $U$ commutes with $H$ and thus it is not effective as it must be.

  • $\begingroup$ I got it. But still I am getting that the two different representations will be linked by some unitary and will not be the same operator. $\endgroup$ Commented Jan 13, 2015 at 9:26
  • $\begingroup$ The operator is unique, the representation is not. $\endgroup$ Commented Jan 13, 2015 at 9:43
  • $\begingroup$ I made an EDIT to make my question clear $\endgroup$ Commented Jan 13, 2015 at 10:50
  • $\begingroup$ so does that mean it is true in general for any hermitian operator that H=UHU† ? $\endgroup$ Commented Jan 14, 2015 at 9:54
  • $\begingroup$ nope, the unitary that performs what you're asking in the EDIT leaves every eigenspace invariant, if you restrict both $U$ and $H$ to an eigenspace, then $H$ becomes a multiple of the identity and therefore the restriction of $U$ commutes with it. Since this is true for every eigenspace, it follows that this particular $U$ commutes with the operator. Any other unitaries need not preserve the eigenspaces. $\endgroup$
    – Phoenix87
    Commented Jan 14, 2015 at 10:04

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