If $\rho=\sum_{i}p_{i}|\psi_{i}\rangle\langle \psi_{i}|$, this ensemble doesn't require $\langle \psi_{i}|\psi_{j}\rangle$=0. Given that $\rho$ is positive semi-definite, by the spectral theorem it can be expressed in diagonal form $$\rho=UDU^{\dagger}, D=\sum_{i}\lambda_{i}|i\rangle\langle i|$$

However, despite having used this notation for a while now, I still find myself confused by the notation of the spectral theorem. Specifically the role the unitaries play. The above states that, $$U^{\dagger}\rho U=U^{\dagger}UDU^{\dagger}U=D$$ which then given $\rho$ in it's diagonal form. However, it can also easily be diagonalised just by calculation of its eigenvalues and eigenvectors, and then re-expression in that basis. Moreover, this just looks like the unitary transformation of $\rho$, which obviosuly isn't going to be the same state. So what are these unitaries then, just the identity operators expanded in the eigenbasis? Or am I meant to interpret them as unitaries whose columns are composed of the states of current basis of $\rho$ expanded in the eigenbasis, ie, not mapping them to their image in another basis, but to themselves expressed in another basis, essentially achieving the same action as the identity?

Edit: It has been pointed out to me that the paragraph wherein I ask my questions is too vague or confusing. So let me try and rephrase. I have a density operator. I have it expressed in matrix form in some basis. I want to change said basis so that it achieves its diagonal form. How would I actually express this as the action of operations on said density operator, given that any such action would just lead to another density operator that is unitarily related, but with a different spectrum? The only way I can see is to rewrite it's entries $|\psi_{i}\rangle = U|\phi_{i}\rangle$ which, to my mind anyway, isn't really the same thing as the action of a unitary operator on $\rho$. To clarify, I am not confused about the action of basis change, but only its representation outside just using the identiy to acieve it, ie $|\psi\rangle=\sum_{i}\langle \phi_{i}|\psi_{i}\rangle |\phi_{i}\rangle$

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – glS
    Commented Sep 29, 2021 at 15:38

3 Answers 3


OK, honestly I did not follow the later part of your post (where you asked the questions) -- it was too confusing. But I suspect that your confusion arises because you were trying to go between abstract bra-ket notation and matrix notation (which entails choosing some basis to express the operators in).

Maybe this will help.

Let $$ \hat{\rho} = \sum_i p_i |\psi_i \rangle \langle \psi_i | $$ be a density operator (I do mean "density matrix", but the name is a bit of a misnomer) where each $|\psi_i \rangle$ is a normalized state of the Hilbert space, but need not be mutually orthogonal.

Since $\hat{\rho}$ is Hermitian, spectral theory says there exists an orthogonal basis $\{ |n\rangle \}_{n=1}^{d}$ (I'm assuming a finite-dimensional space for simplicity), i.e. $\langle n|m\rangle = \delta_{nm}$, which comprise the eigenvectors of $\hat{\rho}$. That is, $\hat{\rho}$ is expressible as \begin{align} \hat{\rho} = \sum_{n=1}^d \lambda_n|n\rangle \langle n| \end{align} where $\lambda_n$ is the (real) eigenvalue of $\hat{\rho}$ associated with the eigenvector $|n\rangle$.

OK, in principle, that is all there is to it.

Qn: Wait, what about all these unitaries $U$ that I oftentimes see quoted? How/where do they come in?

Answer: They arise only if we want to express the operators/spectral theory in some particular matrix representation $\rho$ of the operator $\hat{\rho}$. What I mean is this: Let $\{ |\phi_i\rangle \}_{i=1}^{d}$ be some orthogonal basis of states. Then $$ \rho_{ij} \equiv \langle \phi_i | \hat{\rho} |\phi_j \rangle =\sum_{n=1}^d \langle \phi_i | n\rangle \lambda_n \langle n|\phi_j\rangle = \sum_{n,m=1}^{d} \langle \phi_i|n\rangle (\lambda_n \delta_{nm})\langle m|\phi_j\rangle $$ Since $\{|n\rangle\}, \{|\phi_i\rangle\}$ are both orthogonal basis sets, we find $\langle \phi_i|n\rangle = U_{i,n}$ where $U$ is some $d\times d$ unitary matrix and $U_{i,n}$ is its $(i,n)$ entry. Now $\lambda_n\delta_{nm}$ is interpretable as the $(n,m)$-entry of a matrix $D$, which turns out to be diagonal, and so we can express the above very compactly in terms of matrices and their multiplication: $$ \rho = UDU^\dagger. $$ The spectral theory for matrices thus amounts to saying: should we insist on expressing the abstract operator $\hat{\rho}$ as a matrix $\rho$ in a certain basis, then there exists a unitary matrix $U$ such that $\rho$ can be transformed into a diagonal matrix. Note importantly that a representation $\rho$ (in terms of a table of numbers) of the operator $\hat{\rho}$ depends on the choice of basis, but the operator's action in the abstract Hilbert space is invariant. In particular, if we picked $|\phi_i\rangle = |n\rangle$ the eigenbasis, then $\rho$ is the diagonal matrix $D$ and $U$ is the identity matrix.

Additional remarks in anticipation of potential remnant confusion.

  1. The following is true: Let $\hat{U}$ be any unitary operator acting on the Hilbert space. Then $$ \hat{\rho}=\sum_{n=1}^d \lambda_n \hat{U}\hat{U}^\dagger|n\rangle \langle n|\hat{U} \hat{U}^\dagger \equiv \sum_{n=1}^d \lambda_n \hat{U} |\tilde{n}\rangle \langle{\tilde{n}}|\hat{U}^\dagger = \hat{U}\left( \sum_{n=1}^d \lambda_n |\tilde{n}\rangle \langle \tilde{n} |\right)\hat{U}^\dagger $$ where I defined $|\tilde{n}\rangle = \hat{U}^\dagger|n\rangle$ (which also forms an orthogonal basis). But here I haven't done anything, $\hat{\rho}$ is still the same operator as before.

  2. Let $\hat{U}$ be any unitary operator. Then $$ \hat{U} \hat{\rho} \hat{U}^\dagger \equiv \hat{\tilde{\rho}} $$ defines a different operator than $\hat{\rho}$, but which is unitarily related. So, they do not have the same spectral decomposition (though it is clear they are related).

  3. The following has no meaning: $$ U \hat{\rho} U^\dagger $$ (matrix of numbers multiplying an operator..?), nor $$ \hat{U} \rho \hat{U}^\dagger $$ (operator acting on a matrix of numbers...?), while the following has meaning: $$ U \rho U^\dagger $$ (matrix multiplication), and $$ \hat{U} \hat{\rho} \hat{U}^\dagger $$ (composition of operators). Note $U\rho U^\dagger$ is the matrix representation of $\hat{U}\hat{\rho}\hat{U}^\dagger$, in some basis.

  4. In practice no one but the most fastidious keep the hats for operators, and the symbol $\rho$ is oftentimes in an abuse of notation used for both the abstract density operator and the corresponding density matrix expressed in some basis. To make matters worse, the basis used is often suppressed, though is often assumed to mean the computational basis (for qubits).

  • $\begingroup$ Ok I think this almost completely addresses my question. You are correct, this confusion is arising from notation, which is almost exclusively used to denote an active unitary transformation, and as such alters the spectrum. Given $$ \hat{\rho} = \sum_i p_i |\psi_i \rangle \langle \psi_i | $$ if I was actually trying to express the basis change notationally, without the use of the identity operator, would I just take the entries of the matrix, express them as $|\psi_{i}\rangle=U|n\rangle$ where n is the basis in which it is diagonal in? $\endgroup$ Commented Sep 16, 2021 at 10:47

TL;DR: Active and passive transformations

The dichotomy between the two types of unitary transformations is real and is an example of a division of transformations into active and passive types. This duality is inherent to any use of coordinates and arises from the fact that there is a degree of arbitrariness in the way coordinates are assigned to objects they label.

Example: translation in Euclidean space

For an example of this phenomenon in a more usual setting, consider the three dimensional Euclidean space $\mathbb{E}^3$. By choosing an arbitrary point to serve as the origin and an arbitrary set of mutually perpendicular directions for the axes, we can create a coordinate system which we can use to assign triples of real numbers in $\mathbb{R}^3$ to the points in $\mathbb{E}^3$.

Consider what happens if we translate an object in $\mathbb{E}^3$ by one unit of distance in the positive $x$ direction. The object changes its position from $(x, y, z)$ to $(x + 1, y, z)$. Now, consider what happens if we instead leave the object alone and move our coordinate system by one unit of distance in the negative $x$ direction. The object's new coordinates are once again $(x + 1, y, z)$.

This example demonstrates that a change in numerical coordinates used to describe an object may or may not signify a change in the state of the object. If the change represents an active transformation then the object's state has changed and the new coordinates describe the new state in the same unchanged coordinate system. On the other hand, if the change represents a passive transformation then the object's state remains unchanged, but the coordinate system has changed. In this case, the new coordinates describe the old unchanged state of the object in the new coordinate system.

Example: diagonalization of a density matrix

Elements of a density matrix can be thought of as coordinates that we assign to linear operators on an $n$-dimensional Hilbert space so that we can represent them as matrices in $\mathbb{C}^{n\times n}$. Such a representation is highly convenient, but comes with the caveat demonstrated above in the case of Euclidean space.

Suppose we know that two density matrices $\rho$ and $\sigma$ are related by the equation

$$ \sigma = U\rho U^\dagger\tag1 $$

where $U$ is a unitary matrix. This equation admits two interpretations. In the first, $U$ is an active transformation, e.g. describing the action of a quantum gate. In this case, $\rho$ and $\sigma$ are different states described in the same basis. In the second interpretation, $U$ is a passive transformation, i.e. it describes a basis change. In this case, $\rho$ and $\sigma$ are two descriptions of the same quantum state given in two different bases.

Most of the time, the $U$ in the diagonalization of a density matrix $\rho$ is interpreted as a (passive) transformation that changes the basis to one in which $\rho$ has the diagonal form. However, given a state $\rho$ it is of course also possible to intentionally choose the (active) evolution operator $U$ so that $U\rho U^\dagger$ is diagonal.


The existence of the two types of transformations highlights the fact that a quantum state and its mathematical description using a density matrix are two distinct objects and the mapping between them is mediated by the choice of basis.

  1. A linear operator $A:V\to W$ represents a transformation in some underlying vector space (or more generally, from a vector space to a different one). Let's stick to the case of finite-dimensional spaces for simplicity. Such an operator is not the same as a matrix. A matrix is a way to represent the operator $A$ with respect to a given pair of bases. Given bases $\{v_i\}_i\subset V$ and $\{w_i\}_i\subset W$, you can write the "matrix elements of $A$" as $$A_{ij} \equiv w_i^\dagger A v_j.$$

  2. An equivalent way to write this is using dyadic notation. Assuming the bases to be orthonormal, we have $$A = \sum_{ij} A_{ij} w_i v_j^\dagger,$$ where $w_i v_j^\dagger$ is defined as the linear operator $V\ni x\mapsto w_i \langle v_j,x\rangle$. Note that this is what you usually write in bra-ket notation as $A=\sum_{ij} A_{ij} |w_i\rangle\!\langle v_j|$. Note that there are (infinitely) many ways to write an operator $A$ in this way. You can choose any orthonormal basis for the representation. If the operator turns out to be normal (and $V=W$), then there is a basis with respect to which you can write $$A = \sum_i \lambda_i v_iv_i^\dagger, \tag3$$ for some eigenvalues $\lambda_i\in\mathbb C$. In fact, an operator is normal iff such a representation exists.

  3. Given any pair of orthonormal bases $\{v_i\},\{v_i'\}\subset V$, there is always a unitary operation $U$ connecting them, i.e. such that $Uv_i=v_i'$. This means that a convenient way to express an operator is using a unitary encoding the basis vectors used in the expressions above. For example, you can write (3) as $A=VDV^\dagger$, with $V$ the unitary matrix whose $i$-th column is the vector $v_i$, and $D_{ii}\equiv\lambda_i$.

  4. You can also try to "extract" the diagonal representation of a given normal operator by applying the inverse of the unitary operations above. For example, if $A=VDV^\dagger$, then $V^\dagger A V$ is diagonal.

  5. You can also think of something like $V A V^\dagger$ as representing the action of $A$ in a "rotated basis". By this I mean that you can wonder how the action of $A$ looks like if all vectors are rotated. If you switch from a basis $\{u_i\}$ to a basis $\{v_i\}$, with the two bases connected by some unitary operator $V$, $V u_i=v_i$, then $$\langle v_i, A v_j\rangle = \langle u_i, V^\dagger A V u_j\rangle,$$ and thus $V^\dagger AV$ can be interpreted as the way $A$ acts in the rotated basis.

  • $\begingroup$ For $V^{\dagger}AV$, the columns of the two unitaries in this case wouldn't take on the same meaning as that in the unitary transformation, yes? As in in this case, the columns of $V^{\dagger}$ would be the basis of A expanded in the target basis to which you want to express A in? $\endgroup$ Commented Sep 17, 2021 at 14:05
  • $\begingroup$ @GaussStrife in what context? If $V$ is unitary, both its rows and its columns form an orthonormal basis. You can understand $V$ as saying "change from computational basis to the basis formed by the columns of $V$", or more generally from some basis to some other basis related by $V$. If you write $V=\sum_i v_i e_i^\dagger$, then $V^\dagger AV$ acts in the basis $\{e_i\}$ the same way $A$ acts in the basis $\{v_i\}$. You can think of $V^\dagger AV$ as the representation of $A$ in the transformed basis. $\endgroup$
    – glS
    Commented Sep 17, 2021 at 14:13
  • $\begingroup$ Yes, I understand that the columns and rows form basis states. My main point of confusion is with this notation, and I can't seem to get a clear answer on it, or maybe what I am asking doesn't make sense? If I apply a unitary transformation, I take one basis state to another, and the entries of a column will form the expansion of the other basis state in my current one. If I do a change of basis, I take the column to represent my current basis state, and it's expansion in my target one. The basis of the columns or rows changes, depending on which I am doing, active or passive. Is that correct? $\endgroup$ Commented Sep 17, 2021 at 14:20
  • $\begingroup$ @GaussStrife I guess it just depends how you choose to describe things. If $u_i=U e_i$, and $e_i$ is the canonical basis, then the columns of $U$ are the vectors $u_i$ (assuming you are representing $U$ as a matrix in the standard way). So the "entries of a column" would be the components of the vectors $u_i$ I guess? These are the coefficients of the decomposition of $u_i$ in the canonical basis that was chosen. Is that what you mean with "change of basis"? If you meant instead $U^\dagger AU$, then the columns of $U$ are the basis wrt which you are representing $A$ as a matrix $\endgroup$
    – glS
    Commented Sep 17, 2021 at 14:49
  • $\begingroup$ In the case of $u_{i}=Ue_{i}$, then yes, the columns would represent $u_{i}$, and the entries would be the coefficients weighting each $e_{i}$ when $u_{i}$ is expanded in said basis, and when the matrix multiplication is carried out, you would simply use them as coefficients for $\{e_{i}\}$. For $U^{\dagger}AU$, since this is also a basis change, the columns, unlike in a unitary transformation, are the same basis states, not the image, and when I perform the multiplication, I associate with the results the basis vectors of the basis I wish to express A in, correct? $\endgroup$ Commented Sep 20, 2021 at 11:11

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