# Do the states in a decomposition $\rho=\sum_i p_i |\phi_i\rangle\!\langle \phi_i|$ need to be orthonormal?

On Wikipedia it says:

Let $$\mathcal H_S$$ be a finite-dimensional Hilbert space, and consider a generic (possibly mixed) quantum state $$\rho$$ defined on $$\mathcal H_S$$, and admitting a decomposition of the form $$\rho=\sum_i p_i|\phi_i\rangle \langle\phi_i|,$$ for a collection of (not necessarily mutually orthogonal) states $$|\phi_i\rangle\in\mathcal H_S$$, and coefficients $$p_i\ge 0$$ such that $$\sum_i p_i=1$$. Note that any quantum state can be written in such a way for some $$\{|\phi_i\rangle\}$$ and $$\{p_i\}$$.

1. What is the propery of this set $$\{|\phi_i\rangle\}$$?
• Should it be normalized $$\langle\phi_i|\phi_i\rangle=1$$?
• Should it be orthogonal $$\langle\phi_i|\phi_j\rangle=0$$ for $$i \neq j$$?
• Should it ortho-normal $$\langle\phi_i|\phi_j\rangle=\delta_{ij}$$?
1. Why does Wiki says that $$\{|\phi_i\rangle\}$$ not necessarily mutually orthogonal? If so, how can we assure that $$\sum_i p_i=1$$ in the non-orthogonal basis?

You seem to be confusing the definition for a density matrix and the definition for an orthonormal basis. An orthonormal basis has to have the properties you described. A density matrix is something else. The terms $$p_i$$ can be interpreted as the probability to have the state $$|\psi_i\rangle$$, and this state can be arbitrary.

An example for a 2 level system. $$p_1 = p_2 = 0.5$$ and $$|\psi_1\rangle = (1,0)$$, and $$|\psi_2\rangle = (1,1)/\sqrt 2$$. This gives the perfectly valid density matrix $$\rho = ((3/4, 1/4), (1/4, 1/4))$$, even though the two states are not orthogonal

• thanks, but your example seems strange, if not wrong. You see the trace of $\rho$ is $tr(\rho) \neq 1$. Jan 16 at 23:20
• @МаринаMarinaS I have edited my post and corrected the typo, I mean /4, not /2
– peep
Jan 16 at 23:23
• I will accept it as an answer if no better answer in a week! thanks! Jan 16 at 23:26

Remark: In order to avoid unnecessary mathematical complications, a finite dimensional Hilbert space (dimension $$N$$) will be assumed in the following discussion. The generalization to the infinite dimensional case is (in principle) straightforward.

Any convex linear combination $$\rho= \sum\limits_i p_i |\phi_i \rangle \langle \phi_i |, \quad p_i\ge 0, \, \sum\limits_i p_i=1 \qquad (1)$$ of an arbitrary set of pure states $$| \phi_i\rangle$$ (with $$\langle \phi_i |\phi_i\rangle=1$$) represented by the one-dimensional orthogonal projection operators $$P_i= | \phi_i \rangle \langle \phi_i |$$ describes the density operator $$\rho$$ of a possible mixed state. Note that the unit vectors vectors of the set $$\{ |\phi_i \rangle \}_i$$ do not have to be linearly independent nor do they have to form an orthonormal system. Even a convex linear combination in the form of an integral, $$\rho =\int\limits \! d\alpha \,f(\alpha) |\phi_\alpha \rangle \langle \phi_\alpha|, \qquad f(\alpha) \ge 0, \ \int\! d\alpha \, f(\alpha) =1, \, \langle \phi_\alpha |\phi_\alpha \rangle =1 \qquad (2)$$ with a continuous probability distribution $$f(\alpha)$$ describes a possible density operator.

It is obvious that the representation $$(1)$$ of a given density operator $$\rho$$ is not unique. It is therefore convenient to find a certain "standard representation" for the density operator. As can be seen from $$(1)$$, a density operator is a positive selfadjoint operator ($$\rho= \rho^\dagger \ge 0$$) normalized as $${\rm Tr} \, \rho =1$$. (Alternatively, these properties could be taken as the definition of a density operator.) As a consequence, the spectral theorem guarantees the existence of a complete orthonormal basis $$\{ | \psi_n \rangle \}_{n=1}^N$$ of eigenvectors, $$\rho |\psi_n \rangle = \rho_n |\psi_n \rangle$$ with nonnegative eigenvalues $$\rho_n$$ satisfying $$\sum\limits_{n=1}^N \rho_n=1$$. The spectral representation of the density operator is now given by $$\rho = \sum\limits_{n=1}^N \rho_n | \psi_n \rangle \langle \psi_n |, \quad \rho_n \ge 0, \, \langle \psi_n | \psi_m \rangle = \delta_{nm}, \, \sum\limits_{n=1}^N |\psi_n \rangle \langle \psi_n | = {\mathbf 1}_N. \qquad (3)$$ Note that the representation chosen for the density operator $$\rho$$ of a given mixed state does, of course, not affect the physics. The expectation value of an oberservable $$A$$ in the state described by the density matrix $$\rho$$, given by $$\langle A \rangle_\rho = {\rm Tr} (\rho A)$$, is independent of the chosen decomposition of $$\rho$$.

P.S.: It is a nice homework exercise to find the spectral representation of the density matrix in the example given in the answer by peep!

It feels like point (2) in your question is the one that really confuses you. In a sense, you seem to think that $$\operatorname{tr}\rho=1$$ demands the states $$|\psi_i\rangle$$ to form an orthonormal basis. Perhaps then the best way to approach this is to actually prove that $$\operatorname{tr}\rho = \sum_i p_i$$ regardless of what are the individual states in the decomposition.

Let therefore $$\{|\psi_i\rangle\}$$ be an arbitrary set of vectors, which we assume to be normalized (recall that in QM states are unit rays in a Hilbert space, so we should always normalize our kets). Define $$\rho=\sum_{i}p_i|\psi_i\rangle\langle \psi_i|$$

Let now $$|e_n\rangle$$ be an arbitrary orthonormal basis of the Hilbert space. The trace of $$\rho$$ is $$\operatorname{tr}\rho=\sum_n \langle e_n|\rho |e_n\rangle=\sum_{n,i}p_i\langle e_n|\psi_i\rangle \langle \psi_i|e_n\rangle$$

Now observe that rearranging and summing over $$n$$ first this is:

$$\operatorname{tr}\rho=\sum_i p_i \langle \psi_i|\left(\sum_n |e_n\rangle \langle e_n| \right)|\psi_i\rangle.$$

At this point we use the completeness of the orthogonal basis and we observe that

$$\operatorname{tr}\rho=\sum_i p_i \langle \psi_i|\psi_i\rangle = \sum_i p_i,$$

where the last equality follows from the fact that the $$|\psi_i\rangle$$ are normalized. As such $$\operatorname{tr}\rho=\sum_i p_i$$ without the need for the $$\{|\psi_i\rangle\}$$ to form an orthonormal basis.

Finally, the states themselves cannot assure $$\sum_i p_i=1$$. You can pick any set of (normalized) kets you like and construct any density matrix you like by assembling $$\rho = \sum_i p_i |\psi_i\rangle\langle \psi_i|$$ where $$\{p_i\}$$ are any numbers restricted just to obey $$\sum_i p_i=1$$.

If the kets were initially not normalized, you just need to normalize them before constructing the mixed state.

• "You can pick any set of kets you like" - no. I think this sentence only leads to confusion. Jan 21 at 18:34
• Why? If you pick any set of kets $\{|\psi_i\rangle\}$ and associated numbers $p_i$ obeying $\sum_i p_i =1$ then $\rho = \sum_i p_i |\psi_i\rangle \langle \psi_i|$ is a valid density operator defining a possible mixed state of the system.
– Gold
Jan 21 at 18:45
• No. Pick $p=1$ and $|\psi\rangle\langle \psi|$ not normalized. Jan 21 at 18:50
• But as I wrote, I'm assuming the kets to be normalized. Observe that $$\operatorname{tr}|\psi\rangle \langle \psi| = \sum_n \langle e_n|\psi\rangle \langle \psi|e_n\rangle = \langle \psi|\psi\rangle = 1.$$
– Gold
Jan 21 at 18:51
• Really, I know what you mean; I just wanted to mention that you cannot (as you correctly say also at the beginning of your post) take a set of arbitrary vectors. "You can pick any set of kets you like and construct any density matrix you like" however, leaves, at least IMHO, the wrong expression, which might cause some confusion. Jan 21 at 18:53