# Trouble proving properties of the density matrix

So let's say you've got a Hilbert space that's $$n$$-dimensional with some vector $$\phi$$, such that $$\langle \phi | \phi \rangle = 1$$; let $$|\phi \rangle \langle \phi | = \rho$$. Also say we've got some Hermitian operator $$\hat{T}$$ that denotes some observable $$T$$. I'm trying to show 3 things.

First off, I want to prove that $$\langle T \rangle$$, the average (expected) value observation equals $$\mathrm{Tr}(\rho T)$$. I am very unsure as to how to do this, and have been trying to use the eigenbasis obtained from $$T$$ to express $$\phi$$, but am having a lot of difficulty simplifying and could use some guidance.

Secondly, I'm trying to show that $$\mathrm{Tr}(\rho) = 1$$. This seems like it has to do with the eigenvalues of $$\rho$$ but I seem to be missing some insight.

Finally, I want to prove that the adjoint of $$\rho$$ equals $$\rho$$; in other words that $$\rho$$ itself is a hermitian operator. This too I'm a little confused about; I'm trying to define and use a gram matrix for the inner product definition of the adjoint, with little success.

Would appreciate any help anyone could provide. Thanks!

• $\langle T\rangle = \mathrm{Tr} \rho\, T$ is a definition?! Feb 28, 2022 at 10:35

## 3 Answers

Let $$\mathcal{H}$$ be a finite-dimensional Hilbert space of dimension $$n$$. The trace of an operator $$T$$ on $$\mathcal{H}$$ is defined as $$\mathrm{Tr}(T) := \sum_{k=1}^n \langle e_k | T |e_k\rangle,$$ where $$\{e_k\}_{k=1}^n$$ is an orthonormal basis of $$\mathcal{H}$$. A theorem from linear algebra ensures that the sum on the r.h.s. of this equation is independent of the chosen basis. Using this definition, your questions should be easy to answer:

1. Expectation value of $$T$$ in the state $$\phi$$ with $$\rho= |\phi\rangle\langle\phi |$$: $$\mathrm{Tr}(\rho T) = \mathrm{Tr}(|\phi \rangle\langle\phi | T) = \sum_{k=1}^n \langle e_k | \phi \rangle \langle\phi | T |e_k\rangle = \langle \phi | T | \sum_{k=1}^n \langle e_k | \phi \rangle e_k \rangle = \langle \phi| T |\phi\rangle = \langle T \rangle_\phi.$$ I used Parseval's indentity: $$\phi = \sum_{k=1}^n \langle e_k | \phi \rangle e_k$$.

2. The trace of $$\rho$$: $$\mathrm{Tr}(\rho) = \sum_{k=1}^{n} \langle e_k| \phi\rangle \langle \phi|e_k\rangle = \sum_{k=1}^n |\langle e_k|\phi\rangle|^2 = \|\phi\|^2.$$ If $$|\phi\rangle$$ is normalised, then $$\| \phi\|^2 = \langle \phi | \phi \rangle = 1$$.

3. For hermiticity, we need to show that $$\langle \psi | \rho \varphi \rangle = \langle \rho \psi |\varphi \rangle$$ for all $$\psi,\varphi \in \mathcal{H}$$: $$\langle \psi | \rho \varphi \rangle = \langle \psi | \phi \rangle \langle \phi | \varphi \rangle = \langle \phi \langle \phi | \psi \rangle | \varphi \rangle = \langle \rho \psi | \varphi\rangle.$$ In the second step, I employed that the scalar product is anti-linear in the first component.

As a remark, let me mention that if $$\mathcal{H}$$ is infinite-dimensional (usually the case in quantum mechanics), then the trace of an arbitrary operator $$T$$ is not necessarily defined because the sum $$\sum_{k=1}^{\infty} \langle e_k|T|e_k\rangle$$ may not converge (see trace-class operators).

If $$\rho=|\phi\rangle\langle\phi|$$, then $$tr\{\rho\}=\langle\phi|\phi\rangle=1$$. So that part is easy.

Moreover, $$|\phi\rangle^{\dagger}=\langle\phi|$$ and visa versa. So the Hermitian property of $$\rho$$ is also easy to show.

The only thing is to compute the expectation value of the operator. For that you need to expand $$\rho$$ in terms of the eigenbasis of $$T$$. The expectation values will then be expressed in terms of the expansion coefficients.

Adding to the other answer, if you take it as known that $$\langle T \rangle = \langle \psi | T | \psi \rangle,$$ then you can easily rewrite this as $$\mathrm{Tr}(\langle \psi | T | \psi \rangle) = \mathrm{Tr}(| \psi \rangle \langle \psi | T) = \mathrm{Tr}(\rho T)$$ by cyclicity of trace (since $$x = \mathrm{Tr}(x)$$ when $$x$$ is just a number). Then in particular $$1 = \langle 1 \rangle = \mathrm{Tr}(\rho)$$.

• Note that, for $x\in \mathbb{C}$, $\mathrm{Tr}(x) = x\mathrm{Tr}(1) = nx$. Mar 12, 2022 at 16:18
• @Janik Yes. Here I view $x = \langle \psi | T | \psi \rangle$ as an operator $\mathbb{C} \to \mathbb{C}$, ie a $1 \times 1$ matrix, so $n = 1$. (In contrast, $| \psi \rangle \langle \psi | T$ is an operator from the Hilbert space to itself.) Mar 13, 2022 at 20:36