0
$\begingroup$

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!

$\endgroup$
1
  • $\begingroup$ $\langle T\rangle = \mathrm{Tr} \rho\, T$ is a definition?! $\endgroup$ Feb 28, 2022 at 10:35

3 Answers 3

1
$\begingroup$

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).

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
-1
$\begingroup$

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 $1 = \langle 1 \rangle = \mathrm{Tr}(\rho)$.

$\endgroup$
2
  • 1
    $\begingroup$ Note that, for $x\in \mathbb{C}$, $\mathrm{Tr}(x) = x\mathrm{Tr}(1) = nx$. $\endgroup$
    – Janik
    Mar 12, 2022 at 16:18
  • $\begingroup$ @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.) $\endgroup$ Mar 13, 2022 at 20:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.