# Entropy as a function of a density operator?

I've got this problem about a harmonic oscillator with energy eigenstates |n>, which is prepared in the state

$$|\psi> = \frac{1}{cosh r}\sum_n tanh^nr|n\rangle$$ where r is any real number and the sum is from n=0 to infinity.

I'm asked to calculate the entropy $$S(\rho) = -Tr(\rho ln\rho)$$ if $$\rho = |\psi\rangle\langle \psi|$$.
I'm given the hint that I should write $$S(\rho)$$ in terms of the eigenvalues of $$\rho$$.

So, the first thing I've done is written out my density operator more explicitly:

$$\rho = |\psi\rangle\langle \psi| = \frac{1}{cosh^2r}\sum\tanh^2^nr|n\rangle\langle n|$$

but I'm not even sure about this because, when I've written out the bra I wasn't sure what happens with the complex conjugates of cosh and tanh, so I don't even know if i'm justified in simply squaring them.

Anyway, I come to the hint now, "write S in terms of the eigenvalues of the density operator".
However, when I've been asked to calculate eigenvalues before I've always been given a matrix, say X, and done $$X|\phi\rangle = \lambda|\phi\rangle$$ with $$|\phi\rangle = \left(\begin{array}{c}\phi_1&\phi_2\end{array}\right)$$
to find the eigenvalues and eigenstates.
I'm not sure how to get an eigenvalue equation set up in the first place to find my eigenvalues here, since I'm not told what the matrix representation of $$\rho$$ is.

The one thing I've tried, but I'm not sure about, is saying
$$\rho|\phi\rangle = \lambda|\phi\rangle = \frac{1}{cosh^2r}\sum_n\tanh^2^nr|n\rangle\langle n|\phi\rangle$$

and if I say $$\langle n|\phi\rangle = \delta\_n_\phi$$

then I get $$\frac{1}{cosh^2r}\sum_n\tanh^2^nr|n\rangle = \lambda|n\rangle$$
so I can cancel the |n> off both sides. I'm not too confident in this though...although it does give me infinite numbers of eigenvalues, which is nice...

Assuming all this is somehow correct though, I'm at the trace part and not sure how to work all that into $$Tr(\rho ln\rho) = \sum_j\langle\phi_j|\rho ln\rho|\phi_j\rangle$$.

Not sure how to tackle this...

Last edited:

fzero
Homework Helper
Gold Member
I've got this problem about a harmonic oscillator with energy eigenstates |n>, which is prepared in the state

$$|\psi> = \frac{1}{cosh r}\sum_n tanh^nr|n\rangle$$ where r is any real number and the sum is from n=0 to infinity.

I'm asked to calculate the entropy $$S(\rho) = -Tr(\rho ln\rho)$$ if $$\rho = |\psi\rangle\langle \psi|$$.
I'm given the hint that I should write $$S(\rho)$$ in terms of the eigenvalues of $$\rho$$.

So, the first thing I've done is written out my density operator more explicitly:

$$\rho = |\psi\rangle\langle \psi| = \frac{1}{cosh^2r}\sum\tanh^2^nr|n\rangle\langle n|$$
There are two sums involved:

$$|\psi\rangle\langle \psi| = \frac{1}{cosh^2r}\sum_{n,n'} \tanh^{n+n'}r|n\rangle\langle n'|.$$

but I'm not even sure about this because, when I've written out the bra I wasn't sure what happens with the complex conjugates of cosh and tanh, so I don't even know if i'm justified in simply squaring them.
Remember that r is real.

Anyway, I come to the hint now, "write S in terms of the eigenvalues of the density operator".
However, when I've been asked to calculate eigenvalues before I've always been given a matrix, say X, and done $$X|\phi\rangle = \lambda|\phi\rangle$$ with $$|\phi\rangle = \left(\begin{array}{c}\phi_1&\phi_2\end{array}\right)$$
to find the eigenvalues and eigenstates.
I'm not sure how to get an eigenvalue equation set up in the first place to find my eigenvalues here, since I'm not told what the matrix representation of $$\rho$$ is.
You find the matrix representation by choosing a vector representation for the states $$|n\rangle$$

$$|0\rangle = \begin{pmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix}, |1\rangle = \begin{pmatrix}0\\ 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix}, \ldots$$

Then

$$\rho = \frac{1}{\cosh^2r} \begin{pmatrix} 1 & \tanh r & \cdots \\ \tanh r & \tanh^2 r & \cdots \\ \vdots & \vdots & \cdots \end{pmatrix}.$$

The one thing I've tried, but I'm not sure about, is saying

$$\rho|\phi\rangle = \lambda|\phi\rangle = \frac{1}{cosh^2r}\sum_n\tanh^2^nr|n\rangle\langle n|\phi\rangle$$

and if I say $$\langle n|\phi\rangle = \delta\_n_\phi$$
The eigenvectors are certainly not orthonormal to the $$|n\rangle$$.

Looking at 2x2 and 3x3 matrices of this form, it looks like there's only one nonzero eigenvalue. If you can somehow show this for the general case, you can formally introduce the matrix that diagonalizes $$\rho$$ into the trace to use the hint.