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Tritter is a generalisation of the fifty-fifty beam-splitter to situations where photons can propagate along three paths. It has the following matrix representation: $$T\equiv\left(\begin{array}{ccc} 1 & 1 & 1 \\ 1 & \alpha & \alpha^2 \\ 1 & \alpha^2 & \alpha\end{array}\right),\qquad\text{where }\alpha=\exp(i2\pi/3)$$
(a) Is $T$ unitary/hermitian? (Hint: Note that $1+\alpha+\alpha^2=0$).
(b) Matrix $T$ is written in the following representation: $\vert0\rangle\leftrightarrow(1\,0\,0)^T$, $\vert1\rangle\leftrightarrow(0\,1\,0)^T$ and $\vert2\rangle\leftrightarrow(0\,0\,1)^T$. Write the tritter operator in Dirac notation.

What does the question mean by part b) ? Do I have to make a matrix representation like

$$T = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$

and how does this relate to the question?

Additionally, is there a formula to find an operator in Dirac notation?

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    $\begingroup$ "Tritter" is a new word for me. I've never heard it: it makes me think at once of the short messaging service and a pig with a sore foot. I ken this kind of thing from 3x3, symmetric optical fiber directional couplers and indeed any physical device realizing this trotter operator has to have a threefold, cyclic physical symmetry, or at least simulate such. $\endgroup$ – Selene Routley Mar 22 '18 at 10:20
  • $\begingroup$ @john, I don't know if you got the answer in the end, but it occurs to me that you might simply be confused by the notation. The matrix is called $T$. When they write the basis vectors for $\lvert 0 \rangle, \lvert 1 \rangle, \lvert 2 \rangle$, they use the letter $T$ again, but here it denotes " matrix transpose" -- it has nothing to do with the name of the matrix. If they had been able to typeset column vectors, those superscripts wouldn't be necessary. $\endgroup$ – ostrichCamel Mar 22 '18 at 19:28
  • $\begingroup$ @ostrichCamel yeah, I initially thought that way too but the thing that confuses me the most is when the question says write it in the follow representation, |0⟩↔(1 0 0)^T , |1⟩↔(0 1 0)^T and |2⟩↔(0 0 1)^T, it just means we have to write the operator in 3x 3 matrix but not in 3x3 matrix and must be identity matrix right? $\endgroup$ – john Mar 23 '18 at 8:23
  • $\begingroup$ no, the answer will not look like a matrix. It will be a sum of terms, a continuation of @WetSavannaAnimal aka Rod Vance's answer. the $\lvert \rangle$ represent column vectors and the $\langle \rvert$ represent their conjugate transposes (so complex conjugate row vectors). $\endgroup$ – ostrichCamel Mar 23 '18 at 21:44
  • $\begingroup$ @ostrichCamel but isn't the sum of ket-bra means 3x3 matrix in this case?? $\endgroup$ – john Mar 24 '18 at 4:09
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They are simply defining the states $\lvert 0 \rangle, \lvert 1 \rangle, \lvert 2 \rangle$ as the orthonormal basis in which the matrix has been written. In general, any operator can be written in the form $\sum_{i,j}c_{ij}\lvert i \rangle \langle j \rvert$ where the sum is carried out over a set of basis states, and $c_{ij}$ are the matrix elements in the chosen basis.

Think about how matrix multiplication works and write a sum that has the same effect as the matrix you were given.

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OstrichCamel's hint is certainly correct, but I'd like to say that I can see why you are confused because you're being asked to write the same thing again in a trivially different notation (you simply do as OstrichCamel suggests). You're essentially being asked to tell the subtle difference between a linear operator and its matrix. To help you see what's going on, write an intermediate answer, whereby one defines what happens in Dirac notation by stating the images of the three basis states $|0\rangle$, $|1\rangle$, $|2\rangle$. This intermediate answer is going to be three lines stating these images: the first is:

$$|0\rangle \stackrel{T}{\mapsto}\frac{1}{\sqrt{3}}\left(|0\rangle+|1\rangle+|2\rangle\right)$$

and I'll leave you to finish. So part of the operator (three out of its nine components) is:

$$T = \frac{1}{\sqrt{3}}\left(|0\rangle\langle0| +|1\rangle\langle0|+|2\rangle\langle0|\right) + \cdots$$

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  • $\begingroup$ hi @WetSavannaAnimal aka Rod Vance thanks for commenting so the operator of T can be represented as T=1/sqrt(3) ( |0⟩⟨0|+|1⟩⟨0|+|2⟩⟨0|+ |0⟩⟨1|+|1⟩⟨1|+|2⟩⟨1|+|0⟩⟨2|+|1⟩⟨2|+|2⟩⟨2|) ?? If its true, what is the reason for this??( I'm not sure but can I ask is this a formula for making an operator?) $\endgroup$ – john Mar 22 '18 at 12:00
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    $\begingroup$ @john No, that's not right. Your second image is going to be $|1\rangle \stackrel{T}{\mapsto}\frac{1}{\sqrt{3}}\left(|0\rangle+\alpha\,|1\rangle+\alpha^2\,|2\rangle\right)$, with corresponding three terms in $T$. Can you see what's going on now? $\endgroup$ – Selene Routley Mar 22 '18 at 12:14

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