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In quantum mechanics, bra-ket notation is often used to represent the state vectors of the system. It is also possible to write these state vectors as "actual" vectors, for example if the system at hand is the spin of an electron it would be possible to use the basis vectors $\begin{pmatrix}1\\0\end{pmatrix}$ and $\begin{pmatrix}0\\1\end{pmatrix}$ to do the calculations.
I was wondering about how this short calculation would look like written out more "mathematical". I am intersted in a system with a Hilbert space that can be written as the three-fold tensor product of the Fock space: $\mathcal{H}=\mathcal{F}\otimes\mathcal{F}\otimes\mathcal{F}$. Let $a_i$ denote the annihilation operator for the $i$th subsystem. As an example, I want to calculated the following expectation value and want to see, how the following calculation would look using more stringent notation: $$\begin{align} &\langle0,1,1|a_2^\dagger a_3^\dagger|0,0,0\rangle \tag{1}\label{1} \\ &\qquad= \langle0,1,1|0,1,1\rangle \tag{2}\label{2} \\ &\qquad= \langle0|0\rangle \langle1|1\rangle \langle1|1\rangle \tag{3}\label{3} \\ &\qquad= 1 \tag{4} \end{align}$$ Here, I also want to assume that the state vector of the $i$th subsystem can be written as some abstract vector $\psi_{ik}$ where $k$ is the index running over the components of this vector. I am not interested in finding the components of this vector, but rather in how the above calculation "plays out using $\bigotimes$,$\sum$ and $\prod$". Similarly, the operators are abstract matrices of unknown (uninteresting) components. I am especially interested in the step from (\ref{1}) to (\ref{2}). The two matrices representing the operators have to be applied to the vectors, but at sum point this sum has to be interchanged with a product.

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    $\begingroup$ Can you give an example for the meaning of 'more "mathematical' ? I am not sure what you expect/want as answer. Your last sentence also confuses me. What sum do you mean ? A fairly concise and readable description of tensor product states is given here, ocw.mit.edu/courses/physics/8-05-quantum-physics-ii-fall-2013/… . Is that the kind of notation that you are looking for ? $\endgroup$
    – Hans Wurst
    Jul 8, 2021 at 20:37
  • $\begingroup$ @HansWurst Partially, yes. But what I also mean for example is writing the action of the creation operator as $\sum_j \left(a^\dagger\right)_{ij} \psi_{j}$ instead of $a^\dagger |\psi\rangle$. Basically, I want all of the bra-ket notation removed from the calculation. $\endgroup$
    – schade96
    Jul 8, 2021 at 21:06

2 Answers 2

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I'll keep the Dirac notation for each subsystem, since matrix mechanics is basically defined this way. In this convention, you wish to hyper-formalize $$\begin{align} &\langle0,1,1|a_2^\dagger a_3^\dagger|0,0,0\rangle \tag{1} \\ &\qquad= \langle0,1,1|0,1,1\rangle \tag{2} \\ &\qquad= \langle0|0\rangle \langle1|1\rangle \langle1|1\rangle =1\tag{3} \end{align}$$ expressed more strictly; I don't know why. In any case, here is the dictionary, $$ |0,0,0\rangle \leftrightarrow |0\rangle \otimes |0\rangle \otimes |0\rangle, \\ |1,0,0\rangle \leftrightarrow |1\rangle \otimes |0\rangle \otimes |0\rangle, \\ a_2^\dagger a_3^\dagger \leftrightarrow 1\!\!1 \otimes a_2^\dagger\otimes a_3^\dagger $$ so the ket in (1) to (2) is just $$ a_2^\dagger a_3^\dagger|0,0,0\rangle \leftrightarrow (1\!\!1 \otimes a_2^\dagger\otimes a_3^\dagger) ( |0\rangle \otimes |0\rangle \otimes |0\rangle)= |0\rangle \otimes |1\rangle \otimes |1\rangle , $$ the ket in (2).

For the dot product, you just dot the vectors in each $\cal F$, and of course the tensor product of numbers (scalar products) is the ordinary grade-school product.

Given your last comment, if you wish to chuck Dirac notation, for whatever reason, supplant $\psi(0)$, the ground state, for $|0\rangle$; where I put the energy label in the parenthesis, since subscripts are taken. So the last expression is, instead, $$ (1\!\!1 \otimes a_2^\dagger\otimes a_3^\dagger) ( \psi(0) \otimes \psi(0) \otimes | \psi(0))= \psi(0) \otimes \psi(1)\otimes \psi(1) . $$ The x dependence is implicit, and the dot products involve x-integration, as usual. Since you never dot across differing tensor subspaces, the dot products will involve three integrals, in three different dummy variables. In the coordinate representation, the creation operator is an infinite-dimensional matrix, $(x-\hbar \partial_x)/\sqrt{2\hbar}$, as you know.

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  • $\begingroup$ The indices should have been $2$ and $3$, of course. I corrected that. $\endgroup$
    – schade96
    Jul 8, 2021 at 21:14
  • $\begingroup$ Clear enough? In the last equation, matrix indices are suppressed, but they are obvious. They could be continuous xs in the coordinate representation, as I point out. Are you comfortable with Dirac's book? $\endgroup$ Jul 9, 2021 at 13:12
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We can express states and operators as numeric vectors and matrices after we have picked a discrete basis, for simplicity let the basis be ortho-normal. The state is then represented by a vector containing the expansion coefficients. Lets do this first in 1 dimension.

Let our basis be the set of bosonic Fock states $\{|n\rangle \}$. A general state could be a linear combination of theses basis states. $$\ |\psi\rangle = \sum_{n=0} c_n |n\rangle $$ These Ket vectors are considerd to be elements of a Hilbert space. The numerical representation of this state is with regard to this basis $$ \psi = \left(\begin{matrix}c_0 \\c_1\\ \vdots \end{matrix}\right) $$ This vector lives in the vector space $\mathbb C\times \mathbb C \cdots\times \mathbb C$ and is just a complex number vector. The connection between both space is $$ \psi_l = c_l = \langle l|\psi\rangle $$

A little bit more should be said about bras and dual spaces but for simplicity I'll simply state that we associate them with a conjugate row vector and I'll denote them as $\langle \psi | \rightarrow \bar \psi$.

In a similar manner, we can associate a complex matrix with the creation operator $\hat a^\dagger$, which is defined by the operators so called matrix elements. $$ a^\dagger = \left( \begin{matrix} a^\dagger_{00} & a^\dagger_{01} & \cdots \\ a^\dagger_{10} & a^\dagger_{11} & \dots \\ \vdots & \ddots \end{matrix} \right) $$ The matrix elements are defined by $$ a^\dagger_{nm} = \langle n|\hat a^\dagger|m\rangle $$ The action of the operator can then be calculated by $a^\dagger \psi$, using the complex matrix and the complex number vector. The component wise connection to the Hilbert space notation would be $$ (a^\dagger \psi)_l = \langle l|\hat a ^\dagger |\psi\rangle $$

So far I have only established how a Hilbert Space (that admits a discrete basis) and "numeric space" is connected in 1 dimension.

Lets now look at a tensor product example

$$ \big(\langle 0| \otimes \langle 1| \otimes \langle 1| \big) \hat{\mathbb{I}}_1\otimes \hat a^\dagger_2 \otimes \hat a^\dagger_3 \big(|0\rangle \otimes |0\rangle \otimes |0 \rangle \big) $$

This transfers in the discrete case pretty much one to one to the numeric space, where you just replace each Hilbert space object by its numeric pendant. $$ \big(\bar 0\otimes \bar 1 \otimes \bar 1 \big) (\mathbb{I}_1\otimes a^\dagger_2 \otimes a^\dagger_3) \big(0 \otimes 0 \otimes 0 \big)= \big(\bar 0\otimes \bar 1 \otimes \bar 1 \big) (\mathbb{I}_1 0 \otimes a^\dagger_2 0 \otimes a^\dagger_3 0) \\ =\big(\bar 0\otimes \bar 1 \otimes \bar 1 \big) ( 0 \otimes 1 \otimes 1) =\big(\bar 0\otimes \bar 1 \otimes \bar 1 \big)\cdot ( 0 \otimes 1 \otimes 1) \\ =\big(\bar 0 \cdot 0 \otimes \bar 1 \cdot 1 \otimes \bar 1 \cdot 1 \big)\\ =I\cdot I \cdot I = I $$ In the last line I used roman numerals to express the actual numeric value. I.e $I$ stands for the real number one, since the other real number symbols where already used for the numeric vectors.

We could also write the first line using sums and indices but that gets quite messy and is normally only done if you can't evaluate the operator action analytically,

$$ \big(\bar 0\otimes \bar 1 \otimes \bar 1 \big) (\mathbb{I}_1\otimes a^\dagger_2 \otimes a^\dagger_3) \big(0 \otimes 0 \otimes 0 \big)= \sum_{ijklnm} \bar 0_i \bar 1_k \bar 1_n \mathbb{I}_{ij} a^\dagger_{kl} a^\dagger_{nm}0_j 0_l 0_m $$

If you work out a general form of the matrix elements you could do the calculation numerically. But that would pretty much a pointless hassle in this case since we already picked a basis where we can evaluate the action of the operators trivially, without resorting to the numerical calculation.

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