3
$\begingroup$

In my textbook there is the expression

$$ \left|M \right> = \prod {\left|n_i\right>} \,,$$

where $|A\rangle$ is a ket vector and $\small{\prod{}}$ is (as far as I believe) a symbol of product. What does $\small{\prod{}}$ here mean? I suspect this $\small{\prod{}}$ is a typo and the correct symbol is $\sum$ because I think a value like $|A\rangle |B\rangle$ cannot be defined. But I'm not confident since I'm not familiar with bra-ket notation.

Could anyone please tell me the meaning of the $\small{\prod{}}$?

The equation is Eq.(17.24) (p.199) of "Perspectives on Statistical Thermodynamics" by Yoshitsugu Oono.

$\endgroup$
  • 1
    $\begingroup$ The symbol Π stands for product just like Σ stands for sum. You are multiplying kets because you are constructing a multi-particle state |M> from the single-particle states |ni>. The final ket is supposed to be the state of the full system. To be more strict, a multi-particle state lives on a composite Hlibert space constructed from the tensor product of single-particle Hilbert spaces. To be even more strict, the product symbol is kinda misleading as it represents tensor products and not, as the notation Π suggests, plain products. $\endgroup$ – Panos C. Jul 14 '18 at 13:02
  • 1
    $\begingroup$ In MathJax (and LateX math-mode) you should use \rangle (\langle) to produce the right (left) angle bracket for thinks like kets (bras) because > (<) is an operator and is typeset with extra space around it. Compare "$|A\rangle$" to "$|B>$". (If you need re-sizing brackets use \right> and \left< paired appropriately.) $\endgroup$ – dmckee Jul 14 '18 at 17:17
  • $\begingroup$ @dmckee Thank you. Now I edited my post. Admittedly, it looks more beautiful. (Actually, I wanted to use \ket{} but gave up using | and >.) $\endgroup$ – ynn Jul 14 '18 at 17:57
  • $\begingroup$ MathJax doesn't support any of the packages which provide \ket, so we're stuck doing it the old-school way. In my serious writing I use the physics package which provides nice support for brakets. $\endgroup$ – dmckee Jul 14 '18 at 18:33
  • $\begingroup$ @dmckee I took a little look at the document, and that looks really nice. I'll try it when I write a latex code. $\endgroup$ – ynn Jul 14 '18 at 18:49
3
$\begingroup$

First I explain how a vector works for finite dimensional spaces and then generalize it to the Hilbert spaces in quantum mechanics, starting intuitively and make it a bit more rigorous.

"Multiplication of vectors" happens when you have a direct (tensor) product of vector spaces. For ordinary vectors, let's say 2D vectors in $\mathbb{R}^2$, the direct product of two of them would give a 2nd rank tensor or dyadic (represented by 2x2 matrices) as if you insert the whole vector into the component of another vector: $$\tag{(1)}\mathbf{v} \;\mathbf{u} = \begin{bmatrix} v_1 \mathbf{u} \\ v_2 \mathbf{u} \end{bmatrix} = \begin{bmatrix} v_1 u_1 & v_1 u_2 \\ v_2 u_1 & v_2 u_2 \end{bmatrix}$$ where you can imagine n number of direct products of a vector that constructs an array of a size $2 \times 2 \times \ldots \times 2$, like an n dimensional "cube".

Of course if a "infinite dimensional" Hilbert space is in question it would be more abstract than that but the component-wise representation would have the same structure (n rank arrays). So, the state vectors of the direct product space would be an n-array of infinite numbers, like you place the values inside n dimensional cube but the sides of the cube stretches out to infinity.

For single particle, a generic state would be $|\psi\rangle$ an infinite dimensional vector. If the basis states are represented by a natural number, $n$, then you have infinitely many basis vectors: $$\Big\{ |0\rangle, |1\rangle, |2\rangle, |3\rangle, \ldots \Big\}$$ each of them are vectors, similar to the usual $\hat{i}, \hat{j}, \hat{k}$ in three dimensional space but infinite, not 3. They can be denoted by $|n\rangle$ where n could be any natural number.

Here n is not a variable, it is a label, like instead of writing $\hat{i}, \hat{j}, \hat{k}$, you can write $\hat{e}_i$ where i runs from 1 to 3 and it is a three dimensional vector not 1D. So, bra-ket notation is not like writing the components in paranthesis, it is for labeling the vector itself.

Therefore, the generic state would be $$ |\psi\rangle = a_0 |0\rangle + a_1 |1\rangle+ \ldots = \sum_n a_n |n\rangle, $$ Now, when you take the tensor product of two state vectors, you would get \begin{align} |\Psi\rangle &= |\psi_1 \rangle | \psi_2\rangle \\ &= a_{00} |0\rangle|0\rangle + a_{01} |0\rangle |1\rangle + a_{02}|0\rangle |2\rangle + \ldots + a_{mn} |m\rangle |n\rangle + \ldots \end{align} in terms of the product-space basis. As you can see, each "new" basis is a direct product of two "old" basis. The components of this state vector are $a_{mn}$, just like Eq.(1) for direct product of 2D vectors but it is represented by an infinite dimentional matrix. If you take the direct product of three vector spaces then each component would be $a_{mn\ell}$ like a infinitely wide cube, and so on.


Now, let me be more rigorous.

Let $\mathcal{H}_i$ denote the i-th single particle Hilbert space. Then the Hilbert space of all particles in the system would be the direct product of all Hilbert spaces, \begin{align} \mathcal{H} &= \bigotimes_{i \in I}\mathcal{H}_i \\ & = \mathcal{H}_1 \otimes \mathcal{H}_2 \otimes \ldots \otimes \mathcal{H}_n \end{align}

such that for each vector, $| \Psi \rangle \in \mathcal{H}$, you have \begin{align} |\Psi\rangle &= \prod_{i \in I} |\psi_i\rangle =|\psi_1\rangle |\psi_2\rangle \ldots |\psi_n\rangle \end{align} where $I$ is a finite subset of natural numbers and $| \psi_i \rangle \in \mathcal{H}_i$.

We physicists, most of the time, misuse the notations of mathematical objects. Mathematicians would rather write $\bigotimes$ instead of $\prod$ since it is not a multiplication defined within a structure (i.e., an algebra$^{(*)}$) but rather an external composition. However, since we know that there is no such a multiplication between state vectors, that is, the states do not form an algebra, in quantum mechanics this notation is not ambiguous.

For each $|\psi_i\rangle$ you can have as many as values since $\psi_i$ is a number and it could be real or complex, or just a natural number. For example it could be the momentum eigenstates of the i-th particle, that is, $|p_i \rangle$, and the whole state is representing various momentums of n particles.

In your textbook, the state $|n_i\rangle$ means that it is represented by natural number, n, but this number would be different for each state in the direct product. So, they labeled it by i, for instance, it could be $$ | M \rangle = |1\rangle |3\rangle |15\rangle |4\rangle \ldots $$ Sometimes we just write $|1,3,15, 4, \ldots\rangle$ for short.


$^{(*)}$ An algebra is a vector space equipped with a vector multiplication along with the vector addition that vector spaces have in default. For instance, linear operators in quantum mechanics form an algebra on the complex Hilbert space, $L^2(\mathbb{C})$. But the state vectors do not form an algebra, i.e., you can not multiply two state vectors to get another state vector in the same Hilbert space.

$\endgroup$
  • $\begingroup$ Thank you for your answer and detailed explanation. If, as you say, tensor product of two 2d-vectors makes 2x2 matrix, is the total state vector also represented as matrix, although matrix is not a vector? I thought $|M\rangle = (n_{1x}, n_{1y}, \cdots, n_{Nz})$ (1d-vector) because the new state is characterized by $(n_{1x}, n_{2x}, \cdots, n_{Nz})$, or the total set of quantum number of each particles, but the situation seemes not so simple. $\endgroup$ – ynn Jul 15 '18 at 8:51
  • 1
    $\begingroup$ @ynn Matrices on their own are just an array of numbers. We use matrices as representations here. As a matter of fact, vector components are represented in a row/column matrix (if you agree on the basis) as well, however being vector means to have a algebraic structure as the whole set. That's why I said tensor and said "represented by matrices". On the other hand, $|M\rangle$ is indeed a vector but infinite dimensional. Actually, each $|n_i\rangle$ is also an infinite dimensional vector. I think I have to add some explanation about $|n_i\rangle$ on the answer in order to make it clearer. $\endgroup$ – Oktay Doğangün Jul 15 '18 at 9:02
  • $\begingroup$ @ynn I have made the explanatory edits. Please read the answer from the beginning. $\endgroup$ – Oktay Doğangün Jul 15 '18 at 9:28
  • $\begingroup$ I really appreciate it. Now I understand the meaning of the product and why the new state vector's components are denoted by not just $a_i$ but like $a_{ij}$. (To me, your explanation right before the 1st separator was so clear.) Though I cannot say I've understand tensor product deeply, it must be the fact my understanding has got a further point. $\endgroup$ – ynn Jul 15 '18 at 10:22
5
$\begingroup$

The author means the tensor product of the ket vectors. This is indeed a way of "multiplying" vectors together, although it's subtle because the resulting product vector actually lies in a different vector space than the original ones. His notation is quite nonstandard though, because physicists almost always denote the tensor product of vectors by the symbol $\otimes$, not $\prod$. The more standard notation would be $$|M\rangle = \bigotimes_{i=1}^N |n_i\rangle.$$

$\endgroup$
1
$\begingroup$

The previous answers have explained how a multiplication of vectors can be defined mathematically as a tensor product. I'll give a little more physical information here which is important and wasn't spelled out clearly to me when I learned about this stuff.

In Quantum mechanics states of a system live in a Hilbert space. How do we describe states of a system? Well every system has a number of degrees of freedom. In classical mechanics, for example, a double pendulum would have two degrees of freedom: $\phi_1$ and $\phi_2$, the angles representing the positions of the different pendula. Alternatively you might consider the position of a ball on a 2D plane described by its position variables $x$ and $y$.

Well quantum mechanically it is the same. A system has degrees of freedom. The difference is that these degrees of freedom are quantified not by mere c-number (complex number) variables, but rather by quantum operators. So, for example, $x$ is replace by $\hat{x}$. and $y$ is replaced by $\hat{y}$. $\hat{x}$ and $\hat{y}$ are of course quantum operators. Since they are operators it means they act on Hilbert spaces. In particular, the operator corresponding to each degree of freedom can act on its own Hilbert space. That is, there is a hilbert space $\mathcal{H}_x$ on which $\hat{x}$ acts and a Hilbert space $\mathcal{H}_y$ on which $\hat{y}$ acts.

If only one of the degrees of freedom were present we could describe the total state of the system by giving $\lvert \psi \rangle_x \in \mathcal{H}_x$, the vector in the Hilbert space which represents the state of the system. However, when there are multiple degrees of freedom you must give something that describes the state of each degree of freedom in the system.

I have explained how there is a Hilbert space for each degree of freedom. The tensor product, as nicely explained in previous answers, is the tool used to combine the sub-Hilbert spaces for each degree of freedom into a total Hilbert space for the system. The state of the system is then given entirely by the specification of a vector within the total Hilbert space.

Here is the key point: In general, if a Hilbert space $\mathcal{H}_{tot}$ is a tensor product of smaller Hilbert spaces $\mathcal{H}_1$ and $\mathcal{H}_2$ then any vector in $\mathcal{H}_{tot}$ can be written as a linear combination of tensor products of vectors from $\mathcal{H}_1$ and $\mathcal{H}_2$. The state described in your book is an example of one such state.

In summary: if a quantum system has more than one degree of freedom then states will be described as linear combinations of tensor products of vectors from the Hilbert spaces corresponding to each degree of freedom.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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