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In Cohen-Tannoudji's Quantum Mechanics book the tensor product of two two Hilbert spaces $(\mathcal H = \mathcal H_1 \otimes \mathcal H_2)$ was introduced in (2.312) by saying that to every pair of vectors $$|\phi(1)\rangle \in \mathcal H_1, |\chi(2)\rangle \in \mathcal H_2$$ there belongs a vector $$|\phi(1)\rangle \otimes |\chi(2)\rangle \in \mathcal H$$ In a footnote it stated that the order doesn't matter and that we could also call it $$|\chi(2)\rangle \otimes |\phi(1)\rangle$$ I'm a bit confused, since I though that the order of the tensor product generally matters. What would that expression look like if we picked a basis, say: $$|\phi(1)\rangle = a_1|u_1\rangle + a_2|u_2\rangle + \dotsc$$ $$|\chi(2)\rangle = b_1|v_1\rangle + b_2|v_2\rangle + \dotsc$$

Any help will be appreciated!

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    $\begingroup$ I think the label 1 and 2 are used to keep track of which vector space each state belongs to. $\endgroup$ – Chris2807 Mar 3 '15 at 20:30
  • $\begingroup$ Yes, but does this hold in general for the tensor product: $|\phi(1)\rangle \otimes |\chi(2)\rangle = |\chi(2)\rangle \otimes |\phi(1)\rangle$? $\endgroup$ – jx9845 Mar 3 '15 at 20:35
  • $\begingroup$ @jx9845 No, they're just picking a convention. The original choice of the convention was arbitrary, however. $\endgroup$ – Ryan Unger Mar 3 '15 at 20:37
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    $\begingroup$ You either tag your vectors with a number or else you have to care about the order. $\endgroup$ – jinawee Mar 3 '15 at 20:38
  • $\begingroup$ Is that because we haven't picked a basis yet? And if we picked one, would that change anything? $\endgroup$ – jx9845 Mar 3 '15 at 20:41
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$|\phi(1)\rangle \otimes |\chi(2)\rangle $ is a cumbersome notation to write ket corresponding to $\psi$ function $\phi(\mathbf r_1)\chi(\mathbf r_2)$, where $\mathbf r_i$ refers to coordinates of the $i$-th subsystem. That's why the order of factors in $\otimes$ product does not matter; the resulting ket corresponds to the same $\psi$ function and is thus the same ket.

On the other hand, $|\phi\rangle \otimes |\chi\rangle $ (without labels) is meant to be read according to different convention; here it is commonly understood that the order of factor signifies the sub-system it refers to. So

$|\phi\rangle \otimes |\chi\rangle $ denotes ket corresponding to $\phi(\mathbf r_1)\chi(\mathbf r_2)$ just as $|\phi(1)\rangle \otimes |\chi(2)\rangle $ does, but:

$|\chi\rangle \otimes |\phi\rangle $ denotes ket corresponding to $\chi(\mathbf r_1)\phi(\mathbf r_2)$ which is not the same. This is because different meaning of the $\otimes$ notation is used.

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  • $\begingroup$ Okay, so if one writes $|\chi(2)\rangle \otimes |\phi(1)\rangle$, it is implied that this is a vector in $\mathcal H_1 \otimes \mathcal H_2$ just as $\chi(\mathbf r_2)\phi(\mathbf r_1)$ is in $\mathcal H_1 \otimes \mathcal H_2$, correct? $\endgroup$ – jx9845 Mar 3 '15 at 20:59
  • $\begingroup$ Mathematically, ket $|\psi\rangle$ is not exactly the same thing as the complex function $\psi(\mathbf r)$ it refers to and the kets do not belong to the same space as the functions $\psi$ do. More accurately, kets are from an abstract space of kets $\mathcal H_{kets}$ and normalizable functions $\psi$ are from $L^2$ space of quadratically integrable functions. But don't worry about this, for answering your question, this is largely a distinction without difference. $\endgroup$ – Ján Lalinský Mar 3 '15 at 21:04
  • $\begingroup$ For your examples, I would say in $|\chi(2)\rangle|\otimes|\phi(1)\rangle$ the order is not important because the labels override it. This ket is a member of space $\mathcal H^{(kets)}_{1}\otimes\mathcal H^{(kets)}_{2}$. This space can be also denoted as $\mathcal H^{(kets)}_{2} \otimes \mathcal H^{(kets)}_{1}$. Here labels do not need to override any order, because there is no way order of terms could change the meaning of the product - it is always a ket form of the L2 space that is generated as linear span of the products of functions for the subsystems, so there is not any order. $\endgroup$ – Ján Lalinský Mar 3 '15 at 21:11
  • $\begingroup$ Thank you for your responses. If we are being mathematically rigorous $\mathcal H^{(kets)}_{1}\otimes\mathcal H^{(kets)}_{2}$ is different from $\mathcal H^{(kets)}_{2} \otimes \mathcal H^{(kets)}_{1}$ though, right? So do we simply regard these two vector spaces as the same in physics since they are isomorphic? After all if we start with two particles and want to construct the state of the system, how would we determine which vector space comes first in the tensor product? $\endgroup$ – jx9845 Mar 3 '15 at 21:52
  • $\begingroup$ Even mathematically, these are just different notations for the same resulting space; the order does not matter when $\otimes$ is used to denote product of spaces. $\endgroup$ – Ján Lalinský Mar 3 '15 at 22:46
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You are correct in that the tensor product does not commute in general. The ordering of vectors in some tensor product, say \begin{equation*} |\psi\rangle\otimes|\phi\rangle\otimes|\xi\rangle \equiv |u\rangle \end{equation*} implicitly refers to how the resulting Hilbert space is defined via the tensor product. To the above vectors, let us associate $|\psi\rangle\in H_{\psi}$, $|\phi\rangle\in H_{\phi}$ and $|\xi\rangle\in H_{\xi}$. In this case, we have implicity said that the Hilbert space for our vector $|u\rangle$ is \begin{equation*} |u\rangle\in U\equiv H_{\psi}\otimes H_{\phi}\otimes H_{\xi} \end{equation*} One just need to keep in mind the order of how the underlying Hilbert space is formed. As for a physical description in quantum theory the order is of no importance.

Specifically, in your case when you say \begin{equation*} |\chi(2)\rangle\otimes |\phi(1)\rangle \end{equation*} it just means that the vector belongs to $H_{2}\otimes H_{1}$, rather than $H_{1}\otimes H_{2}$.

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  • $\begingroup$ So if we are being mathematically rigorous, $H_{1}\otimes H_{2} \neq H_{2}\otimes H_{1}$. However, as @Ján Lalinský pointed out the indices override the order of the tensor product so that $|\chi(2)\rangle\otimes |\phi(1)\rangle$ actually means $|\phi(1)\rangle\otimes |\chi(2)\rangle$ and belongs to $H_{1}\otimes H_{2}$, correct? $\endgroup$ – jx9845 Mar 3 '15 at 21:40
  • $\begingroup$ @jx9845 Yes, if you interchange the order of vectors like that, overriding by index as you say, you also change the order of how the smaller spaces are combined with the tensor product. $\endgroup$ – Invoker Mar 3 '15 at 22:25
  • $\begingroup$ So if we combine the states $|\phi\rangle \in H_1$ and $|\chi\rangle\ \in H_2$, how do we know which space the new state is going to be in, that is $H_1 \otimes H_2$ or $H_2 \otimes H_1$? Do we simply pick one and stick with it? $\endgroup$ – jx9845 Mar 3 '15 at 22:31
  • $\begingroup$ That depends on the order of how you combine them. If $|\phi\rangle\otimes |\chi\rangle$ then the vector belongs to $H_{1}\otimes H_{2}$. But as future convetion, form the space first, then consider the vectors inside it. If you from the start set $H\equiv H_{1}\otimes H_{2}$ then the vector in $H$ is expressed as $|\phi\rangle\otimes |\chi\rangle$. And yes, you simply pick one convetion and stick with it. $\endgroup$ – Invoker Mar 3 '15 at 22:39
  • $\begingroup$ A 5-minute rule forces me to enter here; above there's a typo, not "convetion" but rather 'convention'. $\endgroup$ – Invoker Mar 3 '15 at 22:47
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Imagine two different particles, e.g. a proton and an electron, the former described by functions in the Hilbert space $\mathcal H_p$, and the latter by functions in the Hilbert space $\mathcal H_e$. Now assume a base states $\phi_1, \phi_2,...$ in $\mathcal H_p$ and a base $\chi_1, \chi_2,...$ in $\mathcal H_e$. Your pair of particles is (proton, electron), or is (electron, proton)? It doesn't matter, does it? The same with the description of the state of the pir,

$$\Psi = \sum_{i,j} C_{i,j} \ \phi_i \otimes \chi_j = \sum_{i,j} C_{i,j} \ \chi_i \otimes \phi_j . \tag {i}$$

Now, there exists a situation when the order is important: when the functions $\chi$ and $\phi$ look the same, e.g. the eigenfunctions of the spin-$z$ projection operator, $|\uparrow\rangle, |\downarrow\rangle$. In this case instead of mentioning by subscripts to which particle we refer, we assume an order, e.g. in each product of states we write first the state of the electron, and second of the proton: $|\uparrow\rangle \otimes |\downarrow\rangle$ means that the electron has spin-up and the proton spin-down.

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  • $\begingroup$ So is your description in $(i)$ only valid because the basis elements $\phi_k$ and $\chi_l$ are different for all k,l? Isn't $\phi_i \otimes \chi_j$ different from $\chi_i \otimes \phi_j$ though? After all $\phi_i$ and $\phi_j$ are completely different functions and the same goes for the $\chi $'s. $\endgroup$ – jx9845 Mar 3 '15 at 21:28

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