# Mapping a vector to its correspondence in tensor product space [closed]

Lets assume we have two Hilbert spaces $H_1$ and $H_2$. Can you show me how can we map a vector $v\in H_1$ into a vector $v' \in H_1\otimes H_2$?

• I think this is impossible as without sense. At least, may be you could map to a vector $\,v_1\in H_1\,$ the following subspace $\,V_1 \subset H_1\otimes H_2\,$ : \begin{equation} v_1 \longrightarrow V_1 =\lbrace v_1\otimes v_2 \in H_1\otimes H_2 : v_2 \in H_2\rbrace \end{equation} My feeling is that the subspace $\,V_1 \subset H_1\otimes H_2\,$ is a Hilbert space too , but this must be proved. – Frobenius Aug 29 '17 at 7:18
• Note that something like that you ask for could be done with linear transformations. That is, if $\,\mathrm{A_1}\,$ is an operator on $\,H_1\,$ then we could map to it its extension on $\,H_1\otimes H_2\,$ : \begin{equation} \mathrm{A_1} \text{ on } H_1 \longrightarrow \left(\mathrm{A_1}\otimes\mathrm{I_2}\right) \text{ on } H_1\otimes H_2 \end{equation} where $\,\mathrm{I_2}\,$ the identity operator on $\,H_2\,$. – Frobenius Aug 29 '17 at 7:35
• @Frobenius $V_1$ is a Hilbert space isomorphic to $H_2$. – Noiralef Aug 29 '17 at 14:29
• @Noiralef : You are absolutely right. – Frobenius Aug 29 '17 at 15:21
• I rephrased the question. Would you please check it again? – user650585 Aug 29 '17 at 15:27

What you are asking physically is: "My first system $H_1$ is in the state $v$, I don't know the state of my second system $H_2$. What is the state of the total system $H_1 \otimes H_2$?

Of course, this question does not make sense. So, there is no canonical embedding of $H_1$ in $H_1 \otimes H_2$. However, if we fix a state $w \in H_2$, then there is of course the map $$\iota_w: H_1 \to H_1 \otimes H_2, v \mapsto v' = v \otimes w .$$

• In Quantum Mechanics it is said that if their entangled state $\alpha |f_1 e_1\rangle + \beta|f_1 e_2\rangle + \gamma |f_2 e_1\rangle +\delta |f_2 e_2\rangle$ can be written as $\alpha_1\alpha_2 |f_1 e_1\rangle + \beta_1\beta_2|f_1 e_2\rangle + \gamma_1\gamma_2 |f_2 e_1\rangle +\delta_1\delta_2 |f_2 e_2\rangle$ then they are not entangled. I want to know if in such case $\alpha_1 |f_1 e_1\rangle + \beta_1|f_1 e_2\rangle + \gamma_1 |f_2 e_1\rangle +\delta_1 |f_2 e_2\rangle$ is the mapping of one of the original states from $H_1$ to $H_1\otimes H_2$? – user650585 Aug 29 '17 at 15:53
• @user650585 You might want to ask this as a new question. But what you write in the comment / in the question (edit v3) is all wrong, sorry. Entangling two systems does not change the Hilbert space, it is always $H_1 \otimes H_2$. However, if the systems are not entangled, their state will be in the subspace $H_1 \times H_2 \subset H_1 \otimes H_2$, and there is a canonical map $H_1 \times H_2 \ti H_1 \otimes H_2$. – Noiralef Aug 29 '17 at 18:18
• Writing $H_1\times H_2\subset H_1\otimes H_2$ as in your comment is misleading, since the map from one to the other is not injective, so you cannot identify the image with $H_1\times H_2$ itself. – ACuriousMind Aug 29 '17 at 18:21
• @Noiralef : As ACuriousMind pointed out the relation $H_1\times H_2\subset H_1\otimes H_2$ is misleading. If $H_1,H_2$ are finite complex Hilbert spaces of dimensions $r,s$ respectively then in a sense $H_1\equiv\mathbb{C}^r,H_2\equiv\mathbb{C}^s$ so $H_1\times H_2 \equiv \mathbb{C}^{r+s}$ but $H_1\otimes H_2 \equiv \mathbb{C}^{r\cdot s}$ that is the two spaces not even have the same dimension in general. Here accidentally $r=2=s \Longrightarrow r+s=4=r\cdot s$. – Frobenius Aug 29 '17 at 19:17
• True, I was being imprecise in my comment above -- sorry. – Noiralef Aug 29 '17 at 20:42

OP post the following comment complementary to his/her original question :

In Quantum Mechanics it is said that if their entangled state $\alpha |f_1 e_1\rangle + \beta|f_1 e_2\rangle + \gamma |f_2 e_1\rangle +\delta |f_2 e_2\rangle$ can be written as $\alpha_1\alpha_2 |f_1 e_1\rangle + \beta_1\beta_2|f_1 e_2\rangle + \gamma_1\gamma_2 |f_2 e_1\rangle +\delta_1\delta_2 |f_2 e_2\rangle$then they are not entangled. I want to know if in such case $\alpha_1 |f_1 e_1\rangle + \beta_1|f_1 e_2\rangle + \gamma_1 |f_2 e_1\rangle +\delta_1 |f_2 e_2\rangle$ is the mapping of one of the original states from $H_1$ to $H_1\otimes H_2$?

You are confused. A state in $\,H_1\otimes H_2 \,$ \begin{equation} w=\alpha |f_1 e_1\rangle + \beta|f_1 e_2\rangle + \gamma |f_2 e_1\rangle +\delta |f_2 e_2\rangle \in H_1\otimes H_2 \tag{01} \end{equation} is not a product state (also called a separable state) if it could be expressed as \begin{equation} w=\alpha_1\alpha_2 |f_1 e_1\rangle + \beta_1\beta_2|f_1 e_2\rangle + \gamma_1\gamma_2 |f_2 e_1\rangle +\delta_1\delta_2 |f_2 e_2\rangle \tag{02} \end{equation} If so any such state would be a product state and moreover separable in many ways since there exists an infinite number of 8 coefficients

\begin{equation} \left(\alpha_1,\alpha_2,\alpha_3,\alpha_4,\beta_1,\beta_2,\beta_3,\beta_4\right) \tag{03} \end{equation} satisfying the system of equations \begin{equation} \alpha_1\alpha_2=\alpha \, ,\quad \beta_1\beta_2=\beta \, ,\quad \gamma_1\gamma_2=\gamma\, ,\quad \delta_1\delta_2=\delta \tag{04} \end{equation} The state $\,w\,$ in equation (01) is a product or separable state (that is not entagled) if there exist at least a solution $\,\left(\phi_1,\phi_2,\epsilon_1,\epsilon_2\right)\,$ of the system of equations \begin{equation} \phi_1\epsilon_1=\alpha \, ,\quad \phi_1\epsilon_2=\beta \, ,\quad \phi_2\epsilon_1=\gamma\, ,\quad \phi_2\epsilon_2=\delta \tag{05} \end{equation} Then the state $\,w\,$ is a (separable) product state \begin{equation} w=v_1\otimes v_2 =\left(\phi_1f_1+\phi_2f_2\right)\otimes\left(\epsilon_1e_1+\epsilon_2e_2\right) \tag{06} \end{equation} But then why do you want to call the vector $\,w \in H_1\otimes H_2 \,$ the correspodence of $\,v_1 \in H_1$. It has no sense and doesn't help anywhere.

• Thanks for your answer. Regarding your last question what I want to know is if there is a space in which both $v_1$ and $v_1\otimes v_2$ can be represented at the same time? – user650585 Aug 30 '17 at 2:30