Review of Linear Algebra in Quantum Mechanics (IV)
Inner Product
Given a vector space $V$ over the field $\mathbb{C}$, we define an inner product, i.e, a map $(\,,\,): \, V\times V \rightarrow \mathbb{C}$ that satisfies the following axioms:
- $\Big(|\psi\rangle\,,\, |\phi\rangle\Big)=\Big(|\phi\rangle\,,\, |\psi\rangle\Big)^*$
- $\Big(|\psi\rangle\,,\, |\phi\rangle+|\chi\rangle\Big) = \Big(|\psi\rangle\,,\, |\phi\rangle\Big)+\Big(|\psi\rangle\,,\, |\chi\rangle\Big)$
- $\Big(|\psi\rangle\,,\, c|\phi\rangle\Big)=c\Big(|\psi\rangle\,,\, |\phi\rangle\Big)$ for any $c\in\mathbb{C}$
- $\Big(|\psi\rangle\,,\, |\psi\rangle\Big)\rangle \geq 0$, equality holds iff $|\psi\rangle=\mathbf{0}$
Note:
- In the dual vector space, $\langle\, | \,\rangle$ is bilinear. In contrast, here the inner product $(,)$ is linear in its second argument and antilinear in its first argument.
- The definition of inner product leads to the projection theorem, which is the theoretical foundation of linear regression.
Orthonormal Basis
There is a specific basis $|\epsilon_1\rangle, \cdots, |\epsilon_n\rangle$ in the inner product space, named orthonormal basis, that satisfies the property \begin{equation}\Big(|\epsilon_i\rangle\,,\, |\epsilon_j\rangle \Big)=\delta_{ij}\quad\quad\quad\text{for }i, j=1, \cdots, n\,.\tag{1}\end{equation} Once an inner product is introduced, such orthonormal basis can always be constructed from any basis by Gram–Schmidt process for example.
For any two vectors $|\psi\rangle$ and $|\phi\rangle$ that has expansions in the orthonormal basis as $|\psi\rangle=\sum_{i=1}^n\psi_i|\epsilon_i\rangle$ and $|\phi\rangle=\sum_{i=1}^n\phi_i|\epsilon_i\rangle$, their inner product can be computed as \begin{equation}\Big(|\phi\rangle\,,|\psi\rangle \Big)=\sum_{i,j=1}^n \Big(\phi_i|\epsilon_i\rangle\,,\, \psi_j|\epsilon_j\rangle \Big)=\sum_{i,j=1}^n\phi^*_i\psi_j \Big(|\epsilon_i\rangle\,,|\epsilon_j\rangle \Big)=\sum_{i}^n \phi^*_i\psi_i\,,\tag{2}\end{equation} where we use the first three axioms of inner product as well as the property (1) of orthonormal basis. The fourth axiom is also satisfied since \begin{equation} \Big(|\psi\rangle\,,|\psi\rangle \Big)=\sum_{i=1}^n\left|\psi_i\right|^2\geq 0\,.\tag{3}\end{equation} Eq. (3) is the reason that we require the inner product is antilinear in one of the two arguments.
Bra in Quantum Mechanics
We can view inner product as a linear map \begin{equation}\Big(|\psi\rangle\,,\,\cdot\Big):\quad V\rightarrow \mathbb{C}\end{equation}which maps any vector $|\phi\rangle\in V$ to a complex number $\Big(|\psi\rangle,|\phi\rangle\Big)\in \mathbb{C}$. Such linear maps are automatically equipped with addition and scaling operations from the inner product: \begin{eqnarray} \Big(|\psi\rangle+|\phi\rangle\,,\,\cdot\Big)&=&\Big(|\psi\rangle\,,\,\cdot\Big)+\Big(|\phi\rangle\,,\,\cdot\Big)\\ \Big(c\,|\psi\rangle\,,\,\cdot\Big)&=&c^*\,\Big(|\psi\rangle\,,\,\cdot\Big)\end{eqnarray} so that they form a dual vector space.
Since $\Big(|\psi\rangle\,,\,\cdot\Big)$ is an element in the dual vector space, we are able to denote it by the bra \begin{equation}\color{red}{\langle \psi| \equiv \Big(|\psi\rangle\,,\,\cdot\Big)}\,\tag{4}\end{equation} Note:- An inner product space $V$ is naturally isomorphic to its dual space $V^*$. Eq. (4) provides a one-to-one correspondence between each $|\psi\rangle\in V$ and each $\langle \psi|\in V^*$, which does not reply on any basis.
- In quantum mechanics, we can represent a quantum state in both ket $|\psi\rangle$ and bra $\langle\psi|$. Mathematically, ket $|\psi\rangle$ is in the vector space while bra $\langle\psi|$ is in the dual space as shown in Eq. (4).
- An element in the dual space of a general vector space can only be denoted by $\langle f|$ in which $f$ represents a linear map, as shown in previous post. Only the element in the dual space of an inner product space is able to be denoted by $\langle \psi|$ in which $\psi$ is the same symbol as used in the vector $|\psi\rangle$.
Remarks:
- Using the bra notation (4), we can rewrite the orthonormal relation (1) as \begin{equation}\langle \epsilon_i|\epsilon_j\rangle=\delta_{ij}\,.\tag{5}\end{equation}
- Using the bra notation (4), we can rewrite Eq. (2) as \begin{equation}\langle \phi|\psi\rangle=\sum_{i}^n \phi^*_i\psi_i\,,\tag{6}\end{equation} from which we can view \begin{equation}\langle \phi| = \left[\phi^*_1, \phi^*_2, \cdots, \phi^*_n\right]\,,\quad\quad\quad |\psi\rangle =\left[\begin{array}{c}\psi_1\\ \psi_2\\ \vdots \\ \psi_n\end{array}\right]\,\end{equation} as a row and a column within the orthonormal basis. $\langle \phi| \psi\rangle$ is just the ordinary matrix multiplication between a row and a column.
We can view inner product as a linear map \begin{equation}\Big(|\psi\rangle\,,\,\cdot\Big):\quad V\rightarrow \mathbb{C}\end{equation}which maps any vector $|\phi\rangle\in V$ to a complex number $\Big(|\psi\rangle,|\phi\rangle\Big)\in \mathbb{C}$. Such linear maps are automatically equipped with addition and scaling operations from the inner product: \begin{eqnarray} \Big(|\psi\rangle+|\phi\rangle\,,\,\cdot\Big)&=&\Big(|\psi\rangle\,,\,\cdot\Big)+\Big(|\phi\rangle\,,\,\cdot\Big)\\ \Big(c\,|\psi\rangle\,,\,\cdot\Big)&=&c^*\,\Big(|\psi\rangle\,,\,\cdot\Big)\end{eqnarray} so that they form a dual vector space.
Since $\Big(|\psi\rangle\,,\,\cdot\Big)$ is an element in the dual vector space, we are able to denote it by the bra \begin{equation}\color{red}{\langle \psi| \equiv \Big(|\psi\rangle\,,\,\cdot\Big)}\,\tag{4}\end{equation} Note:
- An inner product space $V$ is naturally isomorphic to its dual space $V^*$. Eq. (4) provides a one-to-one correspondence between each $|\psi\rangle\in V$ and each $\langle \psi|\in V^*$, which does not reply on any basis.
- In quantum mechanics, we can represent a quantum state in both ket $|\psi\rangle$ and bra $\langle\psi|$. Mathematically, ket $|\psi\rangle$ is in the vector space while bra $\langle\psi|$ is in the dual space as shown in Eq. (4).
- An element in the dual space of a general vector space can only be denoted by $\langle f|$ in which $f$ represents a linear map, as shown in previous post. Only the element in the dual space of an inner product space is able to be denoted by $\langle \psi|$ in which $\psi$ is the same symbol as used in the vector $|\psi\rangle$.
Remarks:
- Using the bra notation (4), we can rewrite the orthonormal relation (1) as \begin{equation}\langle \epsilon_i|\epsilon_j\rangle=\delta_{ij}\,.\tag{5}\end{equation}
- Using the bra notation (4), we can rewrite Eq. (2) as \begin{equation}\langle \phi|\psi\rangle=\sum_{i}^n \phi^*_i\psi_i\,,\tag{6}\end{equation} from which we can view \begin{equation}\langle \phi| = \left[\phi^*_1, \phi^*_2, \cdots, \phi^*_n\right]\,,\quad\quad\quad |\psi\rangle =\left[\begin{array}{c}\psi_1\\ \psi_2\\ \vdots \\ \psi_n\end{array}\right]\,\end{equation} as a row and a column within the orthonormal basis. $\langle \phi| \psi\rangle$ is just the ordinary matrix multiplication between a row and a column.
No Explicit Form of Inner Product in Quantum Mechanics
It looks like we can compute the inner product between any of the two states $|\phi\rangle$ and $|\psi\rangle$ in quantum mechanics by Eq. (6). But note that in order to apply Eq. (6), we need to first find out an orthonormal basis and then expand the states within this orthonormal basis.
The problem is that there has been no explicit form of inner product in quantum mechanics! We just assume its existence. Mathematically, there is no way to construct a concrete orthonormal basis without knowing an explicit form of inner product, and thus we are not able to get the needed $\phi_i$ and $\psi_i$ in Eq. (6).
The solution is beyond pure mathematics and requires additional physics arguments. This is where quantum mechanics postulates are used. Quantum mechanics postulates tell us that if we measure an observable, the set of resulting quantum states of all the possible outcomes of this measurement form an orthonormal basis. So orthonormal basis in quantum mechanics are not from mathematics but from physics!
For example, if we measure the spin of an electron, there are two possible outcomes: either up or down. Consequently, the electron's state will become either $|\uparrow\rangle$ or $|\downarrow\rangle$ after the measurement. Then according to quantum mechanics postulates, $|\uparrow\rangle$ and $|\downarrow\rangle$ form an orthonormal basis, i.e. $\langle \uparrow|\uparrow\rangle = \langle \downarrow|\downarrow\rangle=1$ and $\langle \uparrow|\downarrow\rangle=0$. Know such relations between $|\uparrow\rangle$ and $|\downarrow\rangle$ is enough to do some meaningful calculations on the electron's spin dynamics, even though we still have no idea on the explicit form of the underlying inner product!
It looks like we can compute the inner product between any of the two states $|\phi\rangle$ and $|\psi\rangle$ in quantum mechanics by Eq. (6). But note that in order to apply Eq. (6), we need to first find out an orthonormal basis and then expand the states within this orthonormal basis.
The problem is that there has been no explicit form of inner product in quantum mechanics! We just assume its existence. Mathematically, there is no way to construct a concrete orthonormal basis without knowing an explicit form of inner product, and thus we are not able to get the needed $\phi_i$ and $\psi_i$ in Eq. (6).
The solution is beyond pure mathematics and requires additional physics arguments. This is where quantum mechanics postulates are used. Quantum mechanics postulates tell us that if we measure an observable, the set of resulting quantum states of all the possible outcomes of this measurement form an orthonormal basis. So orthonormal basis in quantum mechanics are not from mathematics but from physics!
For example, if we measure the spin of an electron, there are two possible outcomes: either up or down. Consequently, the electron's state will become either $|\uparrow\rangle$ or $|\downarrow\rangle$ after the measurement. Then according to quantum mechanics postulates, $|\uparrow\rangle$ and $|\downarrow\rangle$ form an orthonormal basis, i.e. $\langle \uparrow|\uparrow\rangle = \langle \downarrow|\downarrow\rangle=1$ and $\langle \uparrow|\downarrow\rangle=0$. Know such relations between $|\uparrow\rangle$ and $|\downarrow\rangle$ is enough to do some meaningful calculations on the electron's spin dynamics, even though we still have no idea on the explicit form of the underlying inner product!
Appendix
I found a claim presented in the book "Geometry, Topology and Physics" to construct an explicit form of inner product through the dual space. In the section 1.2.1 of the book, the author proposed the following recipe:- Fix a basis $|\alpha_1\rangle, \cdots, |\alpha_n\rangle$ in $V$ and its dual basis $\langle b_1|, \cdots, \langle b_n|$ in $V^*$.
- Make expansions of two vectors $|\psi\rangle$ and $|\phi\rangle$ within the basis: $|\psi\rangle=\sum_{i=1}^n\psi_i|\alpha_i\rangle$ and $|\phi\rangle=\sum_{i=1}^n\phi_i|\alpha_i\rangle$.
- For each $|\phi\rangle$, construct a corresponding linear map through unnatural isomorphism: $\langle f_{\phi}|\equiv \sum_{i=1}^n \phi^*_i\langle b_i|$.
- Define the inner product $\Big(|\phi\rangle, \psi\rangle\Big)$ as $\Big(|\phi\rangle, \psi\rangle\Big)\equiv \langle f_{\phi}|\psi\rangle=\sum_{i=1}^n \phi^*_i\psi_i$.
I believe this approach is problematic for at least three reasons:- Since every vector space has a dual space, any vector space would automatically become an inner product space once following this recipe, which is clearly wrong.
- In quantum mechanics, we are free to change basis. But unnatural isomorphism used in the third step does not hold when making basis transformations.
- The result in the fourth step indicates that $|\psi\rangle$ and $|\phi\rangle$ are expanded within orthonormal basis. Indeed, if we just take $|\phi\rangle$ as $|\alpha_i\rangle$ and $|\psi\rangle$ as $|\alpha_j\rangle$, then the fourth step gives $\langle \alpha_i| \alpha_j\rangle=\delta_{ij}$. However, in the first step, there are no such assumptions on the basis $|\alpha_i\rangle$. For any basis, we are able to find its dual basis: we did a lot of exercises in finding the reciprocal lattice when learning solid state physics! Let alone the orthonormal concept should come after inner product.
This problematic recipe tries to first define the bra $\langle\,|$ by the unnatural isomorphism (the third step) and then define an inner product by the dual space (the fourth step). I believe the correct logic is the opposite: we should first introduce an inner product and then define the bra by the inner product as in Eq. (4).
I found a claim presented in the book "Geometry, Topology and Physics" to construct an explicit form of inner product through the dual space. In the section 1.2.1 of the book, the author proposed the following recipe:
- Fix a basis $|\alpha_1\rangle, \cdots, |\alpha_n\rangle$ in $V$ and its dual basis $\langle b_1|, \cdots, \langle b_n|$ in $V^*$.
- Make expansions of two vectors $|\psi\rangle$ and $|\phi\rangle$ within the basis: $|\psi\rangle=\sum_{i=1}^n\psi_i|\alpha_i\rangle$ and $|\phi\rangle=\sum_{i=1}^n\phi_i|\alpha_i\rangle$.
- For each $|\phi\rangle$, construct a corresponding linear map through unnatural isomorphism: $\langle f_{\phi}|\equiv \sum_{i=1}^n \phi^*_i\langle b_i|$.
- Define the inner product $\Big(|\phi\rangle, \psi\rangle\Big)$ as $\Big(|\phi\rangle, \psi\rangle\Big)\equiv \langle f_{\phi}|\psi\rangle=\sum_{i=1}^n \phi^*_i\psi_i$.
I believe this approach is problematic for at least three reasons:
- Since every vector space has a dual space, any vector space would automatically become an inner product space once following this recipe, which is clearly wrong.
- In quantum mechanics, we are free to change basis. But unnatural isomorphism used in the third step does not hold when making basis transformations.
- The result in the fourth step indicates that $|\psi\rangle$ and $|\phi\rangle$ are expanded within orthonormal basis. Indeed, if we just take $|\phi\rangle$ as $|\alpha_i\rangle$ and $|\psi\rangle$ as $|\alpha_j\rangle$, then the fourth step gives $\langle \alpha_i| \alpha_j\rangle=\delta_{ij}$. However, in the first step, there are no such assumptions on the basis $|\alpha_i\rangle$. For any basis, we are able to find its dual basis: we did a lot of exercises in finding the reciprocal lattice when learning solid state physics! Let alone the orthonormal concept should come after inner product.
This problematic recipe tries to first define the bra $\langle\,|$ by the unnatural isomorphism (the third step) and then define an inner product by the dual space (the fourth step). I believe the correct logic is the opposite: we should first introduce an inner product and then define the bra by the inner product as in Eq. (4).
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