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title: Vector spaces in quantum mechanics
---
# Vector spaces in quantum mechanics
# 4. Vector spaces in quantum mechanics
The lecture on vector spaces in quantum mechanics consists of the following parts:
- [Dirac notation and Hilbert spaces](#dirac-notation-and-hilbert-spaces)
- [4.1. Dirac notation and Hilbert spaces](#41-dirac-notation-and-hilbert-spaces)
- [Inner product of state vectors](#inner-product-of-state-vectors)
- [4.2. Inner product of state vectors](#42-inner-product-of-state-vectors)
- [Matrix representation of ket and bra vectors](#matrix-representation-ket-bra-vectors)
- [4.3. Matrix representation of ket and bra vectors](#43-matrix-representation-ket-and-bra-vectors)
- [A two-dimensional Hilbert space](#two-dimensional-hilbert-space)
- [4.4. A two-dimensional Hilbert space](#44-a-two-dimensional-hilbert-space)
and at the end of the lecture one can find the corresponding [Problems](#problems)
and at the end of the lecture there is a set of exercises
- [4.5. Problems](#45-problems)
---
The contents of this lecture are summarised in the following **videos**:
- [4_vector_spaces_QM_video1](https://www.dropbox.com/s/mnccmpff33pre9r/linear_algebra_04.mov?dl=0)
- [1. Dirac notation and properties of Hilbert spaces](https://www.dropbox.com/s/mnccmpff33pre9r/linear_algebra_04.mov?dl=0)
- [4_vector_spaces_QM_video2](https://www.dropbox.com/s/709bh9j083y7d0s/linear_algebra-06.mov?dl=0)
- [2. Algebra with Dirac notation - bras and kets](https://www.dropbox.com/s/709bh9j083y7d0s/linear_algebra-06.mov?dl=0)
- [4_vector_spaces_QM_video3](https://www.dropbox.com/s/k9plspkonnk3nc0/linear_algebra-07.mov?dl=0)
- [3. Finding expansion coefficients for Dirac notation](https://www.dropbox.com/s/k9plspkonnk3nc0/linear_algebra-07.mov?dl=0)
**Total length of the videos: ~14 minutes**
## Dirac notation and Hilbert spaces
---
## 4.1. Dirac notation and Hilbert spaces
In the previous lecture we reviewed the basic properties of linear vector spaces. We now discuss how the same formalism
In the previous lecture, we reviewed the basic properties of linear vector spaces. Next, we will discuss how the same formalism
can be applied to describe physical states in quantum mechanics.
The state of a physical system in quantum mechanics is represented by a vector belonging to a *complex vector space*.
This vector is known as the *state space* of the system.
Such a physical state of a quantum system is represented by a symbol $|~~\rangle$, known as a *ket*.
This notation is known as the *Dirac notation*, and it is very prominent in the description of quantum mechanics. Note that a *ket* is also refered to as a state vector, *ket* vector, or just state.
This vector space is known as the *state space* of the system.
### Ket
!!! info "Ket"
A physical state of a quantum system is represented by a symbol $$|~~\rangle$$ known as a **ket**.
This notation is known as the *Dirac notation*, and it is very prominent in the description of quantum mechanics.
Note that a *ket* is also referred to as a state vector, *ket* vector, or just a state.
### Hilbert space
The set of all possible state vectors describing a given physical system forms a complex vector space $\mathcal{H}$, which is known as the *Hilbert space* of the system. You can think of the Hilbert space as the space populated by all possible states that a quantum system can be found on. Hilbert spaces inherit a number of the important properties of general vector spaces:
- A linear combination (or superposition) of two or more state vectors $|{\psi_1}\rangle, |{\psi_2}\rangle, |{\psi_3}\rangle,... |{\psi_n}\rangle$, is also a state of the quantum system. Therefore, a linear combination $|{\Psi}\rangle$ of the form
$$|{\Psi}\rangle=c_1|{\psi_1}\rangle+c_2|{\psi_1}\rangle+c_3|{\psi_3}\rangle+...+c_n|{\psi_n}\rangle = \sum_{i=1}^n c_i|{\psi_i}\rangle $$
where $c_1, c_2, c_3, ...$ are general complex numbers will also be a physically allowed state vector of the quantum system.
!!! info "Superposition"
A linear combination (or superposition) of two or more state vectors $|{\psi_1}\rangle, |{\psi_2}\rangle, |{\psi_3}\rangle,... |{\psi_n}\rangle$, is also a state of the quantum system. Therefore, a linear combination $|{\Psi}\rangle$ of the form $$|{\Psi}\rangle=c_1|{\psi_1}\rangle+c_2|{\psi_1}\rangle+c_3|{\psi_3}\rangle+...+c_n|{\psi_n}\rangle = \sum_{i=1}^n c_i|{\psi_i}\rangle$$
where $c_1, c_2, c_3, ...$ are general complex numbers will also be a physically allowed state vector of the quantum system.
- If a physical state of the system is given by a vector $|{\Psi}\rangle$, then the same physical state can also be represented by the vector $c|{\Psi}\rangle$ where $c$ is a non-zero complex number. The reason for this is that the overall normalisation of the state vector *does not change the physics* of the system (or in other words, does not modify the *information content* of the state vector). As we will discuss below, in quantum mechanics it is advantageous to work with *normalised vectors*, that is, whose *length* is one.
We will define in a while what do we mean by length.
- A set of vectors $|{\psi_1}\rangle, |{\psi_2}\rangle, |{\psi_3}\rangle,... |{\psi_n}\rangle$ is said to be *complete* if every state of the quantum system can be represented as a linear combination of them.
In such a case, it becomes possible to express *any* state vector $|{\Psi}\rangle$ of the system's Hilbert space as a superposition of these $n$ vectors,
$$ |{\Psi}\rangle=\sum_{i=1}^n c_i|{\psi_i}\rangle$$
for some specific choice of coefficients $c_i$. The set of vector \{$|{\psi_i}\rangle$\} are then said to *span* the Hilbert space of the quantum system.
!!! info "Normalisation"
If a physical state of the system is given by a vector $|{\Psi}\rangle$, then the same physical state can also be represented by the vector $c|{\Psi}\rangle$ where $c$ is a non-zero complex number. The reason for this is that the overall normalisation of the state vector *does not change the physics* of the system (or in other words, does not modify the *information content* of the state vector). As we will discuss below, in quantum mechanics it is advantageous to work with *normalised vectors*, that is, whose *length* is one.
We will define in a while what do we mean by length.
!!! info "Completeness"
A set of vectors $|{\psi_1}\rangle, |{\psi_2}\rangle, |{\psi_3}\rangle,... |{\psi_n}\rangle$ is said to be *complete* if every state
of the quantum system can be represented as a linear combination of them.
In such a case, it becomes possible to express *any* state vector $|{\Psi}\rangle$ of the system's Hilbert space as a superposition of these $n$ vectors,
$$ |{\Psi}\rangle=\sum_{i=1}^n c_i|{\psi_i}\rangle$$
for some specific choice of coefficients $c_i$. The set of vector \{$|{\psi_i}\rangle$\} are then said to *span* the Hilbert space of the quantum system.
- A set of vectors \{$|{\psi_i}\rangle$\} is said to form a basis for the state space if the set of vectors is *complete* and if in addition they are *linearly independent*. The latter condition means essentially that one cannot express a given basis vector as a linear combination of the rest of basis vectors.
Linear independence can also be expressed as the requirement that if one has that
$$\sum_{i=1}^n c_i |{\psi_i}\rangle=0\;\text{then}\; c_i=0\;\text{for all}\; i$$
!!! info "Basis"
A set of vectors \{$|{\psi_i}\rangle$\} is said to form a basis for the state space if the set of vectors is *complete* and if in addition they are *linearly independent*. The latter condition means essentially that one cannot express a given basis vector as a linear combination of the rest of basis vectors.
Linear independence can also be expressed as the requirement that if one has that
$$\sum_{i=1}^n c_i |{\psi_i}\rangle=0\;\text{then}\; c_i=0\;\text{for all}\; i$$
- The minimum number of vectors needed to form a complete set of basis states is known as the *dimensionality* of the state space. In quantum mechanis you will encounter systems whose Hilbert spaces have very different dimensionality, from the spin-1/2 particle (a $n=2$ vector space) to the free particle (whose state vectors live in an infinite vector space).
!!! info "Dimensionality"
The minimum number of vectors needed to form a complete set of basis states is known as the *dimensionality* of the state space. In quantum mechanics you will encounter systems whose Hilbert spaces have very different dimensionality, from the spin-1/2 particle (a $n=2$ vector space) to the free particle (whose state vectors live in an infinite vector space).
### Bra vectors
We need now to extend a bit the Dirac notation for elements of this vector space. We need to introduce a quantity $\langle{\Psi}|$, known as a *bra vector*,
which represents the *complex conjugates* of the corresponding ket vector. Bra vectors are elements of the vector space $\mathcal{H}^{*}$, which is called the *dual space* of the original Hilbert space $\mathcal{H}$.
We need now to extend the Dirac notation to describe other elements of this vector space. We need to introduce a quantity $\langle{\Psi}|$, known as a *bra vector*, which represents the *complex conjugates* of the corresponding ket vector. Bra vectors are elements of the vector space $\mathcal{H}^{*}$, called the *dual space* of the original Hilbert space $\mathcal{H}$.
If a ket vector is given by $| \Psi\rangle= c_1 |\psi_1\rangle+c_2|\psi_2\rangle$, then the corresponding bra vector will be given by
$$\langle{\Psi}|= c_1^*\langle{\psi_1}|+c_2^*\langle{\psi_2}| \, .$$
As mentioned above, the vector space spanned by all bra vectors $\langle{\Psi}|$ is referred to as the dual space and is represented by $\mathcal{H}^*$. For each ket vector belonging to $\mathcal{H}$, there will exist an associated bra vector belonging to the dual space $\mathcal{H}^*$.
!!! info "Bra vector"
If a ket vector is given by $$| \Psi\rangle= c_1 |\psi_1\rangle+c_2|\psi_2\rangle \, ,$$
then the corresponding bra vector will be given by
$$\langle{\Psi}|= c_1^*\langle{\psi_1}|+c_2^*\langle{\psi_2}| \, .$$
Below we will further discuss the concept of bra vectors when presenting the matrix representation of elements of the Hilbert space.
As mentioned above, the vector space spanned by all bra vectors $\langle{\Psi}|$ is referred to as the dual space and is represented by $\mathcal{H}^*$. For each ket vector belonging to $\mathcal{H}$, there will exist an associated bra vector belonging to the dual space $\mathcal{H}^*$.
Below, we will further discuss the concept of bra vectors when presenting the matrix representation of elements of the Hilbert space.
## Inner product of state vectors
## 4.2. Inner product of state vectors
Assume that $|{\psi}\rangle$ and $|{\phi}\rangle$ are any two state vectors belonging to the
state (Hilbert) space $\mathcal{H}$, then we can define the *inner product*
between them, $\langle{\psi}|{\phi}\rangle$, as follows. This inner product in quantum mechanics is the analog of the
usual scalar product that one encounters in vector spaces and that we reviewed in the previous lecture. As in usual vector spaces, the inner product of two state vectors is an *scalar*, in this case a complex number in general.
between them, $\langle{\psi}|{\phi}\rangle$, as follows.
The inner product in quantum mechanics is the analog of the usual scalar product that one encounters in vector spaces, and which we reviewed in the previous lecture. As in usual vector spaces, the inner product of two state vectors is a *scalar* and in this case a complex number in general.
The value of the inner product $\langle{\psi}|{\phi}\rangle$ indicates the *probability amplitude* (not the probability) of measuring a system characterised by the state $|{\phi}\rangle$ to be in the state $|{\psi}\rangle$. This inner product can also be understood as measuring the *overlap* between the state vectors $|{\psi}\rangle$ and $|{\phi}\rangle$. Then the *probability* of observing the system to be in the state $|\psi\rangle$ given that it is in the state $|\phi\rangle$ will be given by $|\langle \psi | \phi \rangle|^2$. Since the latter quantity is a probability, we know that it should satisfy the condition that $0 \le |\langle \psi | \phi \rangle|^2 \le 1$.
!!! tip "Interpretation of the inner product in quantum mechanics"
1. The value of the inner product $\langle{\psi}|{\phi}\rangle$ indicates the **probability amplitude** (not the probability) of measuring a system, which characterised by the state $|{\phi}\rangle$, to be in the state $|{\psi}\rangle$.
2. This inner product can also be understood as measuring the **overlap** between the state vectors $|{\psi}\rangle$ and $|{\phi}\rangle$.
3. Then the **probability of observing the system to be in the state $|\psi\rangle$** given that it is in the state $|\phi\rangle$ will be given by $$|\langle \psi | \phi \rangle|^2 \, .$$ Since the latter quantity is a probability, we know that it should satisfy the condition that
$$0 \le |\langle \psi | \phi \rangle|^2 \le 1 \, .$$
### Properties of the inner product
The inner product (probability amplitude) $\langle \psi | \phi \rangle$ exhibits the following properties:
- *Complex conjugate*: $\langle \psi | \phi \rangle=\langle \phi | \psi \rangle^*$
- *Distributivity and associativity*: $\langle \psi |\{c_1 |\phi_1\rangle+c_2 |\phi_2 \rangle\}=c_1\langle \psi | \phi_1\rangle+c_2\langle \psi | \phi_2\rangle$
- *Positivity*: $\langle \psi | \psi \rangle\geq0$. If $\langle \psi | \psi \rangle = 0$ then this implies that the state vector $|\psi\rangle=0$ is the null element of the Hilbert space.
- *Orthogonality*: two states $|\psi \rangle$ and $|\phi \rangle$ are said to be *orthogonal* if $\langle \psi | \phi\rangle=0$. By analogy with regular vector spaces, we can think of these two state vectors $|\psi \rangle$ and $|\phi \rangle$ as being *perpendicular* to each other. Note that for a quantum system occupying a certain state, there is a vanishing probability of it being observed in a state orthogonal to it.
The quantity $\sqrt{\langle \psi | \psi \rangle}$ is known as the *length* or the *norm* of the state vector $|\psi\rangle$. You can see from the properties of complex algebra that this length must be a real number.
A physically valid state $|\psi \rangle$ must be normalized to unity, that is $\langle \psi | \psi \rangle=1$. Note that a state that cannot be normalized to unity cannot represent a physically acceptable state.
A set of orthonormal basis vectors $\{|\psi_i\rangle\text{;}\; i=1,2,3,...,n\}$ will have the property $\langle \psi_i |\psi_j \rangle=\delta_{ij}$ where $\delta_{ij}$ is a mathematical symbol known as the *Kronecker delta*, which equals unity if $i=j$ and zero if $i\neq j$.
!!! info "Properties of the inner product"
1. **Complex conjugate:** $\langle \psi | \phi \rangle=\langle \phi | \psi \rangle^*$
2. **Distributivity and associativity:** $\langle \psi |\{c_1 |\phi_1\rangle+c_2 |\phi_2 \rangle\}=c_1\langle \psi | \phi_1\rangle+c_2\langle \psi | \phi_2\rangle$
3. **Positivity:** $\langle \psi | \psi \rangle\geq0 \, .$
If $\langle \psi | \psi \rangle = 0$ then, this implies that the state vector $|\psi\rangle=0$ is the null element of the Hilbert space.
4. **Orthogonality:** Two states $|\psi \rangle$ and $|\phi \rangle$ are said to be *orthogonal* if $\langle \psi | \phi\rangle=0 \, .$
By analogy with regular vector spaces, we can think of these two state vectors $|\psi \rangle$ and $|\phi \rangle$ as being *perpendicular* to each other. Note that for a quantum system occupying a certain state, there is a vanishing probability of it being observed in a state orthogonal to it.
5. **Norm:** The quantity $\sqrt{\langle \psi | \psi \rangle}$ is known as the *length* or the *norm* of the state vector $|\psi\rangle$.
You can see from the properties of complex algebra that this length must be a real number. A physically valid state $|\psi \rangle$ must be normalized to unity, that is $\langle \psi | \psi \rangle=1$. Note that a state that cannot be normalized to unity does not represent a physically acceptable state.
6. **Orthonormality:** A set of orthonormal basis vectors $\{|\psi_i\rangle\text{;}\; i=1,2,3,...,n\}$ will have the property $\langle \psi_i |\psi_j \rangle=\delta_{ij}$ where $\delta_{ij}$ is a mathematical symbol known as the *Kronecker delta*, which equals unity if $i=j$ and zero if $i\neq j$.
From all the above conditions we see that a Hilbert space is a so-called *complex inner product space*, which is nothing but a complex vector space equipped with a inner product. All the vectors belonging to a Hilbert space $\mathcal{H}$ have a finite norm, that is they can be normalized to unity. This normalisation condition is essential is we are to apply the probabilistic interpretation of the state vectors described above.
From all the above conditions, we see that a Hilbert space is a so-called *complex inner product space*, which is nothing else but a complex vector space equipped with a inner product. All the vectors belonging to a Hilbert space $\mathcal{H}$ have a finite norm, which means that they can be normalized to unity. This normalisation condition is essential is we are to apply the probabilistic interpretation of the state vectors described above.
## Matrix representation of ket and bra vectors
## 4.3. Matrix representation of ket and bra vectors
As we have discussed, in quantum mechanics a general state vector $|\psi\rangle$ can be represented in terms of the basis vectors, $\{|\phi_i\rangle;i=1,2,...,n\}$, as follows
$$
|\psi\rangle=\sum_{i=1}^n c_i |\phi_i\rangle \, ,
$$
As we have discussed, in quantum mechanics a general state vector $|\psi\rangle$ can be represented in terms of the basis vectors, $\{|\phi_i\rangle;i=1,2,...,n\}$, as
$$ |\psi\rangle=\sum_{i=1}^n c_i |\phi_i\rangle $$
for some values of the complex coefficients $\{ c_i\}$. To determine the values of these coefficients, we can take the inner product between the bra basis vector $\langle \phi_j|$ and the ket state vector $|\psi\rangle$ and use the orthogonality properties of the basis vectors:
$$
\langle \phi_j|\psi\rangle = \langle \phi_j|\sum_{i=1}^n c_i |\phi_i\rangle = \sum_{i=1}^n c_i\langle \phi_j|\phi_i\rangle = \sum_{i=1}^n c_i\delta_{ij} = c_j \, .
$$
$$ \langle \phi_j|\psi\rangle = \langle \phi_j|\sum_{i=1}^n c_i |\phi_i\rangle = \sum_{i=1}^n c_i\langle \phi_j|\phi_i\rangle = \sum_{i=1}^n c_i\delta_{ij} = c_j \, .$$
Therefore, if we now denote the coefficients $\{ c_i\}$ of the state vector $|\psi\rangle$ by $\{ \psi_i\}$, we have the expansion
$$
|\psi\rangle=\sum_{i=1}^n \psi_i |\phi_i\rangle= \sum_{i=1}^n \left( \langle \phi_i|\psi\rangle \right) |\phi_i\rangle \, .
$$
By analogy with the Euclidean case, we can understand the coefficients $\psi_i$
as the *components* of the state vector $ |\psi\rangle$ along the $n$ directions
spanned by the basis vectors. Here note that in this notation $\psi_i$ is an *scalar* (just a number) and not a vector. Further note that, as opposed to the Euclidean space, the coefficients $\psi_i$ will be in general complex numbers.
$$ |\psi\rangle=\sum_{i=1}^n \psi_i |\phi_i\rangle= \sum_{i=1}^n \left( \langle \phi_i|\psi\rangle \right) |\phi_i\rangle \, .$$
By analogy with the Euclidean case, we can understand the coefficients $\psi_i$ as the *components* of the state vector $ |\psi\rangle$ along the $n$ directions spanned by the basis vectors. Here, note also that in this notation $\psi_i$ is an *scalar* (just a number) and not a vector. Furthermore, note that, as opposed to the Euclidean space, the coefficients $\psi_i$ will generally be complex numbers.
This analogy with the case of ordinary vectors allows us to write the state $|\psi\rangle$ as a *column vector* with respect to the set of basis vectors $\{|\phi_i\rangle;i=1,2,...,n\}$, which are kept implicit:
$$
|\psi\rangle= \begin{pmatrix} \psi_1\\\psi_2\\\psi_3\\\vdots\\\psi_n\end{pmatrix}.
$$
We can also express the basis vectors in this manner. Given that the basis vectors are *orthonormal* among them,
$$ |\psi\rangle= \begin{pmatrix} \psi_1\\\psi_2\\\psi_3\\\vdots\\\psi_n\end{pmatrix} \, . $$
We can also express the basis vectors in this manner. Given that the basis vectors are *orthonormal* among themselves,
the basis state $|\phi_i\rangle$ will have as component in the $j$ direction
$$
(\phi_i)_j=\langle \phi_j|\phi_i\rangle=\delta_{ji} \, ,
$$
$$ (\phi_i)_j=\langle \phi_j|\phi_i\rangle=\delta_{ji} \, ,$$
and thus the vector column expression of the basis vectors will be very simple
$$
|\phi_1\rangle= \begin{pmatrix} 1\\0\\0 \\\vdots\end{pmatrix} \;, \quad
|\phi_2\rangle= \begin{pmatrix} 0\\1\\0 \\\vdots\end{pmatrix} \;, \ldots
$$
Let us show how we can use the matrix representation to evaluate the inner
product (bracket) between two state vectors when expanded in terms of their components in the same basis:
$$
|\psi\rangle=\sum_{i=1}^n \psi_i |\phi_i\rangle \,, \qquad
|\chi\rangle=\sum_{i=1}^n \chi_i |\phi_i\rangle\, .
$$
First of all, we note that we can write the above expansions in the following way
$$
|\psi\rangle=\sum_{i=1}^n |\phi_i \rangle \langle \phi_i | \psi \rangle \, ,
$$
and thus we see that the basis vectors provide a very useful representation of the *identity operator*:
$$
\hat{I} = \sum_i |\phi_i\rangle \langle\phi_i| \, ,
$$
We can insert this identify operator within the bracket to evaluate the inner
product $\langle \chi|\psi\rangle$ between the two state vectors to evaluate the inner product $\langle \chi|\psi\rangle$:
$$
\langle \chi|\psi\rangle=
\langle\chi|\hat{I} |\psi\rangle=\sum_{i=1}^n \langle\chi| \phi_i \rangle \langle\phi_i|\psi\rangle \, .
$$
Next, using that $\chi_i = \langle \phi_i|\chi \rangle$ are the components of the
state vector $|\chi\rangle$ and that $\langle \chi| \phi_n \rangle=(\langle\phi_i|\chi\rangle)^*$,
we have that $\langle \chi |\phi_i\rangle =\chi_i^*$
and therefore the inner product of the two state vectors $|\psi\rangle$
and $|\chi\rangle$ can be expressed in terms of their components
as follows
$$\langle\chi|\psi\rangle=\sum_{i=1}^n\chi_i^*\psi_i.$$
which in the matrix representation of state vectors can also be written as
$$\langle \chi|\psi\rangle=\begin{pmatrix} \chi^*_1 , \chi^*_2 &,\ldots \end{pmatrix}\begin{pmatrix} \psi_1 \\ \psi_2 \\ \vdots \end{pmatrix} \, .$$ Therefore, we can present bra vector $\langle \chi|$ as row vectors and ket vectors as column vector.
The row vector can thus be treated as the *complex conjugate* of the corresponding column vector.
## A two-dimensional Hilbert space
$$ |\phi_1\rangle= \begin{pmatrix} 1\\0\\0 \\\vdots\end{pmatrix} \;, \quad |\phi_2\rangle= \begin{pmatrix} 0\\1\\0 \\\vdots\end{pmatrix} \;, \ldots $$
!!! note "Evaluating the inner product"
Let us show how we can use the matrix representation to evaluate the inner product (bracket) between two state vectors when expanded in terms of their components in the same basis:
$$ |\psi\rangle=\sum_{i=1}^n \psi_i |\phi_i\rangle \,, \qquad |\chi\rangle=\sum_{i=1}^n \chi_i |\phi_i\rangle\, .$$
First of all, we note that we can write the above expansions in the following way
$$
|\psi\rangle=\sum_{i=1}^n |\phi_i \rangle \langle \phi_i | \psi \rangle \, ,
$$
and thus we see that the basis vectors provide a very useful representation of the *identity operator*:
$$
\hat{I} = \sum_i |\phi_i\rangle \langle\phi_i| \, ,
$$
We can insert this identify operator within the bracket to evaluate the inner
product $\langle \chi|\psi\rangle$ between the two state vectors to evaluate the inner product $\langle \chi|\psi\rangle$:
$$
\langle \chi|\psi\rangle=
\langle\chi|\hat{I} |\psi\rangle=\sum_{i=1}^n \langle\chi| \phi_i \rangle \langle\phi_i|\psi\rangle \, .
$$
Next, using that $\chi_i = \langle \phi_i|\chi \rangle$ are the components of the
state vector $|\chi\rangle$ and that $\langle \chi| \phi_n \rangle=(\langle\phi_i|\chi\rangle)^*$,
we have that $\langle \chi |\phi_i\rangle =\chi_i^*$
and therefore the inner product of the two state vectors $|\psi\rangle$
and $|\chi\rangle$ can be expressed in terms of their components
as follows
$$\langle\chi|\psi\rangle=\sum_{i=1}^n\chi_i^*\psi_i.$$
which in the matrix representation of state vectors can also be written as
$$\langle \chi|\psi\rangle=\begin{pmatrix} \chi^*_1 , \chi^*_2 &,\ldots \end{pmatrix}\begin{pmatrix} \psi_1 \\ \psi_2 \\ \vdots \end{pmatrix} \, .$$ Therefore, we can present bra vector $\langle \chi|$ as row vectors and ket vectors as column vector.
The row vector can thus be treated as the *complex conjugate* of the corresponding column vector.
## 4.4. A two-dimensional Hilbert space
As a practical example to illustrate the basic ideas of vector spaces applied to quantum physics presented above, we will consider a quantum system which is fundamental for quantum mechanics and its applications. This system corresponds to the possible states that the intrinsic angular momentum of an electron, known as *spin*, can occupy. As you will see in following courses, the Hilbert space for the electron spin has dimension $n=2$, meaning that we can found an electron *pointing* either in the up direction, denoted by $|+\rangle$, or the down direction, denoted by $|-\rangle$.
@@ -196,11 +204,13 @@ $$
$$
The values of the coefficients $c_+$ and $c_-$ for these examples above are, respectively,
$$
(c_+,c_) = (3,-2i) \, ,\qquad
(c_+,c_-) = (3,-2i) \, ,\qquad
(c_+,c_-) = (i,-4) \, ,\qquad
(c_+,c_-) = (2,5) \, .\qquad
$$
Note however that many other bases are possible, and that the physics of a quantum system do not depend on the basis that we choose.
!!! warning ""
Note however that many other bases are possible, and that the physics of a quantum system do not depend on the basis that we choose.
The bra vectors associated to these ket vectors will be given by
$$ |{\Psi}\rangle= \left( \begin{array}{c}3 \\ -2i \end{array} \right) \, , \qquad
\langle{\Psi}|=\left( 3, 2i \right) \, ,$$
@@ -211,29 +221,30 @@ $$|{\Psi}\rangle= \left( \begin{array}{c}2 \\ 5 \end{array} \right) \, , \qquad
Note however that the above vectors are not normalised (the inner product with themselves is different from unity), and thus
cannot represent physical states. We show below an explicit example of a normalised state vector belonging to this Hilbert space.
We also know how we can evaluate the inner product between any two state vectors belonging to this Hilbert space. If we have two state vectors given by
$$
|\psi\rangle = \frac{1}{\sqrt{2}} \left( \begin{array}{c}1 \\ -i \end{array} \right) \, \quad
|\phi \rangle = \left( \begin{array}{c}0 \\1 \end{array} \right)
$$
then their inner product is
$$
\langle \psi | \phi \rangle =
\frac{1}{\sqrt{2}} \left( 1 , i \right) \left( \begin{array}{c}0 \\1 \end{array} \right) =
\frac{i}{\sqrt{2}}
$$
and the associated probability will be given by
$$
|\langle \psi|\phi\rangle|^2 = \frac{1}{2}
$$
meaning that if I measure the state $| \phi \rangle $, I will have a 50% probability
of finding it in the state $| \psi \rangle$. Recall that probabilities must always be smaller than 1 to make physical sense. Note also that I am using normalised vectors, you can check yourselves that
$$
\langle \psi |\psi \rangle = \langle \phi | \phi\rangle = 1 \, ,
$$
as required to ensure the probabilistic description of the state vector.
## Problems
!!! done "Example: Evaluating the inner product"
We also know how we can evaluate the inner product between any two state vectors belonging to this Hilbert space. If we have two state vectors given by
$$
|\psi\rangle = \frac{1}{\sqrt{2}} \left( \begin{array}{c}1 \\ -i \end{array} \right) \, \quad
|\phi \rangle = \left( \begin{array}{c}0 \\1 \end{array} \right)
$$
then their inner product is
$$
\langle \psi | \phi \rangle =
\frac{1}{\sqrt{2}} \left( 1 , i \right) \left( \begin{array}{c}0 \\1 \end{array} \right) =
\frac{i}{\sqrt{2}}
$$
and the associated probability will be given by
$$
|\langle \psi|\phi\rangle|^2 = \frac{1}{2}
$$
meaning that if I measure the state $| \phi \rangle $, I will have a 50% probability
of finding it in the state $| \psi \rangle$. Recall that probabilities must always be smaller than 1 to make physical sense. Note also that I am using normalised vectors, you can check yourselves that
$$
\langle \psi |\psi \rangle = \langle \phi | \phi\rangle = 1 \, ,
$$
as required to ensure the probabilistic description of the state vector.
## 4.5. Problems
**1)** [:grinning:] *The inner product*