I am not going to give the Postulates as it is in the books or anything, but I just want to postulate and speak its' mathematical importance, in a way I understood.
From the Classical Physics of Lagrangian and Hamiltonian, we know that any system [it can be single particle or multi particle or anything] can be associated with a function so called Lagrangian or Hamiltonian, such that all the information about the system can be extracted from this function using the corresponding Equations of motion.
Mathematically, we assume that the Lagrangian or Hamiltonian function contains all the necessary information we need to describe the system completely.
In the same way, here we assume that "Every Quantum Mechanical System is completely described by an arbitrary State Vector or a Wave function $ \vert{\psi(t)}\rangle $ , read as "ket - psi" is an element of complex linear vector space called Hilbert Space. The State vector contains all the information about the system and it changes only with time.
The state vector is an abstract concept and you can never measure this state vector or imagine it in a physical manner.
From the concept of Vector space, we assume that it is always possible to define a set of vectors which are linearly independent and forms the basis for the Vector Space.
Note: You needn't to panic on hearing the term Vector Space. Your Euclidean space follows the rules of Vector space. Whenever you get in trouble understanding vector space, you can always make a comparison with your 3 dimensional Euclidean space.
The set of basis vectors needn't to be unique, but it is always possible to represent any vector in the Vector space as a linear combination of these basis vectors.
As a consequence, you can imagine this arbitrary state vector as the linear combination of all the basis vectors.
We don't know what are these basis vectors, since there is many possible ways of choosing a set of basis vectors from different possible sets. Let us consider this as a general linear combination.
It is represented as, $$ \vert{\psi(t)}\rangle = \sum_b A_b(t) \vert{\phi_b}\rangle = A_1(t) \vert{\phi_1}\rangle + A_2(t) \vert{\phi_2}\rangle + \ldots.....eq.(1)$$
All the basis vectors are ket vectors, after all left side should be equal to right side. And we can always make the time dependence of ket vector to come into the coefficients.
We already said that, these are in abstract Hilbert space, so we cannot measure anything about them.
To measure anything, we need to make the projection of this abstract quantities in the known space where we could describe the wave function completely.
To measure the projection, we make the dot product of desired known parameter with this abstract Wave function.
So that, the wave function and all its basis vectors are now described using our desired known parameter.
For example, if the desired known parameter is position, then all the Wave function and its basis vectors will be projected into position space (where position is the parameter). And so, the new projected wave function is called "Position Space Wave function".
$$ \vert{\psi(t)}\rangle = \sum_b A_b(t) \vert{\phi_b}\rangle $$
Dotted with x to give the projection in Position space,
$$ \langle{x}\vert{\psi(t)}\rangle = \sum_b A_b(t) \langle{x}\vert{\phi_b}\rangle \,\, \ldots...eq.(2)$$
Now, the new projection of Wave function in Position space, i.e. Position Space wave function is,
$$ \psi(x,t) = \sum_b A_b(t) \phi_b(x) $$
Where $\langle{x}\vert{\psi(t)}\rangle = \psi(x,t)$ and $ \langle{x}\vert{\phi}\rangle = \phi(x)$
If we choose momentum as the desired known parameter, then we can dot momentum with the general wave function. It will result into, $$ \langle{p}\vert{\psi(t)}\rangle = \sum_b A_b(t) \langle{p}\vert{\phi_b}\rangle \,\,\ldots...eq.(3)$$
And the new wave function is called Momentum Space Wave function, $$ \psi(p,t) = \sum_b A_b(t) \phi_b(p) $$
That is all we can do with the first Postulate.
The second Postulate is stated as, "Each dynamical variable that relates to the motion of the particle can be associated with a linear operator".
An operator is called to be linear if it satisfies the condition, $$ \hat{Q}(c_1\psi_1 + c_2\psi_2) = c_1 \hat{Q}\psi_1 + c_2 \hat{Q}\psi_2 \,\,\,....\ldots.eq.(4)$$
With Each operator, it can be associated a linear eigen value equation such that $$ \hat{Q} \psi_i = \lambda_i \psi_i \,\,\,\,\ldots..eq.(5)$$
where $\psi_i $ is called the eigen state and
$\lambda_i$ is called the eigen value.
A linear operator is also an abstract concept, which is represented using a matrix. A linear operator is determined by how it acts on the basis vectors because any vector can be expanded as the linear combination of these basis vectors.
If we know how an operator acts on the basis, then it gives us everything we need to know about the operator on that Vector Space.
Let me represent the basis vectors as $$\vert{e_1}\rangle, \vert{e_2}\rangle, \ldots...$$
Therefore, $$ \hat{Q}\vert{\psi}\rangle = \hat{Q} \vert{e_1}\rangle + \hat{Q} \vert{e_2}\rangle + ... \,\,\ldots...eq.(6)$$
If we represent the linear operators with the matrix, knowing the matrix elements is knowing the operator itself.
Let me take a basis vector $\vert{e_i}\rangle$ in the Hilbert Space. To understand how a linear operator works on this basis vector, we operate it on this basis and it will result some new vector.
For example, you can consider the rotation of the coordinates as an operation that acts on the basis vectors.
Due to linear property, $\hat{Q}\vert{e_i}\rangle$ - the new vector itself can be written again as a linear combination of the basis vectors, represented as $$ Q\vert{e_i}\rangle = \sum_k Q_{kj}\vert{e_k}\rangle $$
You can compare it with the coordinate transformation rules.
Now, the third postulate says that, "Any observable in Quantum Mechanics is a linear Hermitian operator on the Hilbert space, where the eigenvalues are the only possible results of a precise measurement of that observable.
Definition of a Hermitian operator:
$$ \int \psi_i^* (\hat{Q} \psi_i)\, dx = \int (\hat{Q}\psi_i)^* \psi_i \,dx \,\,\,\ldots...eq.(7)$$
Eq.(7) which gives a special property on expanding with eigenvalue eq.(5) as follows,
$$ \int \psi^* (\lambda_i \psi_i)\,dx = \int (\lambda_i \psi_i)^* \psi_i\,dx $$
which gives, $$ \lambda_i \int \psi_i^*\psi_i \,dx = \lambda_i^* \int \psi_i^* \psi_i \,dx $$
$$(\lambda_i - \lambda_i^*) \int \psi_i^*\psi_i \,dx = 0 $$
But $\int\psi_i^*\psi_i \,dx = \int {|\psi_i|}^2 \,dx > 0 $ and it is equal to zero only when $|\psi_i| = 0 $ where wave function itself vanishes and that is not a desirable solution.
So, the only solution is $$ \lambda_i - \lambda_i^* = 0 $$ or $$ \lambda_i = \lambda_i^* $$ It is only possible when $"\lambda_i"$ is a real number. This is a characteristic result of any Hermitian operator, which states that "The eigenvalues of an Hermitian Operator is always a real number".
This is the reason why, eigenvalues of an operator is the only possible results on a precise measurement, because measurement should give a real number.
There are much more things to talk about an operator and important relations like Completeness, Orthogonality, Hermiticity of an operator and etc. It should be dealt separately.
From the Classical Physics of Lagrangian and Hamiltonian, we know that any system [it can be single particle or multi particle or anything] can be associated with a function so called Lagrangian or Hamiltonian, such that all the information about the system can be extracted from this function using the corresponding Equations of motion.
Mathematically, we assume that the Lagrangian or Hamiltonian function contains all the necessary information we need to describe the system completely.
In the same way, here we assume that "Every Quantum Mechanical System is completely described by an arbitrary State Vector or a Wave function $ \vert{\psi(t)}\rangle $ , read as "ket - psi" is an element of complex linear vector space called Hilbert Space. The State vector contains all the information about the system and it changes only with time.
The state vector is an abstract concept and you can never measure this state vector or imagine it in a physical manner.
From the concept of Vector space, we assume that it is always possible to define a set of vectors which are linearly independent and forms the basis for the Vector Space.
Note: You needn't to panic on hearing the term Vector Space. Your Euclidean space follows the rules of Vector space. Whenever you get in trouble understanding vector space, you can always make a comparison with your 3 dimensional Euclidean space.
As a consequence, you can imagine this arbitrary state vector as the linear combination of all the basis vectors.
We don't know what are these basis vectors, since there is many possible ways of choosing a set of basis vectors from different possible sets. Let us consider this as a general linear combination.
It is represented as, $$ \vert{\psi(t)}\rangle = \sum_b A_b(t) \vert{\phi_b}\rangle = A_1(t) \vert{\phi_1}\rangle + A_2(t) \vert{\phi_2}\rangle + \ldots.....eq.(1)$$
All the basis vectors are ket vectors, after all left side should be equal to right side. And we can always make the time dependence of ket vector to come into the coefficients.
We already said that, these are in abstract Hilbert space, so we cannot measure anything about them.
To measure anything, we need to make the projection of this abstract quantities in the known space where we could describe the wave function completely.
To measure the projection, we make the dot product of desired known parameter with this abstract Wave function.
So that, the wave function and all its basis vectors are now described using our desired known parameter.
For example, if the desired known parameter is position, then all the Wave function and its basis vectors will be projected into position space (where position is the parameter). And so, the new projected wave function is called "Position Space Wave function".
$$ \vert{\psi(t)}\rangle = \sum_b A_b(t) \vert{\phi_b}\rangle $$
Dotted with x to give the projection in Position space,
$$ \langle{x}\vert{\psi(t)}\rangle = \sum_b A_b(t) \langle{x}\vert{\phi_b}\rangle \,\, \ldots...eq.(2)$$
Now, the new projection of Wave function in Position space, i.e. Position Space wave function is,
$$ \psi(x,t) = \sum_b A_b(t) \phi_b(x) $$
Where $\langle{x}\vert{\psi(t)}\rangle = \psi(x,t)$ and $ \langle{x}\vert{\phi}\rangle = \phi(x)$
If we choose momentum as the desired known parameter, then we can dot momentum with the general wave function. It will result into, $$ \langle{p}\vert{\psi(t)}\rangle = \sum_b A_b(t) \langle{p}\vert{\phi_b}\rangle \,\,\ldots...eq.(3)$$
And the new wave function is called Momentum Space Wave function, $$ \psi(p,t) = \sum_b A_b(t) \phi_b(p) $$
That is all we can do with the first Postulate.
The second Postulate is stated as, "Each dynamical variable that relates to the motion of the particle can be associated with a linear operator".
An operator is called to be linear if it satisfies the condition, $$ \hat{Q}(c_1\psi_1 + c_2\psi_2) = c_1 \hat{Q}\psi_1 + c_2 \hat{Q}\psi_2 \,\,\,....\ldots.eq.(4)$$
With Each operator, it can be associated a linear eigen value equation such that $$ \hat{Q} \psi_i = \lambda_i \psi_i \,\,\,\,\ldots..eq.(5)$$
where $\psi_i $ is called the eigen state and
$\lambda_i$ is called the eigen value.
A linear operator is also an abstract concept, which is represented using a matrix. A linear operator is determined by how it acts on the basis vectors because any vector can be expanded as the linear combination of these basis vectors.
If we know how an operator acts on the basis, then it gives us everything we need to know about the operator on that Vector Space.
Let me represent the basis vectors as $$\vert{e_1}\rangle, \vert{e_2}\rangle, \ldots...$$
Therefore, $$ \hat{Q}\vert{\psi}\rangle = \hat{Q} \vert{e_1}\rangle + \hat{Q} \vert{e_2}\rangle + ... \,\,\ldots...eq.(6)$$
If we represent the linear operators with the matrix, knowing the matrix elements is knowing the operator itself.
Let me take a basis vector $\vert{e_i}\rangle$ in the Hilbert Space. To understand how a linear operator works on this basis vector, we operate it on this basis and it will result some new vector.
For example, you can consider the rotation of the coordinates as an operation that acts on the basis vectors.
Due to linear property, $\hat{Q}\vert{e_i}\rangle$ - the new vector itself can be written again as a linear combination of the basis vectors, represented as $$ Q\vert{e_i}\rangle = \sum_k Q_{kj}\vert{e_k}\rangle $$
You can compare it with the coordinate transformation rules.
Now, the third postulate says that, "Any observable in Quantum Mechanics is a linear Hermitian operator on the Hilbert space, where the eigenvalues are the only possible results of a precise measurement of that observable.
Definition of a Hermitian operator:
$$ \int \psi_i^* (\hat{Q} \psi_i)\, dx = \int (\hat{Q}\psi_i)^* \psi_i \,dx \,\,\,\ldots...eq.(7)$$
Eq.(7) which gives a special property on expanding with eigenvalue eq.(5) as follows,
$$ \int \psi^* (\lambda_i \psi_i)\,dx = \int (\lambda_i \psi_i)^* \psi_i\,dx $$
which gives, $$ \lambda_i \int \psi_i^*\psi_i \,dx = \lambda_i^* \int \psi_i^* \psi_i \,dx $$
$$(\lambda_i - \lambda_i^*) \int \psi_i^*\psi_i \,dx = 0 $$
But $\int\psi_i^*\psi_i \,dx = \int {|\psi_i|}^2 \,dx > 0 $ and it is equal to zero only when $|\psi_i| = 0 $ where wave function itself vanishes and that is not a desirable solution.
So, the only solution is $$ \lambda_i - \lambda_i^* = 0 $$ or $$ \lambda_i = \lambda_i^* $$ It is only possible when $"\lambda_i"$ is a real number. This is a characteristic result of any Hermitian operator, which states that "The eigenvalues of an Hermitian Operator is always a real number".
This is the reason why, eigenvalues of an operator is the only possible results on a precise measurement, because measurement should give a real number.
There are much more things to talk about an operator and important relations like Completeness, Orthogonality, Hermiticity of an operator and etc. It should be dealt separately.
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