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dirac_notation [2021/02/22 21:13] – created admindirac_notation [2021/02/22 21:15] (current) – [2.ii.5 Basis Independence: The Continuous Case] admin
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 Even when we are dealing with a very concrete Hilbert space like $\mathbb{R}^n$ or $\mathbb{C}^n$, the vector $|\psi \rangle$ should not be thought of as identical with a column vector Even when we are dealing with a very concrete Hilbert space like $\mathbb{R}^n$ or $\mathbb{C}^n$, the vector $|\psi \rangle$ should not be thought of as identical with a column vector
 \[\left ( \begin{array}{c} b_1 \\ b_2 \\ \vdots \\ b_n \end{array} \right ).\] \[\left ( \begin{array}{c} b_1 \\ b_2 \\ \vdots \\ b_n \end{array} \right ).\]
-Instead, the column vector is a //representation// of the vector $|\psi \rangle$ in a specific basis, in this case the standard basis $|e_1 \rangle, |e_2 \rangle, \cdots , |e_n \rangle$.  The vector $|\psi \ket$ itself is basis independent, and has different components when represented in different bases.+Instead, the column vector is a //representation// of the vector $|\psi \rangle$ in a specific basis, in this case the standard basis $|e_1 \rangle, |e_2 \rangle, \cdots , |e_n \rangle$.  The vector $|\psi \rangle$ itself is basis independent, and has different components when represented in different bases.
  
 To understand the relationship between representations in different bases, consider an orthonormal basis $|e_1 \rangle, |e_2 \rangle, \cdots $ (which may have countably infinite dimension).  We can always write $|\psi \rangle$ in this basis as To understand the relationship between representations in different bases, consider an orthonormal basis $|e_1 \rangle, |e_2 \rangle, \cdots $ (which may have countably infinite dimension).  We can always write $|\psi \rangle$ in this basis as
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 The scare quotes are because $\delta(x-x')$ is not really a function, so the value of $\langle x' | x\rangle$ is not a scalar, and hence not an inner product.  As good physicists, we will proceed as if these were well defined vectors and inner products and hope everything will be OK.  The mathematicians in the audience may wish to note that this can be made rigorous in the theory of //rigged Hilbert space//. The scare quotes are because $\delta(x-x')$ is not really a function, so the value of $\langle x' | x\rangle$ is not a scalar, and hence not an inner product.  As good physicists, we will proceed as if these were well defined vectors and inner products and hope everything will be OK.  The mathematicians in the audience may wish to note that this can be made rigorous in the theory of //rigged Hilbert space//.
  
-We can then write the position representation of the vector $|\psi\ket$ as +We can then write the position representation of the vector $|\psi\rangle$ as 
-\[|\psi \ket = \int_{-\infty}^{+\infty} \psi(x) |x\rangle \,\mathrm{d}x,\] +\[|\psi \rangle = \int_{-\infty}^{+\infty} \psi(x) |x\rangle \,\mathrm{d}x,\] 
-where we are thinking of the values of $\psi(x)$ at different values of $x$ as components of a vector in the basis $|x\ket$, just like the components $b_j$ in the decomposition +where we are thinking of the values of $\psi(x)$ at different values of $x$ as components of a vector in the basis $|x\rangle$, just like the components $b_j$ in the decomposition 
-\[|\psi \ket = \sum_j b_j |e_j\ket,\]+\[|\psi \rangle = \sum_j b_j |e_j\rangle,\]
 for the discrete case. for the discrete case.
  
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 \[|\psi \rangle = \int_{-\infty}^{+\infty} |x\rangle\langle x |\psi \rangle \,\mathrm{d}x.\] \[|\psi \rangle = \int_{-\infty}^{+\infty} |x\rangle\langle x |\psi \rangle \,\mathrm{d}x.\]
  
-We can do the same thing for the momentum representation.  We introduce momentum "basis vectors" $|p\vec$ with inner products+We can do the same thing for the momentum representation.  We introduce momentum "basis vectors" $|p\rangle$ with inner products
 \[\langle p' | p\rangle = \delta(p-p'),\] \[\langle p' | p\rangle = \delta(p-p'),\]
-and write the momentum representation of the vector $|\psi\ket$ as +and write the momentum representation of the vector $|\psi\rangle$ as 
-\[|\psi \ket = \int_{-\infty}^{+\infty} \phi(p) |p\rangle \,\mathrm{d}p.\] +\[|\psi \rangle = \int_{-\infty}^{+\infty} \phi(p) |p\rangle \,\mathrm{d}p.\] 
-We will then have $\phi(p) = \langle p | \\psi rangle$ and+We will then have $\phi(p) = \langle p | \psi \rangle$ and
 \[|\psi \rangle = \int_{-\infty}^{+\infty} |p\rangle\langle p |\psi \rangle \,\mathrm{d}p.\] \[|\psi \rangle = \int_{-\infty}^{+\infty} |p\rangle\langle p |\psi \rangle \,\mathrm{d}p.\]