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dirac_notation [2021/02/22 21:14] – [2.ii.3 Basis Independence: The Discrete Case] admindirac_notation [2021/02/22 21:15] (current) – [2.ii.5 Basis Independence: The Continuous Case] admin
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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
 \[|\psi \rangle = \int_{-\infty}^{+\infty} |p\rangle\langle p |\psi \rangle \,\mathrm{d}p.\] \[|\psi \rangle = \int_{-\infty}^{+\infty} |p\rangle\langle p |\psi \rangle \,\mathrm{d}p.\]