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Cosmas Zachos
Cosmas Zachos
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When you represent a periodic function by a Fourier series you do pretty much the same thing: you represent it by sines instead of δ-functions.

For your specific example, you have two alternate resolutions of the identity/ completeness relations, $$ 1\!\!1= \int\!\! dx ~|x\rangle\langle x| \\ 1\!\!1= \sum_a |a\rangle \langle a| $$ where, importantly, $\langle a|x \rangle = \phi_a^* (x)$, your change of basis "matrix".

You then have $$ \psi(x)=\langle x|\psi\rangle = \sum_a \langle x|a\rangle \langle a|\psi\rangle = \sum_a \phi_a(x) ~~\psi_a, ~~\hbox{where}\\ \psi_a=\langle a|\psi\rangle= \int\!\! dx ~ \phi^*_a(x)~ \psi(x), $$ by the above completeness relation. In your expression, you used the same x as a variable, and as a dummy integration variable in your inner product--a disastrous practice.

It is crucial to test-drive this equivalence with the Hermite functions $\psi_n(x)=\langle x|n\rangle$, where the discrete index is the natural number identifying the quantum oscillator energy level, if you have not already done so.

Useful link as per your comment.

When you represent a periodic function by a Fourier series you do pretty much the same thing: you represent it by sines instead of δ-functions.

For your specific example, you have two alternate resolutions of the identity/ completeness relations, $$ 1\!\!1= \int\!\! dx ~|x\rangle\langle x| \\ 1\!\!1= \sum_a |a\rangle \langle a| $$ where, importantly, $\langle a|x \rangle = \phi_a^* (x)$, your change of basis "matrix".

You then have $$ \psi(x)=\langle x|\psi\rangle = \sum_a \langle x|a\rangle \langle a|\psi\rangle = \sum_a \phi_a(x) ~~\psi_a, ~~\hbox{where}\\ \psi_a=\langle a|\psi\rangle= \int\!\! dx ~ \phi^*_a(x)~ \psi(x), $$ by the above completeness relation. In your expression, you used the same x as a variable, and as a dummy integration variable in your inner product--a disastrous practice.

It is crucial to test-drive this equivalence with the Hermite functions $\psi_n(x)=\langle x|n\rangle$, where the discrete index is the natural number identifying the quantum oscillator energy level, if you have not already done so.

When you represent a periodic function by a Fourier series you do pretty much the same thing: you represent it by sines instead of δ-functions.

For your specific example, you have two alternate resolutions of the identity/ completeness relations, $$ 1\!\!1= \int\!\! dx ~|x\rangle\langle x| \\ 1\!\!1= \sum_a |a\rangle \langle a| $$ where, importantly, $\langle a|x \rangle = \phi_a^* (x)$, your change of basis "matrix".

You then have $$ \psi(x)=\langle x|\psi\rangle = \sum_a \langle x|a\rangle \langle a|\psi\rangle = \sum_a \phi_a(x) ~~\psi_a, ~~\hbox{where}\\ \psi_a=\langle a|\psi\rangle= \int\!\! dx ~ \phi^*_a(x)~ \psi(x), $$ by the above completeness relation. In your expression, you used the same x as a variable, and as a dummy integration variable in your inner product--a disastrous practice.

It is crucial to test-drive this equivalence with the Hermite functions $\psi_n(x)=\langle x|n\rangle$, where the discrete index is the natural number identifying the quantum oscillator energy level, if you have not already done so.

Useful link as per your comment.

Source Link
Cosmas Zachos
Cosmas Zachos
  • 66.3k
  • 6
  • 110
  • 248

When you represent a periodic function by a Fourier series you do pretty much the same thing: you represent it by sines instead of δ-functions.

For your specific example, you have two alternate resolutions of the identity/ completeness relations, $$ 1\!\!1= \int\!\! dx ~|x\rangle\langle x| \\ 1\!\!1= \sum_a |a\rangle \langle a| $$ where, importantly, $\langle a|x \rangle = \phi_a^* (x)$, your change of basis "matrix".

You then have $$ \psi(x)=\langle x|\psi\rangle = \sum_a \langle x|a\rangle \langle a|\psi\rangle = \sum_a \phi_a(x) ~~\psi_a, ~~\hbox{where}\\ \psi_a=\langle a|\psi\rangle= \int\!\! dx ~ \phi^*_a(x)~ \psi(x), $$ by the above completeness relation. In your expression, you used the same x as a variable, and as a dummy integration variable in your inner product--a disastrous practice.

It is crucial to test-drive this equivalence with the Hermite functions $\psi_n(x)=\langle x|n\rangle$, where the discrete index is the natural number identifying the quantum oscillator energy level, if you have not already done so.