Sakurai page 54: Is this a Taylor expansion?

In summary, the conversation discusses the derivation of equation (1.7.15) from the book "Modern Quantum Mechanics" by J. J. Sakurai. The equation shows the manipulation of states and wavefunctions in the position space, and it is suggested that the same manipulations can be expressed in terms of appropriate wavefunctions. The conversation also considers the role of complex and hermitian conjugates in the Taylor expansion of a Bra, with the conclusion that they do not come into play due to the real variable of position.
  • #1
omoplata
327
2
From page 54 of 'Modern Quantum Mechanics, revised edition" by J. J. Sakurai.

Obtaining equation (1.7.15),[tex]
\begin{eqnarray}
\left(1- \frac{ip\Delta x'}{\hbar} \right) \mid \alpha \rangle & = & \int dx' \mathcal{T} ( \Delta x' ) \mid x' \rangle \langle x' \mid \alpha \rangle \\
& = & \int dx' \mid x' + \Delta x' \rangle \langle x' \mid \alpha \rangle \\
& = & \int dx' \mid x' \rangle \langle x' - \Delta x' \mid \alpha \rangle \\
& = & \int dx' \mid x' \rangle \left( \langle x' \mid \alpha \rangle - \Delta x' \frac{\partial}{\partial x'} \langle x' \mid \alpha \rangle \right)
\end{eqnarray}
[/tex]How does the last line come about? Is it a Taylor expansion?
 
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  • #2
It looks like a Taylor expansion, indeed.
 
  • #3
Can a Bra be Taylor expanded like this? I guess it is like Taylor expanding a vector? I know that Kets are elements of a vector space, and Bras are the elements of the corresponding dual space.
 
  • #4
If the result is differentiable in some variable, and behaves nicely (insert strict mathematical formulation here): why not?
 
  • #5
omoplata said:
Can a Bra be Taylor expanded like this? I guess it is like Taylor expanding a vector? I know that Kets are elements of a vector space, and Bras are the elements of the corresponding dual space.

It's not quite a bra that is being expanded here. ##\langle x | \alpha \rangle = \psi_\alpha(x)## is the position space wavefunction corresponding to the state ##|\alpha\rangle##. What appears in the expression above is ##\psi_\alpha(x'-\Delta x')##, which is usually a nice enough function to have a well-defined Taylor expansion. It may be the case that you can also manipulate the states themselves in some way, but I expect that if the procedure is to be well-defined, there must be a way to express the same manipulations in terms of appropriate wavefunctions.
 
  • #6
fzero said:
It's not quite a bra that is being expanded here. ##\langle x | \alpha \rangle = \psi_\alpha(x)## is the position space wavefunction corresponding to the state ##|\alpha\rangle##. What appears in the expression above is ##\psi_\alpha(x'-\Delta x')##, which is usually a nice enough function to have a well-defined Taylor expansion. It may be the case that you can also manipulate the states themselves in some way, but I expect that if the procedure is to be well-defined, there must be a way to express the same manipulations in terms of appropriate wavefunctions.

Oh, OK. That's easier to understand.

I was about to ask since all operators [itex]A[/itex] should be [itex]A^{\dagger}[/itex] and all constants [itex]c[/itex] should be [itex]c^*[/itex] in the Bra space, why is it,[tex]
\int dx' \mid x' \rangle \langle x' - \Delta x' \mid \alpha \rangle = \int dx' \mid x' \rangle \left( \langle x' \mid \alpha \rangle - \Delta x' \frac{\partial}{\partial x'} \langle x' \mid \alpha \rangle \right)
[/tex] and not, [tex]
\int dx' \mid x' \rangle \langle x' - \Delta x' \mid \alpha \rangle = \int dx' \mid x' \rangle \left( \langle x' \mid \alpha \rangle - \Delta (x')^* {\frac{\partial}{\partial x'}}^{\dagger} \langle x' \mid \alpha \rangle \right)
[/tex]?
 
  • #7
I mean, I understand that what's given in the book easily follows if you think of it as a wavefunction. But if you think of it as Taylor expanding a Bra, why don't the complex and hermitian conjugates don't come into play?
 
  • #8
omoplata said:
I mean, I understand that what's given in the book easily follows if you think of it as a wavefunction. But if you think of it as Taylor expanding a Bra, why don't the complex and hermitian conjugates don't come into play?
Because position is a real variable?
 

1. What is a Taylor expansion?

A Taylor expansion is a mathematical representation of a function as an infinite sum of polynomials. It is used to approximate the values of a function at a specific point using values of its derivatives at that point.

2. How is a Taylor expansion calculated?

A Taylor expansion is calculated using the derivatives of a function at a specific point and plugging those values into a specific formula. The formula can vary depending on the order of the derivative and the number of terms desired in the expansion.

3. Why is a Taylor expansion important?

A Taylor expansion is important because it allows for the approximation of a function's values at a specific point, which can be useful for solving problems in various fields including physics, engineering, and economics.

4. Is a Taylor expansion always accurate?

No, a Taylor expansion is not always accurate. The accuracy of the approximation depends on the order of the derivative used and the distance from the point of approximation to the point where the function is being evaluated.

5. How is a Taylor expansion related to the Sakurai page 54?

The Sakurai page 54 refers to a specific page in the book "Modern Quantum Mechanics" written by J.J. Sakurai. Page 54 discusses the use of Taylor expansions in quantum mechanics and how they can be applied to solve problems in this field.

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