Shankar Questions About Quantum Mechanics (Schwarz Inequality)

[Dr_Pill]

Hello there,

Im studying QM with Shankar's book.

I'm wrestling myself trough the linear algebra now and I have some questiosn.

Let me start with this one:

I have absolutely no idea where this is coming from or what does it mean.

I don't know how to multiply a ket with an inner product...

Thx in advance

 

[vanhees71]

This is the completeness relation for a bunch of vectors, usually a complete orthonormalized set of Hilbert-space basis vectors, |i\rangle. These usually occur as eigenvectors of an self-adjoint operator.

Usually you need a slight extension of this concept, namely the case for unbounded essentially self-adjoint operators which can have a continuous spectrum (like the position "eigenkets", which are no Hilbert-space vectors but belong to a larger space, i.e., the dual space of the domain of the position and momentum operator). In this case your sum goes over into an integral
|V \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \; |\vec{x} \rangle \langle \vec{x}|V \rangle.
Sometimes also the case occurs, where an operator has both a discrete and a continuous spectrum, e.g., the Hamiltonian for the motion of a particle in the Coulomb potential of another heavy charged particle.

 

[jmcelve]

This is the completeness relation for a bunch of vectors, usually a complete orthonormalized set of Hilbert-space basis vectors, |i\rangle. These usually occur as eigenvectors of an self-adjoint operator.

To expand on vanhees's statement a bit, when we say "completeness," we are talking about a basis (usually an eigenbasis of a self-adjoint operator) for a particular vector space V. If we have a complete basis for a particular space (in this case, |i\rangle collectively spans the vector space V), then we can express any arbitrary vector v (or function f if we're talking about a function space) as a sum of its components multiplied by the basis vectors. For example, if we're working in \Re^3, we can choose as a basis the standard basis vectors: e_1=(1,0,0), e_2=(0,1,0), and e_3=(0,0,1). In Dirac notation, these would typically be |1\rangle, |2\rangle, |3\rangle.

Now given an arbitrary vector in \Re^3, we can write it as a sum of |1\rangle, |2\rangle, |3\rangle with the proper coefficients in front of each basis vector. For example, the vector u=(2,1,0) can be expanded in the form |u\rangle = \sum_{i=1}^{3} | i \rangle \langle i | u \rangle as | u \rangle = 2 | 1 \rangle + 1 | 2 \rangle + 0 | 3 \rangle.

The term \langle i | u \rangle in the sum is the inner product of the basis vector with the arbitrary vector u. This inner product picks out the component of u in the direction of that particular i basis vector. Now this inner product is just a scalar value. The additional | i \rangle is there because we're attempting to expand a vector as a sum of basis vectors multiplied by their respective components in u for each basis vector. So when you multiply an inner product by a ket, you're just scaling a ket. That's all.

I hope this helps!

 

[Dr_Pill]

This is the completeness relation for a bunch of vectors, usually a complete orthonormalized set of Hilbert-space basis vectors, |i\rangle. These usually occur as eigenvectors of an self-adjoint operator.

Usually you need a slight extension of this concept, namely the case for unbounded essentially self-adjoint operators which can have a continuous spectrum (like the position "eigenkets", which are no Hilbert-space vectors but belong to a larger space, i.e., the dual space of the domain of the position and momentum operator). In this case your sum goes over into an integral
|V \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \; |\vec{x} \rangle \langle \vec{x}|V \rangle.
Sometimes also the case occurs, where an operator has both a discrete and a continuous spectrum, e.g., the Hamiltonian for the motion of a particle in the Coulomb potential of another heavy charged particle.

Thanks for the explanation, but it's actually a bit too difficult.
I never saw Hilbert spaces.

I was just wondering what happens if you multipy |i> with <i|V>

Do you get <i| i | V > ?

That is not clear to me.

@ jmcelve, I'll read your post later ( christmas dinner) anyway still thanks for the help.

 

[BruceW]

merry christmas, hohoho

<i|V> is just a number (Since you are taking an inner product). So then to multiply by <i| just means that you have <i| times by a number

 

[andrien]

it is just like writing a vector in terms of it's basis.since the basis are in general infinite in number,it is called hillbert space.

 

[Dr_Pill]

Thx for the help guys.

However now I'm stuck with this proof:

Aplpy axion 1(i), they say, but there is no such axion labelled that way, so I don't even know what axion to apply.

If you can explain 1.3.18 a bit more, because I completely do not understand it, it would be great.

Sorry for the questions ( it's no homework, it's self-study)

Thx in advance.

 

[Doc Al]

Aplpy axion 1(i), they say, but there is no such axion labelled that way, so I don't even know what axion to apply.

While not labeled as such, I assume they mean the axioms bullet-listed on page 8 (in my copy). The first one is skew-symmetry, which is what is invoked here.

 

[Dr_Pill]

But why are there 3 terms in the first step and 4 terms in the second step @1.3.18

Why is this not legit?

 

[Doc Al]

But why are there 3 terms in the first step and 4 terms in the second step @1.3.18

What 3 terms?

 

[Dr_Pill]

What 3 terms?

next to <Z|Z> ?

 

[Doc Al]

next to <Z|Z> ?

I see 4 terms, not 3.

 

[Dr_Pill]

I really don't understand 1.3.18 Doc Al.

Proof in Griffiths is totally different and that, I understand, but this one in Shankar, not a single clue :(.

I don't even understand how you get the <Z|Z> out of 1.3.17, sigh.

 

[Dr_Pill]

Proof Schwarz-Inequality help

Hi there,

I'm reading Shankar, but I'm really stuck on this one:

http://i.imgur.com/SenOx.jpg

The whole equations @ 1.3.18 are complete gibbrish to me.

Maybe somebody can explain this?

 

[mfb]

In the first step, 1.3.17 is used to replace Z, and the second step uses linearity (more precise: sesquilinearity) of the scalar product. With the help of 1.3.19, you can see that this has to be real and at least 0.

 

[Dr_Pill]

In the first step, 1.3.17 is used to replace Z, and the second step uses linearity (more precise: sesquilinearity) of the scalar product. With the help of 1.3.19, you can see that this has to be real and at least 0.

And why is this wrong:

What is <Z| ?

 

[Doc Al]

I really don't understand 1.3.18 Doc Al.

Proof in Griffiths is totally different and that, I understand, but this one in Shankar, not a single clue :(.

I don't even understand how you get the <Z|Z> out of 1.3.17, sigh.

The first line just expresses <Z|Z> in terms of the what Z is defined as in 1.3.17. Then just multiply it out. You get four terms (not three--I don't know where you got three from). Do you understand where each of the four terms come from?

 

[Dr_Pill]

The second step I got (just worked it out), I know get u have to complex conjugate it.

the first step I don't get, how u get <Z|

 

[Dr_Pill]

Now I understand where the 4 vectors come from. It was the formula of antilinearity :)

But I don't know how u get <Z|Z>

I think <Z| = http://i.imgur.com/pOsyZ.jpg ( second line)

I know I'm doing something impossible here, but not what.
Working with brakets first time ever, is very very confusing.In other words: I think you have got to reverse the left Z in <Z|Z> , and tthus also the equation in 1.3.17 , because its a bra, while the right Z is a ket.

 

[Doc Al]

Moderator's note: I merged the two threads. (Once is enough!)​

 

[mfb]

@Dr_Pill: Your <Z| is correct apart from a wrong sign for both parts, but you cannot write <Z|Z> by copying every single character.

Consider a=b+c (real numbers, if you like). Then
a*a != b+cb+c = b+c+cb (wrong)
a*a = (b+c)(b+c)=b^2+bc+cb+c^2 (right)

 

[Dr_Pill]

@ Doc Al, sorry.

@ mgb ok i get it

as for 1.3.19 (almost there)

Is this correct:

I have a feeling its not, how to get rid of the term with |W| ^ 4 in the denominator?

 

[mfb]

$$\frac{<W|W>}{|W|^4}=\frac{1}{|W|^2}$$
This way, you get rid of two terms in your first line and you don't have to "forget" the 2 afterwards ;).

<W|V><V|W>= <V|W>* <V|W> = |<V|W>|²

This can be used to get <V|V> <W|W> ≥ |<V|W>|².
Take the square root, and you are done.

 

[Dr_Pill]

Indeed! Thx Now It's obvious!

<W|W> would be 1 if they were normalized Schrödinger equations, right, guess I was a bit confused.