Understanding Hilbert Space in Quantum Mechanics: A Beginner's Guide

[Sterj]

In quantum mechanic the Hilbert space is often used. I don't study physics, must be said. So, I have a few questions to this space (I can calculate with complex numbers and vectors).

1. What's the different between a bra <p| and a ket |b> vector?
2. What calculation is behind that: <p|b>?
3. Is the Hilbert space a function room or a vector room? Because sometimes it is written |psi> (in Schrödinger equation => wave function) and sometimes it is written that the Hilbert space is a vector room.
4. How can a wave function be 1 (|psi|=1) or is the area under the function equal to 1 (integral)?
5. Is that definition: [A,B]=AB-BA (commutator)?
6. Can somebody solve me an example for this: int[ (psi*(x)) * (f(x)) ] dx
7. Why is the ... (dont know the name in English) value: A=<psi|A|psi>?

Should be easy for you to answer this questions. Thanks.

 

[inha]

I can answer atleast some of those I think

1. I don't think there is a difference besides the name

2. Inner product (integrate the product of p and b's complex conjugate [I don't know if the complex conjugate definition is used outisde of QM] over all space)

3. I don't understand the question but by definition the hilbert space is a vector space with an inner product and it's a complete metric space.

4. I think you mean the the normalizability (the square of the abs. val. integrated over space) here. It has to be one since when you do the integration you count all the probabilities of finding your partcular particle somewhere. It's essential for the probability intrepetation and shuts out unphysical solutions.

5. Yes that's the definition

6. Do you know Fourier transforms? That could be a Fourier transform for example.

7. Expectation value. You should have these <> around the A. I don't know how operator formalism is handled in mathematics but in QM it's rather simple. Straight from the definition of expectation value (check mathworld.wolfram.com for reference) plug in the square of the abs value of the wave function (the probability density) and the operator which's expectation value you're after and you sort of have it there already. If you want I can type out a explicit proof or maybe someone would already have it in a nice format (tex, some image or a pdf?) so it'd look nicer and be easier to read.

 

[Kane O'Donnell]

1. I don't think there is a difference besides the name

No, there is a difference. Suppose your Hilbert space is called H. Then the kets are elements of H, and the bras are elements of the *dual* space H*, which is the space consisting of all bounded linear functionals (maps from H to C, the complex numbers). It turns out that every bounded linear functional looks like an inner product, ie f(h) = <k|h> for some k in H, so to represent the functional f_k that maps h -> <k|h> we can abuse notation and write f_k = <k|, then f_k(h) = <k|h>. The dual space of a Hilbert space is itself a Hilbert space.

In *practice*, new students in Physics are told that the bras are just the *complex conjugates* of the kets, and this is enough to do calculations in elementary QM.

2. Inner product (integrate the product of p and b's complex conjugate [I don't know if the complex conjugate definition is used outisde of QM] over all space)

Yes, they are, any inner product space where the scalars are complex must have that conjugate in there somewhere to make sure the inner product <h|h> is always non-negative.

3. A Hilbert space is both a vector space and a function space at the same time, it is a Banach space (a complete metric space) and an inner product space (vector space + inner product).

5. There are probably thousands of websites that have integrals like this, especially in relation to QM.

7. The above poster is right, the expectation value of an observable A is given by: $$\langle A\rangle = \langle\psi|\hat{A}|\psi\rangle = \langle\psi|\hat{A}\psi\rangle$$, where the hat over the A says that $$\hat{A}$$ is the operator corresponding to the observable A. This definition of the expectation value comes from probability theory, just like the normalisation condition.

Cheerio!


Kane

 

[dextercioby]

No, there is a difference. Suppose your Hilbert space is called H. Then the kets are elements of H, and the bras are elements of the *dual* space H*, which is the space consisting of all bounded linear functionals (maps from H to C, the complex numbers).

AAAAAAAAA:Blasphemy! :grumpy: The dual space comprises all linear functionals defined on H,not only the continuous ones ar the bounded ones...These thig are far more general than u think...

It turns out that every bounded linear functional looks like an inner product, ie f(h) = <k|h> for some k in H, so to represent the functional f_k that maps h -> <k|h> we can abuse notation and write f_k = <k|, then f_k(h) = <k|h>. The dual space of a Hilbert space is itself a Hilbert space.

What is the condition for a functional in the theorem of Riesz??

Kane[/QUOTE]

Daniel.

 

[Kane O'Donnell]

No.

You are right for a general vector space, certainly. However, Hilbert spaces are reflexive, and hence isomorphic to the space of *continuous* linear functionals that form the dual.

Since we're talking about a metric space, continuity is equivalent to boundedness.

The dual of a Hilbert space is therefore the Hilbert space of bounded linear functionals.

This turns up, for example, in the definition of the adjoint of a (possibly unbounded) operator. The domain for such an operator is given by:

$$D_{T^*} = \{ h \in \mathcal{H}: k \mapsto \langle h | Tk \rangle \mbox{ is a bounded linear functional } \}$$​

where I have converted to right-linear braket notation, although my notes are in the usual left-linear math inner prod notation so I may have got the Tk and the h around the wrong way. I'm fairly sure it's ok though.

Regards,

Kane

 

[Kane O'Donnell]

Besides, there's an even more obvious argument. Suppose that the Reisz representation theorem applies to an arbitrary unbounded linear functional $$f:\mathcal{H}\to\mathbb{C}$$. Then RRT implies that there exists $$k \in \mathcal{H}$$ such that $$f(h) = \langle k | h \rangle$$ for all $$h \in \mathcal{H}$$.

But by the Cauchy-Schwarz inequality, $$|\langle k | h \rangle | \leq ||k||~||h||$$. Since $$k$$ is fixed in the representation of the functional, the functional is bounded, contradicting our unbounded hypothesis.

Regards,

Kane O'Donnell

 

[dextercioby]

Besides, there's an even more obvious argument. Suppose that the Reisz representation theorem applies to an arbitrary unbounded linear functional $$f:\mathcal{H}\to\mathbb{C}$$. Then RRT implies that there exists $$k \in \mathcal{H}$$ such that $$f(h) = \langle k | h \rangle$$ for all $$h \in \mathcal{H}$$.

But by the Cauchy-Schwarz inequality, $$|\langle k | h \rangle | \leq ||k||~||h||$$. Since $$k$$ is fixed in the representation of the functional, the functional is bounded, contradicting our unbounded hypothesis.

Regards,

Kane O'Donnell

I'm sorry,but your "proof" is useless.In the previous post you stated that for a topological metric space everywhere continuity is completely equivalent to boundness.
The Theorem of Riesz applies ONLY TO LINEAR FUNCTIONAL EVERYWHERE CONTINUOS IN H,thind which,by virtues of topology,means that the functionals need to be bounded on H.So assuming them unbounded and applying the Riesz Representation Theorem makes no sense,okay...??

So yeah,i was wrong,$$\tilde{H}$$ comprises only bounded (hence everywhere continuous) linear functionals on H.

Daniel.

 

[Kane O'Donnell]

So assuming them unbounded and applying the Riesz Representation Theorem makes no sense,okay...??

Yes, it's called a proof by contradiction. It shows that every linear functional represented by an inner product via the RRT must be bounded. The hypothesis is, to change it's form, "Suppose that the RRT *does* apply to unbounded linear functionals", and then we show a contradiction, which shows that the hypothesis must be false.

Of course, one *shouldn't* technically do a proof like this (it doesn't really prove anything if you know what the hypotheses for the RRT are), but it illustrates precisely why unbounded functionals can't be represented by an inner product.

As you have stated, in a metric space continuity is equivalent to boundedness, so if you've only seen the RRT stated in terms of continuous functionals it would be easy to make the mistake.

No point calling it a topological metric space though, every metric space has a natural topology associated with it, so all metric spaces are topo spaces.

Regards,

Kane

 

[Sterj]

$$\langle A\rangle = \langle\psi|\hat{A}|\psi\rangle = \langle\psi|\hat{A}\psi\rangle$$

Sorry, but can somebody derive me that or tell me how does it come to that? (in google there isn't to find anything and you only can find that that is the expaction value, but not why).

 

[Sterj]

mhh, I mean: <A>=<p|Â|p>

Sorry, but can somebody derive me that or tell me how does it come to that? (in google there isn't to find anything and you only can find that that is the expaction value, but not why).

 

[dextercioby]

Depart from the statistical definition of the average for an observable with mixed spectrum:
$$\langle \hat{A}\rangle =\sum_{n\in I_{d}}a_{n}P_{a_{n}}+\int_{I} d\alpha \ a(\alpha) \ P(\alpha)$$

and apply the third principle (the one with statistical nature of QM) in the case of a pure state $|\psi\rangle$ from the Hilbert space of states...

Daniel.

 

[Kane O'Donnell]

Well, maybe a simpler explanation...

First, remember that the expression $$\langle\psi |\hat{A}|\psi\rangle$$ is just notation, ok - by definition,

$$\langle\psi|\hat{A}|\psi\rangle := \langle \psi|\hat{A}\psi\rangle$$​

The reason they introduce the new notation is because the operator A must be self-adjoint, so $$\langle \hat{A}\psi|\psi\rangle = \langle\psi | \hat{A}\psi\rangle$$, and writing the A in the centre as above is suggestive that the A can act on the bra or the ket, it doesn't matter.

The idea of an expectation value comes from probability theory. If you have a set of outcomes $$\{\lambda_n \}$$ for an experiment, where the probability that the outcome of the experiment will be $$\lambda_i$$ is $$p_i$$ (we must have $$\sum_i p_i = 1$$), then the *average* outcome over a very large number of measurements is:

$$\langle \lambda \rangle = \sum_{i = 1}^{n}\lambda_i p_i$$​

You can also extend this idea to the situation where instead of a finite set of possible outcomes, you have a *continuum* of possible outcomes, which is what the other half of dextercioby's expression above is made up of. The mathematics behind measure theory is quite tricky if you've never seen it before and deserves a bit more discussion that I can put here.

Then, as dextercioby said, we use the postulate of quantum mechanics that ties QM to probability theory - we use the spectral theorem for self-adjoint operators to give the specific details of the interpretation. In the simple case of a wavefunction $$\psi$$ made up of a finite linear combination of eigenfunctions of the Hamiltonian,

$$\psi = a_1\phi_1 + a_2\phi_2+...+a_n\phi_n$$​

(where we have $$\sum_i |a_i|^2 = 1$$) you can see, for exampe, how the quantum mechanical calculation for average energy (expectation value of energy) looks almost exactly like the ordinary probabilistic interpretation - the possible outcomes are the eigenvalues $$\lambda_i$$ for each of the $$\phi_i$$, and the probability that a measurement will yield $$\lambda_i$$ is $$p_i = |a_i|^2$$. You can do the integral calculation yourself (it's quite easy if you start by the definition of the expectation value and use the fact that the $$\phi_i$$ are eigenfunctions of $$\hat{H}$$) but the end result is:

$$\langle H \rangle = \langle\psi|\hat{H}\psi\rangle = \sum_{i = 1}^{n} \lambda_i |a_i|^2 = \sum_{i = 1}^{n}\lambda_i p_i$$​

just like we had above.

Regards,

Kane O'Donnell

 

[Sterj]

thanks for you answers. Now its clearer.

 

[Mike2]

Well, maybe a simpler explanation...

First, remember that the expression $$\langle\psi |\hat{A}|\psi\rangle$$ is just notation, ok - by definition,

$$\langle\psi|\hat{A}|\psi\rangle := \langle \psi|\hat{A}\psi\rangle$$​

The reason they introduce the new notation is because the operator A must be self-adjoint, so $$\langle \hat{A}\psi|\psi\rangle = \langle\psi | \hat{A}\psi\rangle$$, and writing the A in the centre as above is suggestive that the A can act on the bra or the ket, it doesn't matter.

The idea of an expectation value comes from probability theory. If you have a set of outcomes $$\{\lambda_n \}$$ for an experiment, where the probability that the outcome of the experiment will be $$\lambda_i$$ is $$p_i$$ (we must have $$\sum_i p_i = 1$$), then the *average* outcome over a very large number of measurements is:

$$\langle \lambda \rangle = \sum_{i = 1}^{n}\lambda_i p_i$$​

Is there some reason that probabilties enter the picture in the first place? Do probabilities always enter when the eigenvalue problem results in many possible answers?

 

[dextercioby]

Yes,QM is essentially a probabilistic theory.Its probabilistic nature is postulated in its third principle.The reason why they did so,it's a bit more complicated.Schroedinger came up with its wavefunctions,complex square integrable functions,and nobody could explain their physical relevance.Up until Max Born came with the idea that
$$|\Psi(\vec{r},t)|^{2}$$

is nothing but the probability density that a quantum system is found in the point $\vec{r}$ at the moment "t"...

Daniel.

 

[Sterj]

The equation, notation, ... aren't a really problem now cause of your help. I tried to find an example with values for this:

<A>=||(A-<A>)^2|psi>||. I mean it isn't really a problem to solve this with normal values in R, but in the Hilberspace I wouldn't know how to solve this with given values. Sorry, but can you give me a hand? If A is an operator (e.x: d / dx).how can I solve this?

 

[dextercioby]

Where did u get that formula??Did u invent it or did un see it in a book??It looks incorrect.

Daniel.

 

[Sterj]

sorry, it is a little bit incorrect: here the correct one:
d(A) = ||(A-<A>)|p>||
this formula should be the correct one. Can somebody solve this with values?

 

[Kane O'Donnell]

The norm notation ($$||\cdot ||$$) comes from the fact that an inner product space is naturally equipped with a norm:

$$||\psi|| = \sqrt{\langle \psi|\psi\rangle}$$​

In the previous examples we've stated that:

$$\langle \psi|\phi\rangle = \int \psi^{*}\phi$$​

So just substitute into your formula above and you can calculate $$d(\hat{A})$$ for an operator A acting on a function $$\psi$$.

Regards,

Kane O'Donnell

 

[Sterj]

you mean:
d(A) = ||(A-<A>)|p>||

d(A) = ||(A|p>-<A|p>>)||

and then the first part is:
A|p> = a|p>
and the second:
<A|p>> = <a|p>> and thats: integral((ap*)(ap))

Are these first steps correct?

 

[dextercioby]

you mean:
d(A) = ||(A-<A>)|p>||

d(A) = ||(A|p>-<A|p>>)||

and then the first part is:
A|p> = a|p>
and the second:
<A|p>> = <a|p>> and thats: integral((ap*)(ap))

Are these first steps correct?

Nope,they're wrong.Are u trying to compute this animal
$$O=:\langle p|\hat{A}-\langle \hat{A}\rangle_{|p\rangle} \hat{1}|p\rangle$$

??

Prove it is zero...

Daniel.

 

[Sterj]

yes this dexter. Why it is 0. don't this depend on the operator and the function you take? I think this "equation" is a part of the uncertainty principle.

 

[dextercioby]

1.It's because it's the same quantum state $$|p\rangle$$ that the result is zero.It doesn't depend on the operator and on the quantum state,just as long as the latter is kept the same when taking both averages.
2.Yes,that animal comes up when discussing the uncertainty relations.

Daniel.

 

[Sterj]

@dexter: If it is 0 in, my oppinion the uncertainty principle doesn't makes sense, because:
d(A)d(B)=...
and d(A) = 0 = d(B). If it is like you said, no operator and no function can sadisfy this equation (uncertainty p.)

 

[dextercioby]

This fact that i computed is nothing but
$$\langle \Delta \hat{A} \rangle_{|p\rangle}$$

which is indeed zero,but it has nothing to do with the formulation (mathematical expression) of the UP,since the latter involves uncertainties and NOT averages of uncertainties...

Daniel.

 

[Sterj]

@dexter:

"which is indeed zero,but it has nothing to do with the formulation (mathematical expression)"

And what's the meaning of this, if it has nothing to do wiht mathematical expessions?

 

[dextercioby]

It simply means that the average of the uncertainty of any selfadjoint operator is zero and nothing more.It is an immediate side result of the definiton of uncertainty...

Daniel.

 

[Sterj]

If the average of the uncertainty of any selfadjoint operator is zero, where is the uncertaint?
And what sense does this formula make?
d(A) = ||(A-<A>)|p>|| i mean, if it's alway zero?

 

[dextercioby]

Hold on a second.Is that $|p\rangle$ a vector which satisfies the spectral equation
$$\hat{p}|p\rangle=p|p\rangle$$

If,so,pay attention,as it is not normalized...It does not describe a quantum state because it has infinite norm...

Daniel.

 

[Sterj]

oh, yes dexter it is a vector that sadisfies the spectral equation. Must be in the uncertainty p. And there it has to be normalized (length = 1).

d(A) = ||(A-<A>)|p>||

I always thought this "equation" is standing for the variance.

 

[dextercioby]

It cannot be normalized,simply because it's not a state vector.It's not an eigenvector of momentum operator,because the latter does not have eigenvectors,as it doesn't have discrete spectrum...

Pay attention with such details.Mathematical details mean very much in QM.

Daniel.

 

[C*-algebras]

I'm sorry,but your "proof" is useless.In the previous post you stated that for a topological metric space everywhere continuity is completely equivalent to boundness.
The Theorem of Riesz applies ONLY TO LINEAR FUNCTIONAL EVERYWHERE CONTINUOS IN H,thind which,by virtues of topology,means that the functionals need to be bounded on H.So assuming them unbounded and applying the Riesz Representation Theorem makes no sense,okay...??

So yeah,i was wrong,$$\tilde{H}$$ comprises only bounded (hence everywhere continuous) linear functionals on H.

Daniel.

Riesz thm indeed makes great sense. But I guess a more insightful version would be using the GNS construction and Gelfand-Naimark theorem where Hilbert space is constructed through non-commutative structure rather than "postulated" as in Dirac-Neumann axiomatic system.
In this sense, I prefer QM be essentially an ALGEBRAIC theory... prob. somehow becomes a byproduct.